Metrizable

Definition (Metrizable)

A Topological Space $(X,\tau )$ is metrizable if $\exists$ a metric

$$d:X\times X\to {\mathbb{R}}_{\ge 0}$$

which induces the topology $\tau$.

Remark (Non-Hausdorff Spaces are not Metrizable)

Any non-Hausdorff space is clearly non metrizable by Theorem (Metric Spaces are Hausdorff).

Making a Product of Metric Spaces

Suppose we had a bunch of metric spaces $\{(X_i,d_i){\}}_{i\in I}$. Can we metrize the product in a way that induces the product topology?

If $I$ is finite, then this is straightforward.

Theorem (Product Space is Metrizable)

If $(X_i,{\tau }_i{)}_{i\in I}$ are metrizable spaces with metrics $d_i$ then

$$\left(\prod X_i,\text{ product topology}\right)$$

is metrizable by writing

$$d_{\infty }(\underline{x},\underline{y})=\max \{d_i(x_i,y_i)\}$$

where $\underline{x}$ and $\underline{y}$ are “tuples” in the product space.

Proof:

We use the principle: Suppose ${\beta }_1,{\beta }_2$ are two bases. To show that the two topologies are the same, we just need to show that $\forall B_1\in {\beta }_1$, for any $x\in B_1$, there $\exists B_2\in {\beta }_2$ containing $x$ contained in $B_1$, and vice versa.

If $U_1\times U_2\times \cdots \times U_n$ is a basic (part of the basis) open set of the product topology containing some $x$, we can find some

$$B_{{\epsilon }_i}(x_i)\subseteq U_i\text{ for any }\epsilon >0$$

Take the minimum since we have some open set of finite product.

$$\epsilon =\min \{{\epsilon }_1,\dots ,{\epsilon }_n\}$$

We get

$$B_{\epsilon }^{d_{\infty }}(\underline{x})\subseteq U$$

Indeed,

$$B_{\epsilon }^{d_{\infty }}(\underline{x})=B_{\epsilon }(x_1)\times \cdots \times B_{\epsilon }(x_n)$$

which is a basis element for the product topology.

Remark (Different Metric for Product Space)

It would still work if we used

$$d_1(\underline{x},\underline{y})=\sum d_i(x_i,y_i)$$

In fact, these are equivalent metrics (i.e. give the same topology). The proof is slightly different since we have a different $\epsilon$. At a “higher” level, the epsilon balls are no longer a rectangle, but a “diamond” in ${\mathbb{R}}^2$.

Metrizable Infinite Spaces

Naively, we’d define

$$D(\underline{x},\underline{y})=\sup \{d_i(x_i,y_i)\}$$

but this will probably be infinite. A better way is to replace each metric $d_i$ with the bounded metric corresponding to it.

$$\overline{d_i}(x_i,y_i)=\min \{d_i(x_i,y_i)\}$$

We would check $\overline{d_i}$ is a metric and induces the same topology as $d_i$. We then set

$$D(\underline{x},\underline{y})=\underset{i\in I}{\sup }\{\overline{d_i}(x_i,y_i)\}$$

but this gives the uniform topology on the product, which is different from the product topology (in general).

If is uncountable, in general we cannot define metric that induces the product.

Suppose that $I$ is countable. Ultimately, the solution requires us to make the metrics “shrink”. In particular, consider a countable product

$$\prod _{i\in \mathbb{N}}(X_i,d_i)$$

and define

$$D(\underline{x},\underline{y}):=\underset{i\in \mathbb{N}}{\sup }\left\{\frac{\overline{d}(x_i,y_i)}{i}\right\}$$

where the diameter of $x_i$ is now $1/i$ instead of $1$. We have to check two things. One, this is indeed a metric.

• If all the points are the same, the supremum is $0$. And indeed, if two points are different, then at least one element is greater than $0$.
• For the same reasoning, it is symmetric.
• For each $d_i$, we know $$\begin{array}{rl}{\overline{d}}_i(x_i,z_i)/i& \le \overline{d}(x_i,y_i)/i+{\overline{d}}_i(y_i,z_i)/i\\ & \le D(\underline{x},\underline{y})+D(\underline{y},\underline{z})\end{array}$$ Hence ${\sup }_i$ of the LHS is also $\le$ to the RHS, and the triangle inequality is satisfied. Then we need to show it is actually the product topology. We have to show that basic open sets of the $D$ metric on this space can be written as basic epsilon balls in the product topology and conversely.

Now we need to show the topologies coincide. Given $V$ open in the $D-$topology containing $\underline{x}$, we can find $\epsilon >0$ such that

$$B_{\epsilon }^D(\underline{x})\subseteq V$$

Now choose $M$ such that $1/N<\epsilon$ by Archimedean Property and let

$$U=B_{\epsilon }^{d_1}(x_1)\times \cdots \times B_{\epsilon }^{d_n}(x_n)\times \text{whole spaces}$$