In general, we don't have $a^n + b^n \neq (a+b)^n $.

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.

One need to show that $x_n$ is bounded from below; $x_n$ is monotone decreasing. This two show $x_n$ converges to some $x$. Then one need to prove that $x=\sqrt{a}$.

One should start with a Cauchy sequence $\{f_n(x)\}$ in $C(K)$, and construct a function $f(x)$ by taking pointwise limit, define for $x \in [0,1]$, $f(x) = \lim_n f_n(x)$. Thus defined, $f(x)$ is just a function, and may not be continuous, and we don't know yet $f_n \to f$ uniformly or not. Once we show that the convergence is uniform (see solution), then we can use the result that uniform convergence preserve continuity, to conclude that $f$ is a continuous function on $[0,1]$, hence $f \in C(K)$.

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. If one don't specify the extension, I can throw in an open set that equals to $X$, and pick the finite subcover to consist of a single open set, just the added $X$.
• 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 $(0,1) \subset [0,1]$.