## Israel, China, and more

I figured that as interested in Jews and Jewish achievement (and shenanigans) as I am, I should at least learn something real about Israel, which I know little about at the detailed factual level. That part of the world has, predictably, always felt rather remote in my life, though it is in some sense the cradle of civilization. While on the bus with nothing to do, I was just last week, trolling some of my friends on Facebook with some Hebrew I copy pasted. Like, ברוך השם (Baruch HaShem), which literally means “blessed his name.” On that I’m pleased to say that I’m now sort of paying attention to the letters of the Hebrew alphabet when I visit English Wiki pages on Jewish matters with the English transliteration of Hebrew words alongside the Hebrew original. It’s kind of cute that it, like Arabic, reads right to left, a fact I had not known.

Last night, I had the pleasure of going through the history of Israel (particularly its formation) in some degree of detail. So now I know what Haganah, Irgun, Lehi, Palmach are. Interestingly, there was tension between the IDF (headed by Ben-Gurion) and the Irgun (headed by Menachim Begin), which resulted in the Altalena Affair in which a ship containing armaments of the Irgun was ordered to be sunk by Ben-Gurion on high seas by the air force. I was rather surprised there was actually this much discord among the Zionist leaders, as stereotype is of course that Jews are super cohesive. I was also somewhat surprised that the Zionists had the nerve to assassinate Western politicians like Lord Moyne and Folke Bernadotte they did not like who were mediating truces between the Israelis and the Arabs during the 1948 war. Overall, my impression of the war was that neither side had substantial military experience or ability and the war was on a relatively small scale, being in a very small region.

I was not fully aware that American support for Israel really only became substantial following the Six Day War. Around that time, France, which had provided Israel with high end military technology before, had announced an embargo there. Israel’s nuclear weapons was provided to it largely by France as well, and some say that at the time of Six Day War, Israel already had a functional nuke to use as a last resort. Almost certainly, it did in the Yom Kippur War seven years later. Details regarding Israel’s secret nuclear weapons program were revealed to the public via Mordechai Vanunu, who had worked as a technician on classified projects at the Negev Nuclear Research Center, who was eventually caught and shamed for life by Mossad agents.

From this reading, I can better appreciate Israel’s vulnerability due to its small size, in land and in population, the latter especially, that makes it impossible to sustain itself without external aid. This will hold regardless of how advanced it becomes, so even with nuclear ICBMs, they still have much to fear. I’ve seen pro-Jewish sites characterize Israel’s military and survival as a miracle. I’ve also seen that Israel is scared shit of North Korea, which could potentially transfer its nuclear and missile technology to Iran, Syria, etc. Israel’s attitude is of course that the Arab nations cannot obtain nukes at all costs, and Israel will send Mossad agents to assassinate anyone suspected to be assisting them on that, which it has already done many times.

Readers of my blog might know that I write here about that Jew in math I talk with quite a bit, who has some interesting views for sure. As an update there, I didn’t quite expect him to say to me that “Israel surviving is not that impressive.” I would somewhat agree actually given how much support Israel has gotten from the West, which does not apply at all to North Korea, whose survival I would say is much more of a miracle. Only time will tell who will last longer, and I would think that both will remain intact for quite a while.

That guy also tells me that China is very pro-Israel, which I’m not so sure about. China only developed at first secret relations with Israel in the late 70s/early 80s. China at the time was very interested in procuring some high-end Western military tech from Israel, which it did to a significant degree in the 80s, and surely, America is not terribly happy about this. This guy responds with “China” when I ask him how Israel will fare on a weakened America, and I’m not sure how serious he is on that, since I could hardly imagine China actually going to the lengths to rescue Israel under the hypothetical scenario that it is about to be run over.

The aggressive and often overtly biased political attitudes of Jews and Israelis are understandable given how precarious their situation is. They faced life or death and though their situation is much more secure now, they still do, being too small. On this, recalled to me was this physics professor at Washington University of St. Louis whose infamous essay Don’t Become a Scientist I had read and reread, who also has on his web page a collection of political pieces against Iraq and North Korea, with provocative titles such as Anyone Who Bombs Baghdad [when Saddam was in power] Gets My Vote. I haven’t seen yet any mention of Israel and its nukes in his pieces and I sure wonder why. In another one of those, Limiting the Nuclear Club—Iraq, North Korea et al., he characterized Stalin (when it was mostly his USSR that defeated Hitler) very one-sidedly as “a man and system which murdered tens of millions of people with bullets, famine, and prison camps.” On the nuclear club, he also wrote that “fortunately, most of these countries are stable democracies and therefore not aggressors; the two chief exceptions (the Former Soviet Union and China) were successfully contained for many years, and the more powerful of these is making a transition, one hopes successful and irreversible, to democracy,” which again goes to show his blatant bias and lack of rigorous thinking (that he would exhibit in physics) when it comes to politics.

On this, I don’t see why China should be terribly friendly to Jews and Israel when Jews, with their media power and verbal gifts, have done so much to distort modern Chinese history in the West and to smear, sabotage, and peacefully evolve a political system that has worked wonders for China, very plausibly with ulterior motives. I have also seen many Jews support Taiwanese independence, including this guy I talk to. Certain American interests might want to mould the political thinking of Chinese who grow up in America like me (most of whom do not read Chinese), but honestly, I feel like I am too intelligent and politically discerning and realistic to fall for it. I value independent, impartial thinking that is reality grounded, that is cognitively empathetic of interests relations wise, which means the American exceptionalist versions of history and politics don’t work on me, and neither would any such form of exceptionalism in favor of any country or system. I don’t think Chinese who grow up in America will be terribly happy once they realize, as more of them are doing, that the American version of the history and culture of where their parents are from is fraught with glaring inaccuracies and falsehoods motivated by political bias and ill-intent, and elite Jews, who have the most prominent voice in America, can be mostly easily blamed for that. One can even go more extreme and say that the Jews are the main culprit for the shitty and grossly dishonest media in America, with their dominance of press and Hollywood in this country, which they unabashed laud as “free media.” I don’t think this facade can last forever.

Often, one, including myself, is met with the dilemma of whether or not to engage in aggressive and ethically questionable behavior that gives one an advantage, at least in the short term, that goes on record either directly or indirectly in the memories of those alienated by the action as well as oneself.  Even it brings me major gains, I know eventually I might look back and feel shame and regret on the dishonorable means I took to earn them. I like it most when I achieve something based on genuine ability and hard work, as opposed to politicking, striverish behavior, which everyone engages in to some extent. I don’t think lies or deceptive talk can be concealed forever regardless of how much power or media control one has, and in some sense, it is the truth that is the most potent. Additionally, as a Chinese, I am somewhat conscious of how my behavior in every way affects microscopically how Chinese are perceived in general. When I see so many Jews spout nonsense about history and politics, especially parts I am familiar with, it sure doesn’t give me a good view of the group in general, character wise, so as to separate from their objective achievements, especially when that group controls so much of the media where I live, though I am careful to disentangle individuals of the group with the group in general. I do believe that one is to some extent responsible for the actions of one’s group at large. The actions of a nation, of an ethnic group, especially against others, are not just the responsibility of the elites in power who made the decisions but also the ones who allowed those people to come in power. When a nation or people, as a collective, chooses some system or leader or development strategy, they should take some responsibility for the outcomes and “dictatorship” or “democracy” is not an excuse. Fundamentally, what I am describing is actual democracy, as opposed any democratic system by name or by election. If Americans want to elect “democratically” their leaders and their leaders make shitty decisions against their interests and country at large, they should take responsibility for it and blame themselves for choosing such people to elect or blame the election system that is the root of it all. On this, I recall how this guy way smarter than me technically (also of Jewish descent), on my mentioning of a guy I know whose parents were from the USSR whose grandfather could only become a theoretical physics professor in some remote university the name I remember not, was like: “his parents helped destroy the Soviet Union,” followed by that if he, who moved to Israel, were still in Russia, he would be working for MacDonalds, with reference to the economic crisis there in the 90s that was statistically far more murderous and damaging than Stalin’s purges. It was then that occurred to me again that as unpleasant and sad as it may be to accept, Soviets and Russians share collective responsibility for promoting certain wrong people to power in the Soviet era that rendered their nation less competitive and especially for their later letting oligarchs, many of them Jewish, wreck their country irrecoverably, a specific of the generality I had just described. That many of those mega civilization and wealth leechers/destroyers were Jewish tells us more that anti-Semitism is not without reason, and Jews should all take some responsibility for it. Pardon any cultural bias, but this brings to mind a famous quote attributed to Chairman Mao which is “世界上没有无缘无故的爱，也没有无缘无故的恨”，that translates to “the world has no love without reason and no hate without reason,” an obvious reality of human nature that I believe one of high moral character ought to always be cognizant of.

## Urysohn metrization theorem

The Urysohn metrization theorem gives conditions which guarantee that a topological space is metrizable. A topological space $(X, \mathcal{T})$ is metrizable is there is a metric that induces a topology that is equivalent to the topological space itself. These conditions are that the space is regular and second-countable. Regular means that any combination of closed subset and point not in it is separable, and second-countable means there is a countable basis.

Metrization is established by embedding the topological space into a metrizable one (every subspace of a metrizable space is metrizable). Here, we construct a metrization of $[0,1]^{\mathbb{N}}$ and use that for the embedding. We first prove that regular and second-countable implies normal, which is a hypothesis of Urysohn’s lemma. We then use Urysohn’s lemma to construct the embedding.

Lemma Every regular, second-countable space is normal.

Proof: Let $B_1, B_2$ be the sets we want to separate. We can construct a countable open cover of $B_1$, $\{U_i\}$, whose closures intersect not $B_2$ by taking a open neighborhoods of each element of $B_1$. With second-countability, the union of those can be represented as a union of a countable number of open sets, which yields our desired cover. Do the same for $B_2$ to get a similar cover $\{V_i\}$.

Now we wish to minus out from our covers in such a way that their closures are disjoint. We need to modify each of the $U_i$s and $V_i$s such that they do not mutually intersect in their closures. A way to do that would be that for any $U_i$ and $V_j$, we have the part of $\bar{U_i}$ in $V_j$ subtracted away from it if $j \geq i$ and also the other way round. This would give us $U_i' = U_i \setminus \sum_{j=1}^i \bar{V_j}$ and $V_i' = V_i \setminus \sum_{j=1}^i \bar{V_j}$.     ▢

Urysohn’s lemma Let $A$ and $B$ be disjoint closed sets in a normal space $X$. Then, there is a continuous function $f : X \to [0,1]$ such that $f(A) = \{0\}$ and $f(B) = \{1\}$.

Proof: Observe that if for all dyadic fractions (those with least common denominator a power of $2$) $r \in (0,1)$, we assign open subsets of $X$ $U(r)$ such that

1. $U(r)$ contains $A$ and is disjoint from $B$ for all $r$
2. $r < s$ implies that $\overline{U(r)} \subset U(s)$

and set $f(x) = 1$ if $x \notin U(r)$ for any $r$ and $f(x) = \inf \{r : x \in U(r)\}$ otherwise, we are mostly done. Obviously, $f(A) = \{0\}$ and $f(B) = \{1\}$. To show that it is continuous, it suffices to show that the preimages of $[0, a)$ and $(a, 1]$ are open for any $x$. For $[0, a)$, the preimage is the union of $U(r)$ over $r < a$, as for any element to go to $a' < a$, by being an infimum, there must be a $s \in (a', a)$ such that $U(s)$ contains it. Now, suppose $f(x) \in (a, 1]$ and take $s \in (a, f(x))$. Then, $X \setminus \bar{U(s)}$ is an open neighborhood of $x$ that maps to a subset of $(a, 1]$. We see that $x \in X \setminus \overline{U(s)}$, with if otherwise, $s < f(x)$ and thereby $f(x) \leq s' < f(x)$ for $s' > s$ and $U(s') \supset \overline{U(s)}$. Moreover, with $s > a$, we have excluded anything that does not map above $a$.

Now we proceed with the aforementioned assignment of subsets. In the process, we construct another assignment $V$. Initialize $U(1) = X \setminus B$ and $V(0) = X \setminus A$. Let $U(1/2)$ and $V(1/2)$ be disjoint open sets containing $A$ and $B$ respectively (this is where we need our normality hypothesis). Notice how in normality, we have disjoint closed sets $B_1$ and $B_2$ with open sets $U_1$ and $U_2$ disjoint which contain them respectively, one can complement $B_1$ to derive a closed set larger than $U_2$, which we call $U_2'$ and run the same normal separation process on $A_1$ and $U_2'$. With this, we can construct $U(1/4), U(3/4), V(1/4), V(3/4)$ and the relations

$X \setminus V(0) \subset U(1/4) \subset X \setminus V(1/4) \subset U(1/2)$,

$X \setminus U(1) \subset V(3/4) \subset X \setminus U(3/4) \subset V(1/2)$.

Inductively, we can show that we can continue this process on $X \setminus V(a/2^n)$ and $X \setminus U((a+1)/2^n)$ for each $a = 0,1,\ldots,2^n-1$ provided $U$ and $V$ on all dyadics with denominator $2^n$ to fill in the ones with denominator $2^{n+1}$. One can draw a picture to help visualize this process and to see that this satisfies the required aforementioned conditions for $U$.     ▢

Now we will find a metric for $\mathbb{R}^{\mathbb{N}}$ the product space. Remember that the base for product space is such that all projections are open and a cofinite of them are the full space itself (due to closure under only finite intersection). Thus our metric must be such that every $\epsilon$-ball contains some open set of the product space where a cofinite number of the indices project to $\mathbb{R}$. The value of $x - y$ for $x,y \in \mathbb{R}$ as well as its powers is unbounded, so obviously we need to enforce that the distance exceed not some finite value, say $1$. We also need that for any $\epsilon > 0$, the distance contributed by all of the indices but a finite number exceeds it not. For this, we can tighten the upper bound on the $i$th index to $1/i$, and instead of summing (what would be a series), we take a $\sup$, which allows for all $n > N$ where $1/N < \epsilon$, the $n$th index is $\mathbb{R}$ as desired. We let our metric be

$D(\mathbf{x}, \mathbf{y}) = \sup\{\frac{\min(|x_i-y_i|, 1)}{i} : i \in \mathbb{N}\}$.

That this satisfies the conditions of metric is very mechanical to verify.

Proposition The metric $D$ induces the product topology on $\mathbb{R}^{\mathbb{N}}$.

Proof: An $\epsilon$-ball about some point must be of the form

$(x_1 - \epsilon/2, x_1 + \epsilon/2) \times (x_2 - 2\epsilon/2, x_2 + 2\epsilon/2) \times \cdots \times (x_n - n\epsilon/2, x_n + n\epsilon/2) \times \mathbb{R} \times \cdots \times \mathbb{R} \times \cdots$,

where $n$ is the largest such that $n\epsilon < 1$. Clearly, we can fit into that an open set of the product space.

Conversely, take any open set and assume WLOG that it is connected. Then, there must be only a finite set of natural number indices $I$ which project to not the entire space but instead to those with length we can assume to be at most $1$. That must have a maximum, which we call $n$. For this we can simply take the minimum over $i \leq n$ of the length of the interval for $i$ divided by $i$ as our $\epsilon$.     ▢

Now we need to construct a homeomorphism from our second-countable, regular (and thereby normal) space to some subspace of $\mathbb{R}^\mathbb{N}$. A homeomorphism is injective as part of definition. How to satisfy that? Provide a countable collection of continuous functions to $\mathbb{R}$ such that at least one of them differs whenever two points differ. Here normal comes in handy. Take any two distinct points. Take two non-intersecting closed sets around them and invoke Urysohn’s lemma to construct a continuous function. That would have to be $0$ at one and $1$ at the other. Since our space is second-countable, we can do that for each pair of points with only a countable number. For every pair in the basis $B_n, B_m$ where $\bar{B_n} \subset B_m$, we do this on $\bar{B_n}$ and $X \setminus B_m$.

Proposition Our above construction is homeomorphic to $[0,1]^{\mathbb{R}}$.

Proof: Call our function $f$. Each of its component functions is continuous so the entire Cartesian product is also continuous. It remains to show the other way, that $U$ in the domain open implies the image of $U$ is open. For that it is enough to take $z_0 = f(x_0)$ for any $x_0 \in U$ and find some open neighborhood of it contained in $f(U)$. $U$ contains some basis element of the space and thus, there is a component (call it $f_n$) that sends $X \setminus U$ to all to $0$ and $x_0$ not to $0$. This essentially partitions $X$ by $0$ vs not $0$, with the latter portion lying inside $U$, which means that $\pi_n^{-1}((0, \infty)) \cap f(X)$ is strictly inside $f(U)$. The projections in product space are continuous so that set must be open. This suffices to show that $f(U)$ is open.     ▢

With this, we’ve shown our arbitrary regular, second-countable space to be homeomorphic to a space we directly metrized, which means of course that any regular, second-countable space is metrizable, the very statement of the Urysohn metrication theorem.

## Jewish pro-Americanism

In America, people often bring up what they view as China’s suppression of free expression. I personally dislike strongly the usage of “free expression,” because it is meaninglessly vague. And there is no such thing as free expression in the strictest sense of it. Especially when you are in a job dealing with a boss who can fire you, which is why politics is generally supposed to be a no-no in the workplace, discussion wise. People avoid it out of prudent protection of their careers. One naturally feels at disease when what one wishes to express is such that is unwelcome or hostile in the environment of one’s residence. In such case, one feels that his or her right of free expression is being beaten down. This is very much the case in America right now, in many places.

I’ll say that overall I would feel that China is actually more free in expression overall. Go on the Chinese internet and people can discuss certain matters honestly in a manner unimaginable on the American internet. It helps much that it is for the most part a ethnically homogeneous society, unlike in America, where you have to often be very sensitive to the background of the person you’re talking to (another peril of our cherished diversity I guess). This excepts a few in some sense politically taboo topics like Tiananmen, which people with some interest in the matter might discuss say eating out, just not publicly online. There are also the other two Ts, Taiwan and Tibet. From what I know, Tibet is seldom on the minds of people in China and neither is Taiwan really. In all honestly, people in China have, for the most part, way more interesting things to think about politically than any of these three Ts.

Back to the title of this article, I would say that I am somewhat surprised and also amused at how many highly educated American Jews express openly some diehard belief in American exceptionalism, in particular its “freedom and democracy.” There are plenty of prominent Jewish voices and even actors (like Kissinger) in American foreign policy, one of whom, Amitai Etzioni, I learned about yesterday, seeing that he has written Security First: For a Muscular, Moral Foreign Policy and Avoiding War with China: Two Nations, One World along with many articles on mass media channels like CNN. It’s kind of funny that a guy who fought for Israel against Arabs in 1948 as a teenager (according to Wiki) has such a high position in the BS field of geopolitical strategy in America, as a director and professor in something policy at George Washington University. I won’t name more names but I’ve seen many.

This is not surprising until I think about the situation more carefully. A cynic has a every reason to view Jews in their elite to desire infinite world power and control for America, the country where they exert the most political and economic influence, which their homeland Israel depends on much. On the other hand, they know that anti-Semitism is still very real in Russia, which is not that powerful anymore. I’ll say that I listen to some Soviet songs written by very talented Jewish composers and I admire much the brilliance and work of genius Soviet scientists with Jewish blood too many to name. However, let’s just say that in the Soviet era, Soviet Jews were kept out of political and economic power and more or less confined in the arts and sciences in which they excelled. On that, I’ll say that Trotsky sure left a bad mark for Jews in the Soviet Union. The second most powerful country in the world now, China, Jews basically have zero chance in. Plus, many of them might be aware that in China, one can talk about the grossly disproportionate economic, cultural, and political power possessed by American Jews without any fear of repercussions. That is basically an openly acknowledged fact among Chinese who engage in business. So this highly talented but very small subgroup which has made so many enemies only has America and to a lesser extent its puppet Europe to cling to.

What is rather ironic is that recently, Jews have arguably contributed much to America’s decline. Let’s just say that the Iraq War (which Israel very likely supported) and the financial crisis and recession (Goldman-Sachs is run by Jews) did no good for America, weakening greatly its international position. Those might have put Jews more in favor in America in terms of their control of the economy and their political influence, but of course, that only really counts if America is actually powerful.

Don’t know for sure what Jews were thinking with all that, but if they wanted to play genuine zero-sum games for their own favor, they’d want to strengthen America as much as they can (provided they maintain reasonable level of control over it) and weaken its adversaries, where they have little chance of gaining power without bringing about a coup that replaces the regime with a pro-American one. Of course, support for Israel is always desired, but Israel too small can never sustain itself, which means leeching off America (or some other giant) is an absolute necessity. It’s fair to say that Jews have boosted America to some extent by promoting immigration of high-end talent to work for American companies, whose smart kids will also, by virtue of growing up there, become American culturally and inevitably stay there. It’s also fair to say that Jews have tried hard to bring American culture and products into the rest of the world (to further integrate the rest of world into the American-led world order) with some success. The most glaring failure there is highlighted by that the Chinese government could not be convinced to let in Google and Facebook, which has contributed to a boom of indigenous Chinese tech companies, like Baidu and Tencent. China back a few decades ago seemed puny (with very low GDP per capita and lack of many advanced technologies), but now it is, for the most part, a superpower rivaling the US. With this, China is much more confident and is seeking more create an alternative system that challenges America and thereby Jews. Jewish anti-Chinese (often disguised as anti-communism) sentiment explained. It also hurts that the position of North Korea, which Israel, which itself has nukes, views as a major threat, having survived, whereas Iraq and Libya did not, is more secure the more powerful China becomes. With this, any fantasy of Jewish-led American world domination is ever more a fantasy.

I’ve seen much contempt for China and Chinese among Jews. There are all these stereotypes that Jews are creative and Chinese are not, with Jews 625 times more likely to win a Nobel Prize than an Asian person. It is so much engrained in the culture of stereotypes that I used to sort of believe it myself. Of course, when one looks more closely, one sees that those Nobel Prizes (which may have bias towards certain groups themselves) are mostly awarded to those already in old age, which means it takes not only time but also that a place has been developed and advanced for quite a while. I was rather surprised on seeing how many Nobel Prizes have been awarded to Japanese (mostly working in Japan) in the 21st century. That rate is comparable or close to the rate at which Americans win Nobels if one excludes the BS prizes of peace, literature, and economics and immigrants. Jews can be dismissive of China’s ability to innovate and they even tie it irrelevantly to its political system, in particular its great firewall. They are contradicting themselves. Anyone in the right mind knows that the political system doesn’t affect science research at all so long as the research is adequately economically supported and not disrupted. Ask yourselves why Jews were so successful in science in the “totalitarian” Soviet Union.

I have especially seen contempt among Jews for China’s political system, which some of them see as menacing and threatening. The faked moral superiority will not fool anyone who is not delusional. Everyone acts for his own interests for some degree or another, including China, including America, including Israel, including Russia. To back off from pursuing what is best for oneself under soft pressures and political deception is nothing but a sign of weakness. Anyone strong of heart, including the genuinely loyal Chinese party members, working in all arenas, know the importance of conviction and dedication and not letting it go amidst distraction and enticement.

Anyone with the slightest of political consciousness is aware that most always extraordinary talent is not enough though surely it can overcome initial disadvantages. For instance, being born into a rich, well-educated family is always an advantage. It gives you more material resources to develop your talents and more importantly, the access to connections which often make or break careers. To study and pursue excellence is a privilege that implies that the problem of basic material necessity has already been solved. In this regard, I shall comment that Jewish preeminence in intellectual and artistic pursuits is arguably as much a product of the superior economic, cultural, and social conditions they have accumulated over the past generations as it is of their superior raw talent. It is the former that turns the latter to fruit at higher rates. There is also that Jews, with their verbal talents (and perhaps certain personality characteristics too) combined with their being part of Western culture essentially, are excessively good and willing at self-promotion, which explains why they excel more in softer fields than in hard ones. I don’t want to sound too unpleasantly incisive here, but every time a smart Jew takes a scarce opportunity or position, it deprives a talented white or Asian from developing herself, her gifts, and her career.

America has, needless to say, provided Jews with too much opportunity. It is not often that a group can obtain a disproportionate number of positions of money and power in a powerful, advanced country where they are an immigrant minority with a different culture. America has virtually handed much of itself, its vast resources, to this group, and thus, this group must put America above all, so long its control of it is maintained.

From my experience, white Americans are often too nice and too naive. They don’t know how to scheme and deceive and are often oblivious when it is done to them. America is a wealthy, resource-rich country (per capita) that has not had a war at home since 150 years ago, unlike most of the rest of the world, after all, so there is less of a need to. On the other hand, Jews, aside from having a much higher IQ, have been met with some form of persecution for centuries and even millennia as a minority within Gentile society and necessarily developed such instincts, useful in moneylending, for their own survival that eventually enabled them to dominate more and more of the upper echelons of society with their form of shrewdness. Chinese, being from a densely populated place with little arable land and mountains abound, who have for the last century dealt successfully with powerful opponents, in one case the most powerful country in the world, trying virtually everything to make their country fail, can also easily see through shenanigans. Being much better positioned economically now, Chinese are also more equipped to fight back against them when it is in their interest to do so.

When dealing with anyone or any group, always expect them to place their interests first regardless of how they appear on the surface. As a special case of that, Jewish pro-Americanism is not Jewish pro-Americanism but Jewish pro Jewish domination of a powerful America that not necessarily pro-American at heart, as evidenced by the decline of America in aggregate over the past couple decades, especially following the 2008 financial crisis.

## Riemann mapping theorem

I am going to make an effort to understand the proof of the Riemann mapping theorem, which states that there exists a conformal map from any simply connected region that is not the entire plane to the unit disk. I learned of its significance that its combination with the Poisson integral formula can be used to solve basically any Dirichlet problem where the region in question in simply connected.

Involved in this is Montel’s theorem, which I will now state and prove.

Definition A normal family of continuous functions is one for which every sequence in it has a uniformly convergent subsequence.

Montel’s theorem A family $\mathcal{F}$ on domain $D$ of holomorphic functions which is locally uniformly bounded is a normal family.

Proof: Turns out holomorphic alongside local uniform boundness is enough for us to establish local equicontinuity via the Cauchy integral formula. On any compact set $K \subset D$, we can find some $r$ for which for every point $z_0 \in K$, $\overline{B(z_0, 2r)} \subset D$. By local boundedness we have some $M>0$ such that $|f(z)| \leq M$ in all of $B(z_0, 2r)$. Thus, for any $w \in K$, we can use Cauchy’s integral formula, for any $z \in B(w, r)$. In that, the radius $r$ versus $2r$ is used to bound the denominator with $2r^2$.

\begin{aligned} |f(z) - f(w)| &= \left| \oint_{\partial B(z_0, 2r)} \frac{f(\zeta)}{\zeta - z} - \frac{f(\zeta)}{\zeta - w}d\zeta \right| \\ & \leq |z-w| \oint_{\partial B(z_0, 2r)} \left| \frac{f(\zeta)}{(\zeta - z)(\zeta - w)} \right| d\zeta \\ & < \frac{|z-w|2\pi r}{2\pi 2r^2} M. \end{aligned}

This shows it’s locally Lipschitz and thus locally equicontinuous. To choose the $\delta$ we can divide our $\epsilon$ by that Lipschitz constant alongside enforcing less than $2r$ so as to stay inside the domain.

With this we can finish off with the Arzela-Ascoli theorem.     ▢

Now take the family $\mathcal{F}$ of analytic, injective functions from simply connected region $\Omega$ onto $\mathbb{D}$ the unit disk which take $z_0$ to $0$. On this we have the following.

Proposition If $f \in \mathcal{F}$ is such that for all $g \in \mathcal{F}$, $|f'(z)| \geq g'(z)$, then $f$ surjects onto $\mathbb{D}$.

Proof:  We prove the contrapositive. In order to do so, it suffices to find for any $f$ that hits not $w \in \mathbb{D}$, $f = s \circ g$, where $s, g$ are analytic with $g(z_0) = 0$ and $s$ is a self-map on $\mathbb{D}$ that fixes $0$ and is not an automorphism. In that case, we can deduce from Schwarz lemma that $|s'(0)| < 1$ and thereby from the chain rule that $g'(z_0) > f'(z_0)$.

Recall that we have automorphisms on $\mathbb{D}$, $T_w = \frac{z-w}{1-wz}$, for all $w \in \mathbb{D}$ and that their inverses are also automorphisms. Let’s try to take $0$ to $w$, then $w$ to $w^2$ via $p(z) = z^2$, and finally $w^2$ to $0$. With this, we have a working $s = T_{w^2} \circ p \circ T_w^{-1}$.     ▢

Nonemptiness of family

It is not difficult to construct an analytic injective self map on $\mathbb{D}$ that sends $z_0$ to $0$. The part of mapping $z_0$ to $0$ is in fact trivial with the $T_w$s. To do that it suffices to map $\mathbb{D}$ to $\mathbb{C} \setminus \overline{\mathbb{D}}$ as after that, we can invert.

Since $D$ is not the entire complex plane, there is some $a \notin D$. By translation, we can assume that $a = 0$. Because the region is simply connected, there is a path from $0$ to $\infty$ outside the region, which means there is an analytic branch of the square root. For any $w$ that gets hits by that, $-w$ does not. By the open mapping theorem, we can find a ball centered at $-w$ that is entirely outside the region. With this, we can translate and dilate accordingly to shift that to the unit disk.

Construction of limit to surjection

We can see now that if we can construct a sequence of functions in our family that converges to an analytic one with the same zero at $z_0$ with maximal derivative (in absolute value) there, we are finished. Specifically, let $\{f_n\}$ be a sequence from $\mathcal{F}$ such that

$\lim_n |f'_n(z_0)| = \sup_{f \in \mathcal{F}} \{|f'(z_0)|\}$.

This can be done by taking functions with sufficiently increasing derivatives at $z_0$. With Montel’s theorem on our obviously locally uniformly bounded family, we know that our family is normal, and thus by definition, we can extract some subsequence that is uniformly convergent on compact sets. Now it remains to show that the function converged to is analytic and injective.

The injective part follows from a corollary of Hurwitz’s theorem, which we now state.

Hurwitz’s theorem (corollary of) If $f_n$ is a sequence of injective analytic functions with converge uniformly on compact sets to $f$, then $f$ is constant or injective.

Proof: Recall that Hurwitz’s theorem states that if $f$ has a zero of some multiplicity $m$ at some point $z_0$, then for any $\epsilon > 0$, we will, past some $N$ in the index of the sequence, have $m$ zeros within $B(z_0, \epsilon)$ for all $f_n, n > N$, provided $f$ is not constantly $0$. For any point to see that a non-constant $f$ can hit it only once, it suffices to do a translation by that point on all the $f_n$s to turn it into a zero, so that the hypothesis of Hurwitz’s theorem, which in this case, bounds the number of zeros above by $1$, with the $f_n$s being injective, can be applied.     ▢

To show analyticity, we can use Weierstrass’s theorem.

Weierstrass’s theorem Take $\{f_n\}$ and supposed it converges uniformly on compact sets to $f$. Then the following hold:
a. $f$ is analytic.
b. $\{f'_n\}$ converges to $f'$ uniformly on compact sets.

Proof: This is a more standard theorem, so I will only sketch the proof. Recall the definition of compact as possessing the every cover has finite subcover property. This is so powerful, because we can for any collection of balls centered at every point of the cover, find a finite of them that covers the entire space, and finiteness allows us to take a maximum or minimum of finite $N$s or $\delta$s to uniformize some limit.

We can do the same here. For every $z$ on a compact set, express $f_n$ as integral of $\frac{f_n}{\zeta - z}$ via Cauchy’s integral formula on some ball centered at $z$. Uniform convergence of $\frac{f_n}{\zeta - z}$ on the boundary to $\frac{f}{\zeta-z}$ allows us to put the limit inside the integral to give us $f$, as represented via Cauchy’s integral formula. The same can be done for the $\{f'_n\}$

Again we can use two radii as done in the proof of Montel’s theorem to impose uniform convergence on a smaller ball.     ▢

Finally, our candidate conformal map to $\mathbb{D}$ satisfies that $f(z_0) = 0$. If not, convergence would be naught at $z_0$ since $f_n(z_0) = 0$ for all $n$.

This gives us existence. There is also a uniqueness aspect of the Riemann mapping theorem that comes when one imposes $f'(z_0) \in \mathbb{R}$. This is very elementary to prove and will be left to the reader.

## Arzela-Ascoli theorem and delta epsilons

I always like to think of understanding of the delta epsilon definition of limit as somewhat of an ideal dividing line on the cognitive hierarchy, between actually smart and pseudo smart. I still remember vividly struggling to grok that back in high school when I first saw it junior year, though summer after, it made sense, as for why it was reasonable to define it that way. That such was only established in the 19th century goes to show how unnatural such abstract precise definitions are for the human brain (more reason to use cognitive genomics to enhance it 😉 ). At that time, I would not have imagined easily that this limit definition could be generalized further, discarding the deltas and epsilons, which presumes and restricts to real numbers, as it already felt abstract enough. Don’t even get me started on topological spaces, nets, filters, and ultrafilters; my understanding of them is still cursory at best, but someday I will fully internalize them.

Fortunately, I have improved significantly since then, both in terms of experience and in terms of my biological intelligence, that last night, I was able to reconstruct in my head the proof of the Arzela-Ascoli theorem, which also had been once upon a time mind-boggling. Remarkably, that proof, viewed appropriately, comes naturally out of just a few simple, critical observations.

The statement of the theorem is as follows.

Arzela-Ascoli theorem Let $F$ be a family of functions from $I = [a,b]$ to $\mathbb{R}$ that are uniformly bounded and equicontinuous. Then there is a sequence of $f_n$ of elements in $F$ that converges uniformly in $I$.

The rationals are dense in the reals and make an excellent starting point. Uniform boundedness enables us employ Bolzano-Weierstrass to construct a sequence of functions convergent at any rational in $I$. With a countable number of such rationals, we can iteratively filter this sequence to construct one that converges at every rational in $I$. Essentially, we have an enumeration of the rationals in $I$ and a chain of subsequences based on that, where in the $n$th subsequence is convergent at the first $n$ rationals in the enumeration, and Bolzano-Weierstrass is applied onto the results of the application of the functions of that subsequence on the $n+1$th rational to yield another subsequence. Take, diagonally, the $n$th function in the $n$th subsequence, to derive our desired uniformly convergent sequence, which we call $\{f_i\}$.

To show uniform convergence, it suffices to show uniform Cauchyness, namely that for any $\epsilon > 0$, there is an $N$ such that $n,m > N$ implies $|f_m(x) - f_n(x)| < \epsilon$ for all $x \in I$. By compactness, open neighborhoods of all rationals of $I$, as an open cover, has a finite subcover. Each element of the subcover comes from some rational of $I$ and across that finite subset of $I$ we can for any $\epsilon > 0$ take the max of all the $N$s for convergence. This means that so long as our neighborhoods are sufficiently small, we can for any point $x \in I$ have some $x'$ that is the point of focus of the neighborhood of our finite subcover containing $x$ and thereby connect $f_m(x)$ to $f_m(x')$ by equicontinuity and use our maximum $N$ (over finite elements) to connect that to $f_n(x')$ and use equicontinuity again to connect that to $f_n(x)$. Thus, triangle inequality over three of $\epsilon/3$ suffices.

More explicitly, equicontinuity-wise, we have for every $x$ some open neighborhood of $U_x$ such that $s,t \in U_x$ implies that $|f(s) - f(t)| < \epsilon$.

## My awesome roommate

The guy that is the topic of this post himself did up to high school, as far as I know, in Hong Kong, so we have some more in common than usual culturally I guess. He was just telling me about how he had read 矛盾论, which I haven’t even read, at least not in detail, myself. He was saying, on the putative connection between scientific talent and Marxism, perhaps how dialectical materialism is inherently a very scientific way of thinking. I myself know basically nothing about dialectical materialism and even think it’s kind of high verbal low math bullshit, but I can tell that the materialist side of it is very scientific in its very nature, and similarly, dialectics is a very analogies/relationships way of thinking, which is something that high IQ people are by definition good at. Surely, there is much more I can learn from this guy, especially about Chinese language and culture and politics.

On this, I am reminded of another amateur (but professional, or better, level for sure) Marxist scholar, who is genuinely encyclopedic in his historical and cultural knowledge, in particularly a perceptive quote of him that made a deep impression on me:

Europe has always been in rebellion against itself, and continues to be so.  There was nothing but futility in the attempt by superficially Westernised Chinese to be authentically Westernised Chinese by being imitative and reverential of the current embodiment of those values.  You could only be an authentically Westernised Chinese by being a rebel against the current embodiments of Western values, at least in as far as they hampered China or seemed to be irrelevant.  And that’s why Mao was China’s best Westerniser to date, despite his very limited experience of the mundanities of Western life.

As I’ll detail in a future article, visitors to the Chinese Communist bases at Bao’an and later Yen’an noticed that these were the only Chinese in China who behaved more or less as Westerners would have behaved in a similar situation.  Other Chinese might speak good English, wear Western suits and sometimes show considerable knowledge of Western culture: but it was all imitation and the inner core was different and ineffective.  Western-trained engineers and geologists who returned to China kept their distance from hands-on practical work, because anything resembling manual labour would have lost them status in the eyes of Chinese intellectuals.  They were imprisoned by a tradition stretching back to Confucius and beyond.  Only a few broke these ancient taboos, mostly the Communists and some scattered left-wingers in the weak middle ground.  And it was the modernised Chinese in the Communist Party who chose to raise up Mao as the prime teacher of this new understanding.

I remember when my obsessively talented Russian friend once said to me that sometimes he feels like he’s another Pavel Korchagin, I thought he was ridiculous. Well, I’ll be equally ridiculous and say that I feel like I very much exhibit what Gwydion described in Mao that is “authentically Westernized Chinese,” which is very much the antithesis of what I see in most ABCs, despite being half an ABC myself.

If only more people could be like me…

## Path lifting lemma and fundamental group of circle

I’ve been reading some algebraic topology lately. It is horrendously abstract, at least for me at my current stage. Nonetheless, I’ve managed to make a little progress. On that, I’ll say that the path lifting lemma, a beautiful fundamental result in the field, makes more sense to me now at the formal level, where as perceived by me right now, the difficulty lies largely in the formalisms.

Path lifting lemma:    Let $p : \tilde{X} \to X$ be a covering projection and $\gamma : [0,1] \to X$ be a path such that for some $x_0 \in X$ and $\tilde{x} \in \tilde{X}$,

$\gamma(0) = x_0 = p(\tilde{x_0}). \ \ \ \ (1)$

Then there exists a unique path $\tilde{\gamma} : [0,1] \to \tilde{X}$ such that

$p \circ \tilde{\gamma} = \gamma, \qquad \tilde{y}(0) = \tilde{x_0}. \ \ \ \ (2)$

How to prove this at a high level? First, we use the Lebesgue number lemma on an open cover of $X$ by evenly covered open sets to partition $[0,1]$ into intervals of length $1/n < \eta$, with $\eta$ the Lebesgue number, to induce $n$ pieces of the path in $X$ which all lie in some open set of the cover. Because every open set is evenly covered, we for each piece have a uniquely determined continuous map (by the homeomorphism of the covering map plus boundary condition). Glue them together to get the lifted path, via the gluing lemma.

Let $\mathcal{O}$ be our cover of $X$ by evenly covered open sets. Let $\eta > 0$ be a Lebesgue number for $\gamma^{-1}(\mathcal{O})$, with $n$ such that $1/n < \eta$.

Let $\gamma_j$ be $\gamma$ restricted to $[\frac{j}{n}, \frac{j+1}{n}]$. At $j = 0$, we have that $p^{-1}(\gamma_0([0,\frac{1}{n}]))$ consists of disjoint sets each of which is homeomorphic to $\gamma_0([0, \frac{1}{n}])$, and we pick the one that contains $\tilde{x_0}$, letting $q_0$ denote the associated map for that, to $\tilde{X}$, so that $p \circ (q_0 \circ \gamma_0) = \gamma_0$, with $\tilde{\gamma_0} = q_0 \circ \gamma_0$.

We continue like this for $j$ up to $n-1$, using the value imposed on the boundary, which we have by induction to determine the homeomorphism associated with the covering projection that keeps the path continuous, which we call $q_j$. With this, we have

$\tilde{\gamma_j} = q_j \circ \gamma_j$.

A continuous path $\tilde{\gamma}$ is obtained by applying to the gluing lemma to these. That

$p \circ \tilde{\gamma} = \gamma$

is satisfied because it is satisfied on sets the union of which is the entire domain, namely $\{[\frac{j}{n}, \frac{j+1}{n}] : j = 0,1,\ldots,n-1\}$.

A canonical example of path lifting is that of lifting a path on the unit circle to a path on the real line. To every point on the unit circle is associated its preimage under the map $t \mapsto (\cos t, \sin t)$. It is not hard to verify that this is in fact a covering space. By the path lifting lemma, there is some unique path on the real line that projects to our path on the circle that ends at some integer multiple of $2\pi$, call it $2\pi n$, and that path is homotopic to the direct path from $0$ to $2\pi n$ via the linear homotopy. Application of the projection onto that homotopy yields that our path on the circle, which we call $f$, is homotopic to the path where one winds around the circle $n$ times counterclockwise, which we call $\omega_n$.

Homotopy between $f$ and $\omega_n$ is unique. If on the other hand, $\omega_n$ were homotopic to $\omega_m$ for $m \neq n$, they we could lift the homotopy onto the real line, thereby yielding a contradiction as there the endpoints would not be the same.

This requires a homotopy lifting lemma. The proof of that is similar to that of path lifting, but it is more complicated, since there is an additional homotopy parameter, by convention, within $[0,1]$, alongside the path parameter. Again, we use the Lebesgue number lemma, but this time on grid $[0,1] \times [0,1]$, and again for each grid component there is a unique way to select the local homeomorphism such that there is agreement with its neighboring components, with the parameter space in common here an edge common to two adjacent grid components.

With that every path on the circle is uniquely equivalent by homotopy to some unique $\omega_n$, we have that its fundamental group is $\mathbb{Z}$, since clearly, $\omega_m * \omega_n = \omega_{m+n}$, where here, $*$ is the path concatenation operation.

## Oleg

Oleg is one of my ubermensch Soviet (and also part Jewish) friends. He has placed at (or at least near) the top on the most elite of math contests. He is now a math PhD student with an advisor even crazier than he is, who he says sometimes makes him feel bad, because he has done too little math research wise. However, this persona alone is not that rare. Oleg’s sheer impressiveness largely stems from that on top of this, he is a terrific athlete, extremely buff and coordinated, enough that he can do handstand pushups, to the extent that he regards such as routine. Yes, it is routine for a guy contending for a spot on a legit gymnastics team, but you wouldn’t expect this from a math nerd huh?

Today, I was talking to him and some others about gym. In particular, I was saying how I could at one point do 10 pullups but dropped down to 2 after a long hiatus. The conversation went as follows:

Me: Oleg I’m back to 5 pull-ups now
Oleg: that’s good although make sure you’re doing them for real
i still don’t believe you could do 10 but then dropped down to 2
Me: Oh I’m very sure they’re full pullups
Okay maybe it was 8
Oleg: i’d like to see evidence
Me: Alright I’ll have someone videotape me do pullups today in gym

And so I did.

Later, Oleg suggested something pretty funny:

i still think you should get tattoos and gain 25 lb of muscle, that would be hilarious