Compactness of subset of $c_{0}$ Let $x\in c_{0}$ and $X=\{ y\in c_{0} :|y_{n}|\leq |x_{n}|\}$.
I am ask to prove that $X$ is compact in $c_{0}$ but I don't know how I should do that.
 A: Since $c_0$ is a metric space, the notions of compactness and sequential compactness are the same. Hence, we use the latter. Take a sequence of $y$'s in $X$. Since $y_1\leq x_1$ for all $y\in X$, we can use the the Bolzano-Weierstrass theorem to get a subsequence of $y$'s such that their first element converges. Using this subsequence, we can further extract a subsequence such that the 2nd element also converges and so on. Finally, we can take the diagonal sequence for which all elements converge. We define the limit sequence to be the pointwise limit of the diagonal sequence. To show that our original sequence converges to this in the $\ell^\infty$ norm, use the fact that $x_n\rightarrow 0$ as $n\rightarrow \infty$. Since, $y_n\leq x_n$, the limit sequence also belongs to $c_0$. Truncate the limit sequence at point $N$ such that it is smaller in magnitude than $\epsilon$ after that. Now, go as far into the diagonal sequence as needed to ensure that the first $N$ terms of the diagonal sequence do not differ much from the first $N$ terms of the limit sequence.
A: It may be helpful to see $c_0$ as the subspace of $C(\mathbb N_\infty)$ (where $\mathbb N_\infty$ is the one-point compactification of $\mathbb N$) with $x(\infty)=0$.  Then apply the Arzela-Ascoli theorem.
Clearly $X$ is uniformly bounded (by $\|x\|$).  Now you just need to show that $X$ is equicontinuous and closed (both of these are fairly easy to show).
