# Urysohn Metrization Theorem: show that $F$ is continuous.

This is quite some text for an actaully small question...

Let $X$ be a second countable regular space with countable basis $\mathcal{B}=\{B_n\}_{n=1}^{\infty}$. We can easily prove $X$ is normal. Consider the collection of all ordered pairs $(i,j)$ of integers for which $\bar{B_i}\subset B_j$. By Urysohn's Lemma, there is for each such pair $(i,j)$ a Urysohn function $f: X \rightarrow [0,1]$ such that $f(\bar{B_i})=0$, $f(X\setminus B_j)=1$.

Let $\mathcal{F}$ denote such a collection of Urysohn functions having one member for each ordered pair $(i,j)$ for which $\bar{B_i}\subset B_j$. Since $\mathcal{F}$ is countable, then it can be indexed by the set of positive integers.

Define a function $F: X \rightarrow H$ from $X$ into Hilbert space $H$ by $$F(x) = \left( f_1(x), \frac{f_2(x)}{2}, \frac{f_3(x)}{3}, \dots\right), x \in X$$

Thus the coordinates of $F(x)$ are determined by the values of the members of $\mathcal{F}$ at x; each value $f_n(x)$ is divided by $n$ to insure that $F(x)$ is a member of $H$:

$$\Sigma_{n=1}^{\infty}\left( \frac{f_n(x)}{n} \right)^2 \leq \Sigma_{n=1}^{\infty} 1/n^2$$

so the sum of the squares of the coordinates of $F(x)$ is a convergent series of real numbers.

Show that $F$ is continuous.

The reason I ask this is because I have read several proofs and saw different approaches. How would I complete the proof if I'd begin like:

Let $x \in X$ and let $B(F(x),\epsilon)$ be an open ball in $H$ with positive $\epsilon$. We want to show that there exists $V\subset X$ such that $x\in V$, $F(V)\subset B(F(x),\epsilon)$.

Choose $N > 0$ s.t. $\Sigma_{n=N+1}^{\infty} \frac{1}{n^2} < \frac{\epsilon^2}{2}$ . We have that every $f_ n$, $1 \leq n \leq N$, is continuous. So there exists an open $V_n$ such that for every $y \in V_n$ $$|f_n(x) - f_n(y)| < \frac{\epsilon}{\sqrt{2N}}$$

$V = \cap_{n=1}^{N} V_n$ is an open set which contains x.

What would be the calculation to show that $F(V)$ is is a subset of $B(F(x), \epsilon)$.

Let $y$ be in $V$. Then $F(y) = \left(\frac{f_n(y)}{n}\right)_{n\ge 1}$, and one needs to show that it lies in $B(F(x),\epsilon)$, so consider $d(F(y), F(x))$, where $d$ is the sum of squares metric on Hilbert space. Then
\begin{align*} d\big(F(y), F(x)\big)^2 &= \sum_{n=1}^{\infty} \left(\frac{f_n(y)}{n} - \frac{f_n(x)}{n}\right)^2\\ &=\sum_{n=1}^N \left(\frac{|f_n(y) - f_n(x)|}{n}\right)^2 + \sum_{n=N+1}^{\infty} \left(\frac{|f_n(y) - f_n(x)|}{n}\right)^2\\ &<N\left(\frac{\epsilon}{\sqrt{2N}}\right)^2 + \sum_{n=N+1}^{\infty} \frac{1}{n^2}\\ &=\frac{\epsilon^2}{2} + \frac{\epsilon^2}{2}\\ &=\epsilon^2 \end{align*}
as $y \in V$ implies $y \in V_n$ for all $n \le N$, and this means we can bound every term in the first sum by $\frac{\epsilon}{\sqrt{2N}}$ (and forget the fractional terms that are $< 1$ anyway, the extra $N$ comes from the number of terms), while in the second sum we estimate the differences of the function values $|f_n(x) - f_n(y)|$ by 1, as all these values lie in $[0,1]$, and then use the fact that $N$ was chosen to have a small tail of the series. Taking square roots shows that indeed $F(y)$ is the ball around $F(x)$ and we are done.
• With $f(y) = \sum_{n=1}^{\infty} \frac{f_n(y)}{n}$, you're referring to some Urysohn function $f:X\rightarrow [0,1]$? Or the embedding $F:X\rightarrow H$? I am guessing the latter, but then what do you mean with the sum of all Urysohn functions $\frac{f_n}{n}$? Because, $F(y) = \left( f_1(y), \frac{f_2(y)}{2}, \frac{f_3(y)}{3}, \dots\right)$? Or am I seeing thing the other way around.. – onimoni Feb 6 '13 at 23:36
• @omar: I’m sure enough that Henno meant $F(y)=\left\langle\frac{f_n(h)}n:n\in\Bbb Z^+\right\rangle$ that I’ve made the change. – Brian M. Scott Feb 7 '13 at 0:00