Compact and Open subsets of $\ell^p$ Let $(a_n)_1^{\infty}$ be a sequence of positive real numbers. Consider $$ A = \{x \in \ell^p : |x_n| < a_n \ \ \forall n\}$$ $$ B = \{x \in \ell^p : |x_n| \leq a_n \ \ \forall n\} $$
I'm interested to know under what conditions imposed on $a_n$ would $A$ be an Open Set and likewise what conditions would make $B$ Compact
I was able to show the special case of $\ell^{\infty}$ where $B$ is Compact $\iff a_n \in c_0$
Similiarly I think $B$ is Compact in $\ell^{p}$ if  $a_n \  \in \ell^{p}$ for $p < \infty$ 
However I'm not sure if the converse is true. Thoughts on this and on $A$ would be much appreciated
Some info is available here at page 4 and 5, Proposition #2 and Proposition #3 http://www.math.mcgill.ca/jakobson/courses/ma354/snarski-analysis3.pdf

The notes you provided shows that $A$ is open if and only if $1\le p<\infty$ and $\inf a_n>0$. You are correct that if $p=\infty$, $B$ is compact if and only if $a_n\to0$ (this is also shown in the notes). You are also correct that if $1\le p<\infty$, $B$ is compact if and only if $(a_n)\in\ell^p$. I'll sketch a proof.
Assume $(a_n)\in\ell^p$ and suppose $(x^k)_k$ is a sequence in $B$, so for each $k$, $(x^k_n)_n\in B$. For each $n$, the sequence $(x^k_n)_k$ is bounded in $\mathbb C$ (or $\mathbb R$ if we are only dealing with real sequences, but it really doesn't matter). By using a diagonal method, we can find a single subsequence $(k_j)_j$ such that $x_n^{k_j}\to x_n$ as $j\to\infty$ for all $n$. This implies $|x_n|\le a_n$ and so $x\in\ell^p$, hence $x\in B$. It suffices to show that $x^{k_j}\to x$ in $\ell^p$, that is, to show
$$\lim_{j\to\infty}\sum_{n=1}^\infty|x_n^{k_j}-x_n|^p=0.$$
Since $(a_n)\in\ell^p$, this follows from the dominated convergence theorem, proving $B$ is compact.
Now suppose $(a_n)\notin\ell^p$. Define $x_n=(a_1,a_2,\ldots,a_n,0,0,\ldots)\in\ell^p$. Then clearly $x_n\in B$ for all $n$ but $\|x_n\|_p\to\infty$, so $B$ is not bounded and hence not compact.
