## 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.

## 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.

## Another characterization of compactness

The canonical definition of compactness of a topological space $X$ is every open cover has finite sub-cover. We can via contraposition translate this to every family of open sets with no finite subfamily that covers $X$ is not a cover. Not a cover via de Morgan’s laws can be characterized equivalently as has complements (which are all closed sets) which have finite intersection. The product is:

A topological space is compact iff for every family of closed sets with the finite intersection property, the intersection of that family is non-empty.

## Grassmannian manifold

We all know of real projective space $\mathbb{R}P^n$. It is in fact a special space of the Grassmannian manifold, which denoted $G_{k,n}(\mathbb{R})$, is the set of $k$-dimensional subspaces of $\mathbb{R}^n$. Such can be represented via the ranges of the $k \times n$ matrices of rank $k, k \leq n$. On application of that operator we can apply any $g \in GL(k, \mathbb{R})$ and the range will stay the same. Partitioning by range, we introduce the equivalence relation $\sim$ by $\bar{A} \sim A$ if there exists $g \in GL(k, \mathbb{R})$ such that $\bar{A} = gA$. This Grassmannian can be identified with $M_{k,n}(\mathbb{R}) / GL(k, \mathbb{R})$.

Now we find the charts of it. There must be a minor $k \times k$ with nonzero determinant. We can assume without loss of generality (as swapping columns changes not the range) that the first minor made of the first $k$ columns is one of such, for the convenience of writing $A = (A_1, \tilde{A_1})$, where the $\tilde{A_1}$ is $k \times (n-k)$. We get

$A_1^{-1}A = (I_k, A_1^{-1}\tilde{A_1})$.

Thus the degrees of freedom are given by the $k \times (n-k)$ matrix on the right, so $k(n-k)$. If that submatrix is not the same between two full matrices reduced via inverting by minor, they cannot be the same as an application of any non identity element in $GL(k, \mathbb{R})$ would alter the identity matrix on the left.

I’ll leave it to the reader to run this on the real projective case, where $k = 1, n = n+1$.