Are $l^p$ spaces compact? Given the sequence space $l^p$ where it represents the set of sequences $d=(d_n)_{n\in\mathbb{N}}$ s.t 
$$\sum_{n=1}^{\infty} |d_n|^p<\infty$$
Is the metric space $(l^p,q)$ compact? Where $q$ is a metric I apologise if this a a dumb question but because the terms a bounded I assume intuitivly it's true but I'm not sure how one would prove it. this is the first I'm taking a math course in years.
 A: No.  The only compact normed linear space is $\{0\}$.  Moreover, $l^p$ is infinite-dimensional so it is not even locally compact.
A: Firstly, normed vector spaces are metric spaces, and compactness in metric spaces is equivalent to sequential compactness. So you only need to look at converging subsequences to check compactness.
Note that if $e_{1}\in \ell^{p}$ is the vector $e_{1}=(1,0,0,...)$, then defining $x_{n}=n e_{1}$ for all $n\in\mathbb{N}$ we get a sequence $(x_{n})_{n=1}^{\infty}\subseteq \ell^{p}$ with $\|x_{n}-x_{m}\|=|n-m|\geq 1$ for all $n\neq m$, so it can not have a convergent subsequence. Hence $\ell^{p}$ is not compact. This same argument can be used for any norm space with a non zero vector.
Regarding the follow up question: $\ell^{p}$ is separable. The subset of $\ell^{p}$ consisting of all sequences with finitely many non-zero rational elements is countable and dense. Note however that $\ell^{\infty}$ is not separable. You can for example consider the set of all the characteristic functions on $2^{\mathbb{N}}$. On each element of this subset you can draw an open ball so that all these balls are disjoint, and note that $2^{\mathbb{N}}$ is uncountable.
