Suppose $(M, \rho)$ is separable, and that $\rho(x,y)\leq 1$ for all x and y in M. Let $x_n$ be a countable dense set of M. Define the Hilbert cube $H^{\infty}$ as the collection of all real sequences $y_n$ were $\lvert y_n \rvert \leq 1$ for all n. Define a metric $d$ on this space by $d(x, y) = \sum_{n=1}^{\infty} 2^{-n}\lvert x_n – y_n\rvert$. Then define $f : M \rightarrow H^{\infty}$ by $f(x) = \rho(x, x_n)_{n=1}^{\infty}$. The goal is to show $f$ is a homeomorphism onto its image. I have shown it's 1-1 and continuous, but the book I'm going through then says that I need to show that $\forall \epsilon > 0$ $\exists \delta > 0$ such that $\rho(x,y) < \epsilon$ whenever $d(f(x), f(y)) < \delta$, and I'm stuck here. Any help would be appreciated.
Prove that a mapping from a separable space to the Hilbert cube is a homeomorphism
real-analysis
Related Solutions
Let's work on completeness.
Let $(x_j)$ be a Cauchy sequence with respect to the given metric in the Hilbert cube.
Let's think of $x_j$ as the $j^{\text{th}}$ row of a matrix, and $x^k$ as the $k^{\text{th}}$ column, consisting of all the $k^{\text{th}}$ entries of the elements of the sequence (which are themselves sequences). The $k^{\text{th}}$ entry of the $j^{\text{th}}$ element is $x_{jk}$.
A sketch of the first part of the proof is:
- assume the rows form a Cauchy sequence w.r.t. your $\sup$ metric
- see the individually each column is Cauchy w.r.t. the Euclidean metric on $\mathbb{R}$.
- use the fact that $\mathbb{R}$ is complete to establish a limit for each column
- establish a putative limit which has its elements as the limit of each column. $$\begin{bmatrix} x_{00} & x_{01} & x_{02} & \cdots \\ x_{10} & x_{11} & x_{12} & \cdots \\ \vdots & \vdots & \vdots & \vdots \\ x^*_0 & x^*_1 & x^*_2 & \cdots \\ \end{bmatrix} $$ I'll call our putative limit sequence $x^* = (x_0^*, x_1^*, \ldots)$. You should be able to fill out these details.
Now we just have to show that this limit $x^*$ is legitimate. This means verifying two things:
- $x^*$ is actually an element of the Hilbert cube. This is straightforward, but do verify it.
- $(x_j)$ actually converges to $x^*$ with respect to the $\sup$ metric given. Right now we only know that each column converges w.r.t. to the Euclidean metric in $\mathbb{R}$.
Let's attack (2). Fix $\epsilon > 0$. We can take, for each $i$, $N_i$ such that $n > N_i$ implies $|x^*_i - x_{ni}| < \epsilon$. This is the definition of convergence in $\mathbb{R}$ applied to each column. Now the important thing is to finally use the definition of the Hilbert cube, which implies that there exists $N^*$ such that $i > N^*$ implies $|x^*_i - x_{ni}| < \epsilon$, independent of $n$. This is because the entries of a point in the Hilbert cube get really small, indeed by definition we have $|x_{ni} \leq \frac{1}{2^i}|$ independent of $n$, therefore $|x^*_i - x_{ni}| \leq \frac{1}{2^i}$, independent of $n$. Having established this, set $N = \max_{i \leq N^*}N_i$. You can confirm that $n > N$ implies $|x^*_i - x_{ni}| < \epsilon$ for all $i$. Hence we have $d(x^*, x_n) = \sup_i|x_i^* - x_{ni}| < \epsilon$ for $n > N$. That's convergence!
The important thing is that we can construct an infinite class of something (here the $N_i$) but only worry about finitely many of them due to the Hilbert cube's property that entries of its elements get really small.
See if you can get total boundedness using this as inspiration, it's a nice problem and the flavor is similar!
Hint (if you want):
try to explicitly cover the Hilbert cube with $\epsilon$ balls of size $\epsilon = 2^{-k}$ for $k= 1,2,3$. Notice for each $k$ how you can ignore all of the indices $i>k$. What's the pattern for how many balls it requires? (you can count exactly). Generalize this construction for all $k$. Now given any $\epsilon > 2^{-k}$, your construction with balls of size $2^{-k}$ suffices, because if a finite number of balls covers a space and all you do is inflate them (keeping them centered as before), they'll always still cover the space.
Just a few hints: Let $A:=\{x\in H^\infty:\ |x_k|<1, k=1,\ldots, 2000\}$. Now fix $x\in A$. You need to find an $r>0$, such that the Ball $B_r(x):=\{y\in H^\infty:\ d(x,y)<r\}\subset A$. Let us pick an arbitrary $r>0$ and a $y\in B_r(x)$. Then $$d(x,y)=\sum_n 2^{-n}|x_n-y_n|<r.$$ Hence $|x_n-y_n| < C\cdot r$ for some constant $C>0$ and $n=1,\ldots,2000$. Now try to use $$|y_n|\leq |x_n-y_n|+|x_n|$$ to find an $r>0$ (you will have to choose it small), such that $|y_n| <1$ for $n=1,\ldots,2000$.
Best Answer
If $f(x^{j}) \to f(x)$ in the range of $f$ then $\rho (x^{j},x_n) \to \rho (x,x_n)$ for each $n$ because convergence in $H^{\infty}$ implies convergence of each coordinate. Now $\rho (x^{j},x) \leq \rho (x^{j},x_n)+\rho (x_n,x)$. Given $\epsilon >0$ choose $n$ such that the second term is less than $\epsilon$ and the let $j \to \infty$. Thus $f(x^{j}) \to f(x)$ implies $x^{j} \to x$ which means $f^{-1}$ is continuous.