Let $H^\infty$ be the set of real sequences such that each element in each sequence has $|a_n|\leq 1$. The metric is defined as $$d(\{a_n\}, \{b_n\}) = \sum_{n=1}^\infty \frac{|a_n - b_n|}{2^n}.$$ Prove that $H^\infty$ is a compact metric space.

To prove this, I want to show that every sequence in $H^\infty$ has a convergent subsequence. I know that if we have a sequence $\{\{a_n\}^{(k)}\}$ in $H^\infty$, then for all $i$, the real sequence $\{a_i^{(k)}\}$ has a convergent subsequence, since it is bounded by 1. So we can get a convergent subsequence $\{a_1^{(k_j)}\}$, and then a convergent subsequence of $\{a_2^{(k_j)}\}$, and continue taking subsequences of subsequences until we have a convergent subsequence of $(a_1, a_2, a_3, ..., a_n)^{(k)}$ with $n$ some positive integer if we stop taking subsequences at the $nth$ subsequence; this gives a sequence $\{x_n\}$ where $x_n$ is the limit of the $n$th convergent subsequence of $\{a_n^{(k)}\}$. Ideally, we could show that the sequence in $H^\infty$ converges to $\{x_n\}$.

I know that if we have the $nth$ subsequence of $\{\{a_i\}^{(k)}\}$ defined in the way described above, then for any $\epsilon > 0$ there exist $N_1, ..., N_n$ such that if $k\geq \max_{i\leq n}\{N_i\}$, then for $1\leq i\leq n$, $|a_i^{(k)} - x_i| < \epsilon/2n$. By choosing $n$ sufficiently large that $\sum_{i=n+1}^\infty |a_i^{(k)} - x_i|/2^i\leq \sum_{i=n+1}^\infty 1/2^{i-1} < \epsilon/2$, we can ensure that $d(\{a_n\}^{(k)}, \{x_n\}) < \epsilon$. But the problem here is that for each $\epsilon$, we end up choosing a different convergent subsequence of the first $n$ terms (since we need to choose $n$, which determines how many subsequences of subsequences we take). Any idea on how to proceed?


One way is to take first a subsequence $(a_1^i)_i$ such that the first coordinate converges. Now take a subsequence of this subsequence $(a_2^i)_i$ such that the second coordinate converges. Proceeding this way you obtain a sequence of nested subsequences $(a_{k+1}^i)_i \subset (a_k^i)_i$. Now take the sequence $(a_k^k)_k$. By the last property every coordinate of this sequence converges, and this is equivalent to convergence in your metric. (Note that for each $k,i$ that $a_k^i$ is a sequence of real numbers so we have three layers of sequences...). By the way this is called a diagonal argument, and it's a really useful trick to have available.

Another way to do this would be to note that your space is basically $\prod_{n=1}^{\infty} [0,1]$ with the product topology, so an appeal to Tychonoff's theorem gives the conclusion.

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    $\begingroup$ +1. However, the point of this exercise is to prove (a special case of) Tychonoff's theorem, so appealing to it would seem circular. (The proof of the general version of the theorem is also based on this idea, though the machinery of nets / ultrafilters / Zorn's lemma / transfinite induction tends to obscure it.) $\endgroup$ – Nate Eldredge Aug 1 '12 at 23:05
  • $\begingroup$ I don't fully understand your notation. Is $(a^k_k)_k$ the $k$th element of the $k$th subsequence of the sequence in $H^\infty$? I'm not sure what the three layers of sequences are. I know that we need to find a convergent subsequence of the sequence in $H^\infty$ (that is, a sequence of real number sequences that converges to some element of $H^\infty$), but I don't understand how to find such a subsequence. $\endgroup$ – thobanster Aug 1 '12 at 23:36
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    $\begingroup$ @thobanster: For fixed $k,i$ $a_k^i$ is an element in $H^\infty$. (The layers I'm talking about is that these elements are sequences and we take sequences of these elements and then a sequence of subsequences of the sequence of elements in $H^\infty$) $\endgroup$ – Jose27 Aug 1 '12 at 23:39
  • $\begingroup$ I get it now, thanks :). $\endgroup$ – thobanster Aug 1 '12 at 23:59

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