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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 $B(F(x), \epsilon)$.

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1 Answer 1

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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.

Essentially, in an infinite sum (like the Hilbert metric, or the standard metric on the product of countably many metric spaces) we can often concentrate on the first part, as the final part is insignificant (can be made as small as we like), when the sums converge, of course, as they do here.

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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.. –  omar Feb 6 '13 at 23:36
1  
@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

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