In my last post (found here) I mentioned going through the metrization theorem. Today, I’ll actually go through and prove the result.

We recall a few definitions ahead of time. A topological vector space is a vector space \(X\) (over a field \(F\) which is either \(\mathbb{R}\) or \(\mathbb{C}\)) equipped with a topology \(\tau\) such that the singletons are closed sets and scalar multiplication \(F \times X \rightarrow X\), \((\alpha, x) \mapsto \alpha x\) and addition \(X \times X \rightarrow X\), \((x,y) \mapsto x+y\) are continuous.

A topological vector space \((X,\tau)\) (where we generally drop the \(\tau\) and just call \(X\) a topological vector space) is said to be metrizable if there exists a metric \(d : X \times X \rightarrow \mathbb{R}_{\geq 0}\) such that the topology of open balls generates \(\tau\). Recall as well that the metric \(d\) generates the topology \(\tau\) if the open balls \(B_\epsilon(x) = \{ y : d(x,y) < \epsilon\}\) is a base for the topology. This should be sufficient to get us started for the theorem. A neighborhood of the origin is an open set containing the 0 vector, and a local base is a set of neighborhoods of the origin such that for every open set \(U\) which contains the origin, we have that there is an open set \(V\) in our local base with \(V \subset U\). In essence, we can write every open set \(U\) containing the origin as a union of sets from our local base.

Metrization theorem statement

Let \((X, \tau)\) be a topological vector space. Then \(X\) is metrizable if and only if \(X\) has a countable local base.

To prove this, we will actually break it up into two parts. There is a very trivial direction (the forward direction) and a not so trivial direction (the backwards direction, or the converse). However, while we prove the converse, we’ll actually get a lot more for free.

Proof of metrization theorem

Before getting too far, we remark that if we can find a local base (meaning a base for the set of neighborhoods centered at the origin), then we have a base for the entire topology. This follows by noting that the translation map

\[T_\gamma : X \rightarrow X\]

given by

\[T_\gamma(x) = x + \gamma\]

is a homeomorphism. So if we have a local base \(\mathcal{B}(0)\), we can shift it via this homeomorphism to get a base centered at a different point, and then taking the union of translates of these open sets we get a base for the entire topology. In essence, we reduce the problem of looking at the whole topology to looking at open neighborhoods at the origin.

Now, assuming that \(X\) is metrizable, we get a metric \(d\) which is compatible with our topology, meaning that the open balls \(B_\epsilon(0)\) are a local base. Notice that we can refine this local base to the open balls \(B_{1/n}(0)\) for \(n\) a natural number greater than or equal to 1. This is a countable local base, and so we are done.

For the backwards direction, we need to get a little trickier. Note that a set \(U \subset X\) is said to be balanced if we have that, for all \(\alpha \in F\) our underlying field with \(\mid\alpha\mid = 1\), we get \(\alpha U \subset U\).

Theorem

If \(X\) is a topological vector space with a countable local base, then there is a metric \(d\) on \(X\) such that:

(a) \(d\) is compatible with the topology of \(X\).

(b) the open balls centered at \(0\) are balanced, and

(c) \(d\) is invariant, meaning

\[d(x+z, y+z) = d(x,y).\]

Proof

Step 1: Every neighborhood of 0 contains a balanced neighborhood of 0.

Let \(U\) be a neighborhood of 0 in \(X\). Scalar multiplication is continuous, so there is a \(\delta > 0\) and a neighborhood of 0 \(V\) so that \(\alpha V \subset U\) whenever \(\mid \alpha\mid < \delta\). Taking the union over all \(\alpha V\), we have a balanced neighborhood of 0.

As a consequence of this, for every local base we can find a local balanced base.

Step 2: For every neighborhood of 0 in \(X\), say \(W\), there is a neighborhood \(U\) of \(0\) which is symmetric and which satisfies \(U + U \subset W\).

We have 0 + 0 = 0, and addition is continuous, so we can find neighborhoods \(V_1, V_2\) so that \(V_1 + V_2 \subset W\). Taking

\[U = V_1 \cap V_2 \cap (-V_1) \cap (-V_2),\]

we have the desired set.

Step 3: By assumption, we have a countable local base \(\{U_n\}\). By steps 1 and 2, we can refine this to get a countable local base \(\{V_n\}\) where

\[V_{n+1} + V_{n+1} + V_{n+1} + V_{n+1} \subset V_n.\]

Step 4: Consider the dyadic rational numbers; i.e. consider the set \(D\) consisting of rational numbers \(r\) of the form

\[r = \sum_{n=1}^\infty c_n(r) 2^{-n},\]

where the digits \(c_j(r)\) are either 0 or 1, and only finitely many are 1. Note that \(0 \leq r < 1\).

Put \(A(r) = X\) if $r \geq 1$, and for \(r \in D\) define

\[A(r) = c_1(r) V_1 + c_2(r) V_2 + \cdots.\]

Note that each of these sums are actually finite.

Step 5: Define the function

\(f(x) = \inf \{ r : x \in A(r)\}\) for \(x \in X\). Note that \(f(x) \leq 1\) for all \(x \in X\), so that the infimum is finite.

Step 5: We show that if \(r,s\) and \(r+s\) are in \(D\) and \(c_n(r) + c_n(s) \neq c_n(r+s)\) for some \(n\), then for the smallest such \(n\) where this happens we have that \(c_n(r) = c_n(s) = 0\) and \(c_n(r+s) = 1\). This is just a consequence of the fact that

\[r = \sum c_n(r) 2^{-n},\]

\(s = \sum c_n(s) 2^{-n}\),

and so if \(c_n(r) = c_n(s) = 1\), then we have that their sum is 2. Since these are dyadic numbers, this shifts the coefficient to \(2^{-n+1}\). Thus, we must have the inequality happens when \(c_n(r) = c_n(s) = 0\) and \(c_n(r+s) = 1\).

Step 6: We establish

\[A(r) + A(s) \subset A(r+s)\]

for all \(r,s \in D\). If \(r +s \geq 1\), then we are done, since \(A(r+s) = X\). Thus, we consider \(r+ s < 1\). Put \(\alpha_n = c_n(r)\), \(\beta_n = c_n(s)\), and \(\gamma_n = c_n(r+s)\). Let N be the smallest integer for which \(\alpha_N + \beta_N \neq \gamma_N\). Then by Step 5, we see \(\alpha_N = \beta_N = 0\) and \(\gamma_N = 1\), giving us

\[A(r) \subset \alpha_1 V_1 + \cdots + \alpha_{N-1}V_{N-1} + V_{N+1} + V_{N+2} + \cdots\] \[\subset \alpha_1 V_1 + \cdots + \alpha_{N-1}V_{N-1} + V_{N+1} + V_{N+1},\]

where here the last subset is due to the construction in Step 3. We have an analogous result for \(A(s)\), and we see that since \(\alpha_n + \beta_n = \gamma_n\) for \(n < N\), we get

\[A(r) + A(s) \subset \gamma_1 V_1 + \cdots + \gamma_{N-1}V_{N-1} + V_N \subset A(r+s).\]

Step 7: Since every \(A(s)\) contains 0, we see that Step 6 says

\[A(r) \subset A(r) + A(t-r) \subset A(r + (t-r)) = A(t).\]

Thus, \(\{A(r)\}\) is a POSET by inclusion.

Step 8: We claim that Step 7 gives us that

\[f(x+y) \leq f(x) + f(y).\]

We assume without loss of generality that \(f(x) + f(y) < 1\), since otherwise this trivially follows. Fixing \(\epsilon > 0,\) there exists r and s dyadic rational numbers such that \(f(x) < r, f(y) < s, r + s < f(x) + f(y) + \epsilon.\) So \(x \in A(r), y \in A(s)\) and Step 6 implies that \(x + y \in A(r+s)\). So

\[f(x+y) \leq r + s < f(x) + f(y) + \epsilon.\]

Step 9: Since the \(A(r)\) are balanced, we get \(f(x) = f(-x)\), and \(f(0) = 0\). So \(d(x,y) = d(y,x).\) If \(x \neq 0\), then \(x \notin V_n = A(2^{-n})\) for some \(n\), so \(f(x) \geq 2^{-n} > 0\). Hence, we see that \(d(x,y) = 0\) if and only if \(x = y\). Step 9 establishes the triangle inequality, since

\[d(x,y) = f(x-y) = f(x - z + z - y) \leq f(x - z) + f(z-y) = d(x,z) + d(y,z).\]

Finally, translation invariance follows from \(d(x+z,y+z) = f(x+z-y-z) = f(x-y) = d(x,y).\)

Step 10: We need to show that this generates the topology. The open balls centered at \(0\) are the sets

\[B_\delta(0) = \{x : d(x,0) < \epsilon\} = \{x : f(x) < \delta\} = \bigcup_{r < \delta} A(r).\]

To prove this is a local base, it suffices to show that for small enough \(\delta\), we have \(B_\delta(0) \subset V_n\). This just follows from choosing \(\delta < 2^{-n}.\) Each \(A(r)\) is balanced, so \(B_\delta(0)\) is balanced, and so we have all the desired properties. Q.E.D.