Is the space of all sequences on $[0,1]$ sequentially compact? We are in the space of sequences on [0,1] under the metric $d(x,y) = \Sigma^{\infty}_{i=1} 2^{-k} \mid x_k - y_k \mid $
The hint I've been given is to use a diagonal argument.
I'm thinking to take a sequence of sequences in the space, and by Bolzano-Weierstrass each of these sequences must have a convergent subsequence. If I take the first term of the first sequence, the second of the second etc. then that seems to be a 'diagonal' argument as told to use in the hint - but then I only have one sequence. Am I not supposed to find a sequence of sequences which converges?
 A: Hint : Consider a sequence $(x^k)_{k \in \mathbb{N}}$ of elements of $[0,1]^{\mathbb{N}}$ (that is, each $x^k$ is a sequence $(x^k_n)_{n \in \mathbb{N}}$ of real numbers between $0$ and $1$).
Then $(x^0_n)_{n \in \mathbb{N}}$ is a bounded sequence of real numbers, so there exists $\varphi_0 : \mathbb{N} \rightarrow \mathbb{N}$ such that $(x^0_{\varphi_0(n)})_{n \in \mathbb{N}}$ converges.
Now $(x^1_{\varphi_0(n)})_{n \in \mathbb{N}}$ is a bounded sequence of real numbers, so there exists $\varphi_1 : \mathbb{N} \rightarrow \mathbb{N}$ such that $(x^0_{\varphi_0 \circ \varphi_1(n)})_{n \in \mathbb{N}}$ converges.
Now $(x^2_{\varphi_0 \circ \varphi_1(n)})_{n \in \mathbb{N}}$ is a bounded sequence of real numbers, so there exists $\varphi_2 : \mathbb{N} \rightarrow \mathbb{N}$ such that $(x^2_{\varphi_0 \circ \varphi_1 \circ \varphi_2(n)})_{n \in \mathbb{N}}$ converges.
And so on, you can construct $\varphi_0, \varphi_1, ..., \varphi_n, ...$ such that for all $k \in \mathbb{N}$, the sequence $(x^k_{\varphi_0 \circ ... \circ \varphi_k(n)})_{n \in \mathbb{N}}$ converges.
Finally, let $\varphi : \mathbb{N}  \rightarrow \mathbb{N}$ defined for all $n \in \mathbb{N}$ by $\varphi(n) = \varphi_0 \circ ... \circ \varphi_n(n)$ (here is the diagonal argument !). You can show that $\varphi$ is strictly increasing, and that the sequence $(x^{\varphi(n)})_{n \in \mathbb{N}}$ converges.
A: Let $(x_i^{n})$ be a sequence in this space. The diagonal argument gives you integers $n_1,n_2,..$ such that $\lim_{k \to \infty} x_i^{n_k}$ exists for each $i$. Call this limit $x_i$. We would like to show that $(x_i^{k})$ tends to $(x_i)$. Given $\epsilon >0$ choose $N$ such that $ \sum\limits_{k=N+1}^{\infty} \frac 1 {2^{k}} <\epsilon$. Split the sum in $d((x_i^{k}), (x_i))$ into the sum from $1$ to $N$ and the sum from $N+1$ to $\infty$. Can you finish?
