# Abstract Nonsense

## Thoughts about separation (Urysohn’s Metrization theorem)

We now come up to one of the most important results of normality.

Although not explicitly stated every metric space has all the separation properties discussed up until now. Consequently, it is sometimes readily obvious when a space is not metrizable. For example, the topological space $\left(\{a,b,c\},\left\{\varnothing,\{a\},\{a,b,c\}\right\}\right)$ is not metrizable for the simple fact that it’s not $T_1$. Conversely,every discrete space is metrizable under the discrete metric. That said, it is, in general, much harder to prove that a space is metrizable than to prove it is not. It turns that a sufficient (but not necessary) condition for a space to be metrizable is that it’s second countable and normal.

Before we give the actual theorem we prove an elementary theorem.

Theorem: Let $X$ be a topological space, then $X$ is metrizable if and only if it is homeomorphic to a subspace of a metric space $M$.

Proof:

$\implies$: If $\left(X,\mathfrak{J}\right)$ is metrizable then there exists some metric space $\left(X,d\right)$ such that the open sets induced by $d$, denoted $O_d$, are equal to $\mathfrak{J}$. Thus, if define $\iota_X:\left(X,\mathfrak{J}\right)\mapsto \left(X,d\right)$ as the identity function. This is clearly bijective, so it remains to show that it’s bicontinuous.

So, let $U$ be open in $\left(X,d\right)$ then $U\in O_d$ and so by previous discussion $U\in\mathfrak{J}$. But, since $\iota^{-1}\left(U\right)=U$ we see that it’s preimage is open. Thus, it is continuous.

Conversely, if $U$ is open in $\left(X,\mathfrak{J}\right)$ then $U\in\mathfrak{J}\implies U\in O_d$ and so $\iota\left(O\right)=O$ is open.

The conclusion follows.

$\Longleftarrow$: Suppose that $\left(X,\mathfrak{J}\right)\approx E$ which is a subspace of $\left(M,d\right)$. We know then that there exists some $f:E\mapsto \left(X,\mathfrak{J}\right)$ which is a homeomorphism. So, since for each $x,y\in X$ we have that $x=f(e_x),y=f(e_y)$ for some $e_x,e_y\in E$ define $d':X\times X\mapsto\mathbb{R}$ by $d'(x,y)=d(e_x,e_y)$. This is clearly a metric on $X$. The conclusion follows. $\blacksquare$

With this theorem in mind, to show that every second countable normal space is metrizable we must merely show that it’s homeomorphic to a subspace of a metric space. This process is commonly known as embedding.

Before we begin we need some preliminary results about a very common metric space.

$\mathbb{R}^{\infty}$ or $\ell^2$: The book I learned from refers to this space as $\mathbb{R}^\infty$ but those with an analysis background you will recognize this as the more commonly coined $\ell^2$.

So, let us construct this metric space and make some comments on it.

We define the underlying set of $\mathbb{R}^{\infty}$ to be the set of all $f:\mathbb{N}\mapsto\mathbb{R}$ such that $\displaystyle \sum_{n=1}^{\infty}\left|f(n)\right|^2<\infty$. In words “it is the set of all square summable sequences”.

We define $d:\mathbb{R}^{\infty}\times\mathbb{R}^{\infty}\mapsto\mathbb{R}$ by $\displaystyle \sum_{n=1}^{\infty}\left|f(n)-g(n)\right|^2$.

The positive-definiteness and symmetry are obvious, and the triangle inequality follows from Minkowski’s inequality.

Thus, $\left(\mathbb{R}^{\infty},d\right)$ (from now on just denoted as $\mathbb{R}^{\infty}$) is a metric space. We now may ask what some of it’s qualities are?

Well, not super-relevant but for those more accustomed with functional analysis, it is obvious that $\mathbb{R}^{\infty}$ is a Hilbert space, and thus a Banach space.

Also, it is second countable. There is a way to finagle this around to use the earlier result that the countable product of second countable spaces is second countable, but there is an easier way. Let $\mathbb{Q}_{\mathcal{Z}}^{\infty}$ the set of all eventually zero rational sequences. It is clear that this is countable since $\displaystyle \mathbb{Q}_{\mathcal{Z}}^{\infty}\simeq\bigcup_{n=1}^{\infty}\bigcup_{\bold{q}\in\mathbb{Q}^n}\bold{q}$.

It is also clear that it’s dense in $\mathbb{R}^{\infty}$. For, given any $\varepsilon>0$ and any $f \in\mathbb{R}^{\infty}$  we may find some $N\in\mathbb{N}$ such that $N\leqslant j\implies \sum_{n=j}^{\infty}|f(n)|^2<\frac{\varepsilon}{2}$ (this is just the Cauchy criterion). But, since $\mathbb{Q}$ is dense in $\mathbb{R}$ we may find some $q_k\in\mathbb{Q}$ such that $\left|q_k-f(k)\right|<\frac{\sqrt{\varepsilon}}{2\sqrt{N}},\text{ }1\leqslant k\leqslant N$. Thus, if we let $\displaystyle q_n=\begin{cases} q_k & \mbox{if} \quad 1\leqslant k\leqslant N \\ 0 & \mbox{if} \quad k>N \end{cases}$ then $q_n\in\mathbb{Q}_{\mathcal{Z}}^{\infty}$ and

$\displaystyle d(q_n,f(n))=\sum_{n=1}^{\infty}\left|q_n-f(n)\right|^2=\sum_{n=1}^{N}\left|q_n-f(n)\right|^2+\sum_{n=N+1}^{\infty}|f(n)|^2<\varepsilon$

One last note. Since, as was mentioned above, $\mathbb{R}^{\infty}$ is a Hilbert space, it is common to use the $\|\cdot\|$ norm notation instead of the more metric space $d(,)$ notation.

It turns out that second countability and normality is all a topological space requires to be embedded into $\mathbb{R}^{\infty}$. This is proved in the following theorem.

Theorem (Urysohn’s Metrization Theorem): Let $X$ be a second countable normal space, then $X$ may be embedded into $\mathbb{R}^{\infty}$ and thus it is metrizable.

Proof: Let $\mathfrak{B}$ be the countable base for $X$. Define $\mathfrak{U}\subseteq\mathfrak{B}\times\mathfrak{B}$ to be $\mathfrak{U}=\left\{(B_j,B_k):\overline{B_j}\subseteq B_k\right\}$. It is clear that $\mathfrak{U}$ is countable and so $\mathfrak{U}=\left\{U_n\right\}_{n\in\mathbb{N}}$. For each $U_n\in\mathfrak{U}$ we have that $U_n=\left(B_{j_n},B_{k_n}\right)$ such that $\overline{B_{j_n}}\subseteq B_{k_n}$. Thus, by $X$‘s normality and Urysohn’s lemma we may find some $\varphi_n:X\mapsto [0,1]$ such that $\varphi\left(\overline{B_{j_n}}\right)=0,\text{ }\varphi\left(X-B_{k_n}\right)=1$. We are thus provided with a countable collection of mappings $\left\{\varphi_n\right\}_{n\in\mathbb{N}}$.

So, define $F:X\mapsto\mathbb{R}^{\infty}$ by $\displaystyle x\mapsto \prod_{n=1}^{\infty}\left\{\frac{\varphi_n(x)}{n}\right\}$. There does need to be some justification for why $\displaystyle \prod_{n=1}^{\infty}\left\{\frac{\varphi_n(x)}{n}\right\}$ is square summable. But, this follows directly from the fact that

$\displaystyle 0\leqslant\sum_{n=1}^{\infty}\left|\frac{\varphi_n(x)}{n}\right|^2\leqslant\sum_{n=1}^{\infty}\frac{1}{n^2}<\infty$

Also, it’s injective. This is a little less obvious though. To see it, let $x\ne y$ then by $X$‘s normality there exists some basic open $B$ such that $x\in B,\text{ }y\notin B$. But, by normality there exists some basic open $B'$ such that $x\in\overline{B'}\subseteq B$. It follows that there exists some $\varphi_N\in\left\{\varphi_n\right\}_{n\in\mathbb{N}}$ such that $\varphi_N(x)=0,\varphi_N(y)=1$. Thus,

$\pi_N F(x)\ne \pi_N F(y)\implies F(x)\ne F(y)$ .

It is clear then that $F:X\mapsto F(X)\subseteq \mathbb{R}^{\infty}$ is a  map. It remains to show that it’s bicontinuous.

To prove that $F$ is continuous it suffices to show that given any $x\in X$ and $\varepsilon>0$ there exists some neighborhood $H$ of $x$ such that $y\in H\implies \left\|F(x),F(y)\right\|<\varepsilon$ . Since an infinite series of functions converges uniformly if it’s terms are bounded by the terms of an infinite series of constants (this is just the Weirstrass M-test), it is easy to see that there exists some $N\in\mathbb{N}$ such that given any $y\in X$ we have

$\displaystyle \left\|F(y)-F(x)\right\|^2=\sum_{n=1}^{\infty}\left|\frac{\varphi_n(y)-\varphi_n(x)}{n}\right|^2<\sum_{n=1}^{N}\left|\frac{\varphi_n(y)-\varphi_n(x)}{n}\right|^2+\frac{\varepsilon^2}{2}$

But, by the continuity of each $\varphi_n$ there exists some $H_n$ of $x$ such that

$\displaystyle y\in H_n\implies \left|\frac{\varphi_n(y)-\varphi_n(x)}{n}\right|^2<\frac{\varepsilon^2}{2N}$

Thus, if we let $H=H_1\cap\cdots\cap H_N$, it is clear that $y\in H\implies \left\|F(x)-F(y)\right\|<\varepsilon$ by previous comment.

It remains to show that $F^{-1}:F(X)\mapsto X$ is continuous. Now, it suffices to show that given any $x\in X$ and some basic open neighborhood $B$ of $x$, that there exists some $\varepsilon>0$ such that $\left\|F(y)-F(x)\right\|<\varepsilon\implies y\in B$. We know that $B$ is the second coordinate of some pair $U_N=\left(B',B\right)$ such that $x\in B'\subseteq \overline{B'}\subseteq B$. So, if we choose $\varepsilon<\frac{1}{2N}$, then we see that

$\displaystyle \left\|F(y)-F(x)\right\|<\varepsilon\implies\sum_{n=1}^{\infty}\left|\frac{\varphi_n(y)-\varphi_n(x)}{n}\right|^2$

$<\frac{1}{4N^2}\implies \left|f_N(y)-f_N(x)\right|<\frac{1}{2}$

. Since $x\in B'$, $f_N(x)=0$, and so $\left|f_N(y)\right|M\frac{1}{2}$. But, since $f_N(X-B')=1$ we see that $y\in B$. The conclusion follows. $\blacksquare$

This is an amazing theorem but note that we said earlier that it is sufficient but not necessary for a second countable normal space to be metrizable. We end this by proving a quick theorem and then giving an example of a normal space which is not metrizable.

Remark: It is clear that there are metric spaces which aren’t second countable. Take any uncountable discrete space for example.

Theorem: Let $X$ be a compact Hausdorff space, then $X$ is metrizable if and only if it’s second countable.

Proof:

$\implies$: If it is metrizable, then it is homeomorphic to a compact metric space. But, as we discussed earlier every compact metric space is separable and since separability and second countability imply each other in metric spaces it follows that the homeomorphic image of $X$ is second countable. But, since second countability is clearly invariant under homeomorphisms it follows that $X$ is second countable.

$\Longleftarrow$: Conversely, if it’s second countable and compact Hausdorff it is second countable and normal, and thus by the above theorem it is metrizable. $\blacksquare$

We now give an example of a space which is normal but not metrizable.

Example: Let $X_j=[0,1]$ under the usual topology for each $j\in[0,1]$ and define $\displaystyle X=\prod_{j\in[0,1]}X_j$ under the product topology. Since the product of compact Hausdorff spaces is compact Hausdorff it follows that $X$ is compact Hausdorff, and thus normal. But, consider $\left\{O_j\right\}_{j\in\mathcal{J}}$ where $\pi_j(O_j)=\left(\frac{1}{4},\frac{1}{2}\right)$ and $\pi_k(O_j)=X_k,\text{ }k\ne j$. This is clearly a class of open sets in $X$ and so $\displaystyle O=\bigcup_{j\in[0,1]}O_j$ is an open set which clearly has $\left\{O_j\right\}_{j\in\mathcal{J}}$ as an open cover. But, since every second countable space is Lindelof there must be a countable subocover of this cover. But, upon close inspection it has  no proper subcover, and consequently no countable subcover. It follows that $X$ is not Lindelof and thus not second countable. And so appealing to the previous theorem we see that it cannot be metrizable.