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. These two conditions together show $x_n$ converges to some $x$. Then one need to prove that $x=\sqrt{a}$. Missing any of the three steps would make one lose points.

There are other methods to prove this problem, such as creating an auxillary sequence $y_1 = x_1, y_{n+1} = (y_n + \sqrt{a})/2$, and show that $x_n \leq y_n$ and $y_n \to \sqrt{a}$.

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 $f \in C(K)$.

If a continuous function $f: (0,1) \to \R$ is unbounded, then it means either $\lim_{x \to 0} |f(x)| = \infty$ or $\lim_{x \to 1} |f(x)| = \infty$, since the value of $f(p)$ is finite for any $p \in (0,1)$. For example, consider the function $f(x) = 1/x + 1/(1-x)$, it is an example of unbounded function (and it is not uniformly continuous).

A common mistake is to say, assume $f$ is unbounded, then there exists a $p \in (0,1)$, such that $\lim_{x \to p} f(x)=\infty$. That is not what 'unbounded' mean.

Some approach is like this,

• Consider the interval $[0,1]$, take a global max of $f(x)$ on $[0,1]$. assume it is at $x=b$.
• For any $u \in (f(0), f(b))$, which is also $(f(1), f(b))$, there exists a $b_1(u) \in (0,b)$ and $b_2(u) \in (b,1)$, such that $f(b_1(u)) = u$, $f(b_2(u))=u$.
• So far the two sentences are correct. But the problem is that, as $u \to f(b)$, it is not true that $b_1(u) \to b$ and $b_2(u) \to b$. (imagine $f(x)$ has a 'plateau' instead of a 'peak' near $x=b$). And it is not true that, as one move $u$, the function $b_1(u)$ and $b_2(u)$ varies continuously.

This is an interesting direction, but needs more careful argument to make it work.

• It is tempting to consider $f(x) = f(0)+\int_0^x f'(t) dt$, however, we don't know if $f'(t)$ is integrable or not. (-1 point)
• Even if we assume $f'(x)$ is integrable, it is wrong to say $\int f'(x) dx \leq \int 2 dx$. One need to use definite integral.

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 $(0,1) \subset [0,1]$.

• Write $\int_1^2 \frac{A(x)}{B(x)} dx = \frac{\int_1^2 A(x) dx}{\int_1^2 B(x) dx}$.