Given $\sum_n a_n$ converge, and $a_n > 0$

• It is wrong to conclude that $\limsup (a_{n+1}/a_n)) < 1$.
• It is wrong to conclude $a_n$ is decreasing.

It is tempting to consider $f(x) = \int_0^x f'(t) dt$, however, we don't know if $f'(t)$ is integrable or not.

Given $A, B$ compact subset of $X$, one need to show that $A \cap B$ is compact.

• If one wants to prove using definition, then one needs to start with an arbitrary open cover of $A \cap B$. Note that this may not be an open cover of either $A$ or $B$.
• It is a good idea to extend the above open cover to an open cover of $A$. For example, by add in the open set $B^c$. Some answer are vague about how to do this extension.
• Some answer also write, compact set is equivalent to being closed and bounded. That's false for general metric space $X$.
• Some answer write, “any subset of a compact set is compact”. False, for example $(0,1) \subset [0,1]$.