# Proving that a metric space is compact

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.
• 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. – thobanster Aug 1 '12 at 23:36
• @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 $H^\infty$) – Jose27 Aug 1 '12 at 23:39