How to prove a set of sequences is compact?

Question

Can anyone give me some clues on how to solve this problem? I think I need to construct a sequence of sequences and show the convergent.

Answer 1

For (a): You can look at the following sequence: \begin{equation} (1,0,0,\ldots)\\ (0,1,0,\ldots)\\ (0,0,1,\ldots)\\ \vdots \end{equation}

For (b): First note that for every $\epsilon>0$ and any two elements $y=(y_1,y_2,\ldots)$ and $z=(z_1,z_2,\ldots)$ we have $\sum_{n=N+1}^{+\infty}|y_n-z_n|<\epsilon$ for sufficiently large $N$.

Now if you have sequence $x^1,x^2,\ldots$ in $C$ you can observe that \begin{equation} x^n_j\in \left[-\frac{1}{2^j},\frac{1}{2^j}\right]\ n\in\mathbb N,j\in\mathbb N. \end{equation} Thus every sequence $\{x^n_j\}_{n=1}^{+\infty}$ has convergent subsequence.

You can couple this with what we noted at the begining.

For (c): Look at the following sequence: \begin{equation} (1,0,0,0\ldots)\\ (1,1/2,0,0,\ldots)\\ (1,1/2,1/3,0\ldots)\\ \vdots \end{equation}

Answer 2

For part a I want to point out an important result from Functional Analysis.

The closed unit ball in a normed linear space is compact iff it is finite dimensional vector space. (This is a standard result and I will allow you to find the solution yourself).

Now in a normed linear space (which is a metric space) sequential compactness and compactness are the same. Can you see why this solves part a?