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I am looking for a way to prove that the Hilbert Cube is compact as a subset of $\ell^2$ using Tychonoff's Theorem. As I've found, this can be done using the fact that C is homeomorphic to $[-1,1]^{\mathbb{N}}$ which is, by Tychonoff, compact when endowed with the metric $d(x,y)=\displaystyle\sum_{j\in\mathbb{N}}\dfrac{1}{2^j}|x_j-y_j|$.

It seems to me that one possible homeomorphism is $h: [-1,1]^{\mathbb{N}} \rightarrow C$ defined by $h(x)=(x_1,\dfrac{x_2}{2},...,\dfrac{x_i}{i},...)$, but I'm not being able to prove its continuity, how can I do it? I checked Here but I don't get the proof of the continuity of $F$ as it doesn't use the metric. I'm sure someone who knows general topology can easily translate it to metric-terms, I'd appreciate it very much.

Thanks in advance!

PS: With "traditional" Hilbert Cube I mean $C=\{ \{x_n\}_{n\in\mathbb{N}}: 0\leq x_n \leq \dfrac{1}{n}\}$.

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V. Jiménez
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  • Can you change the first metric to $d(x,y)=\sum_{j=1}^\infty\frac1{j^2}|y_j-x_j|$? – Lutz Lehmann Nov 03 '16 at 13:00
  • @LutzL You mean the metric on $[0,1]^{\mathbb{N}}$? I don't think so, cause that seems to me the core of the problem. But I guess that if I can justify that change, surely I could. – V. Jiménez Nov 03 '16 at 13:48

2 Answers2

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Hint: If $x_n \to x$ in $[-1,1]^{\mathbb N},$ then $x_n(k) \to x(k)$ for each $k.$

zhw.
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  • Thanks! Is it an equivalence in this particular case, isn't it? If so, I think I got it, but I'm not very sure how to prove it is indeed an equivalence, as I think that's not true in general. – V. Jiménez Nov 03 '16 at 22:17
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One way to homeomorpically embed $C$ into Hilbert space is to let $F((x_i)_i)=((2^{-i}x_i)_i).$ Then $\|F(x)-F(y)\|\leq d(x,y)$ so $F$ is Lipschitz-continuous.

DanielWainfleet
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  • That's pretty concise and cristal clear, thanks! I'm not sure, though, how to show that $F$ thus defined is surjective, it seems intuitively so but I can't work out the details. – V. Jiménez Nov 03 '16 at 22:26
  • Surjective to what? If $D={f(x): x\in C}$ then $f:C\to D$ then $f$ is a continuous bijection from one compact Hausdorff space to another so $ f:C\to D$ is a homeomorphism. ($D$ is compact because it's the continuous image of $C.$) – DanielWainfleet Nov 03 '16 at 22:44
  • Yeah! I mean the fact that $F([-1,1]^{\mathbb{N}})=C$. – V. Jiménez Nov 03 '16 at 22:46
  • I was using $C$ to mean $ [-1,1]^{\mathbb N} ,$ not the "traditional"$C,4 and mapping it to a subset of Hilbert space. – DanielWainfleet Nov 03 '16 at 22:55
  • Oh, got it. But then I honestly don't see how that solves my problem, does it? – V. Jiménez Nov 03 '16 at 23:02