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NOTE: I have no notion of product spaces, so, a proof using the more basic principals would be fantastic. If it helps, I have that a space is compact iff it is sequentially compact iff it is complete and totally bounded?

We let $X = [0,1]^\mathbb{N}$. We wish to show that the metric space $(X,d)$ is compact but the space $(X,d_1)$ is not compact, given the following distance functions $d$ and $d_1$:

$$d(a,b) = \sum_{n=1}^{\infty} \frac{\lvert a(n) - b(n) \rvert}{2^n}$$

$$d_{1}(a,b) = \sup{\{\lvert a(n) - b(n) \rvert | n \in \mathbb{N} \}}$$

I am struggling to see why the first is compact but the second is not, however, I think it may be possible to show that the second isn't either complete or totally bounded which would imply that it is not compact. I'm not sure about showing that the first is compact.

Thanks for your help!

Lauren

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  • $\begingroup$ To clarify: does $[0, 1]^{\mathbb{N}}$ refer to the decimals between 0 and 1, inclusive, written out as strings of digits? $\endgroup$ – Sam Birns Nov 18 '15 at 15:20
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    $\begingroup$ Sequences of numbers in the closed interval $[0,1]$ $\endgroup$ – Lauren Hayes Nov 18 '15 at 15:21
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$d$ induces the product topology (so Tychonoff applies), while $d_1$ sees $X = [0,1]^\mathbb{N}$ as a large subset of the unit ball of $\ell_\infty$. The sequence of unit vectors (one coordinate 1, rest 0) in $X$ cannot have a convergent subsequence, e.g.

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  • $\begingroup$ Hi, I haven't heard of Tychonoff or the concept of product topology, do you have a more basic proof by any chance? Thanks a lot, Lauren! $\endgroup$ – Lauren Hayes Nov 18 '15 at 18:41
  • $\begingroup$ Do you mean $\ell_\infty$ instead of $\ell_1$? $\endgroup$ – Jose27 Nov 18 '15 at 18:46
  • $\begingroup$ Indeed, I will edit it. $\endgroup$ – Henno Brandsma Nov 18 '15 at 19:02
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$(X, d_1)$ is not compact because \begin{align*} 100000 &\dots \\ 010000 &\dots \\ 001000 & \dots \\ \end{align*} is a sequence with no convergent subsequence.

Let $a_n$ be a sequence in $(X, d)$ where each $a_i = (x_{1i}, x_{2i}, \dots, x_{ni}, \dots)$ for the numbers $x_{ij}$ in the sequence. Since $([0, 1],d)$ is sequentially compact, we can pick a subsequence of $a_i$ such that the sequence $(x_{11}, x_{12}, \dots)$ converges. Similarly, we can pick a subsequence of this subsequence such that the sequence $(x_{2_{n_1}}, x_{2_{n_2}}, \dots)$ converges. Visually, we have the array \begin{align*} x_{11} x_{21} x_{31} &\dots \\ x_{12} x_{22} x_{32} &\dots \\ x_{13} x_{23} x_{33} &\dots \end{align*} where each column converges. The only concern with this method would be accidentally throwing out too many terms (because we're throwing infinitely many terms away with each iteration) so take the subsequence of the diagonal sequence of this array. This converges by construction and it's still a sequence, so we're done.

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  • $\begingroup$ The sequence you used for $d_1$ doesn't converge to zero (or to anything else). But since it has no convergent subsequence, it still serves your purpose of showing that the space isn't compact. (If you had a convergent sequence whose terms are all at distance 1 from each other, then $d_1$ wouldn't even be a metric.) $\endgroup$ – Andreas Blass Nov 18 '15 at 15:51
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    $\begingroup$ Thanks for catching that. Also I realized my proof that $(X,d)$ is compact is flagrantly wrong. Can I salvage it, or should I just delete my answer? $\endgroup$ – Sam Birns Nov 18 '15 at 15:54
  • $\begingroup$ What do you mean by "take the subsequence of the diagonal sequence of this array"? Thanks for your help! Lauren $\endgroup$ – Lauren Hayes Nov 18 '15 at 22:19

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