Examples and counter-examples in Real analysis - check my answers please I've been given some practice examples, without solutions in preparation for an upcoming exam, and was hoping I could get them double checked here.
For each of the following, either give an example or say why such an example
is impossible.
(a) A set A with exactly one limit point.
Ans: Take the set $A = \{\frac{1}{n} : n \in \mathbb{N}\}$ (the limit point is $0$)
(b) A compact set $K$ and a sequence $(a_n)$ in $K$ such that $(a_n)$ does not converge.
Ans: Take the sequence $(1,0,1,0,1,0,1,0,1,...)$
(c) A monotone sequence that does not converge.
Ans: $(1,2,3,4,5,...)$
(d) Two open sets $U, V$ contained in $[0, 1]$ such that $U ∩ V$ is not open.
Ans: Let $U = (0,0.5)$ and $V = (0.5,0.6)$
^question here, is a set which is not open, imply a closed set? I'm 99.5% sure the answer is no, and that we can also have a set neither open nor closed such as $(a,b]$ - can anyone confirm?
(e) A continuous function $f : [0, 2) → \mathbb{R}$ that has no maximum value.
Ans: $f(x) = \cfrac{1}{x-2}$
(f) A continuous function $g : [0, 1] → \mathbb{R}$ and a Cauchy sequence $(x_n)$ in $[0, 1]$  such that $(g(x_n))$ does not converge.
Ans: impossible, cauchy sequence is convergent by definition, and since $g$ is continuous on its domain, we cannot have a divergent $g(x_n)$.
(g) A function $u : \mathbb{R} → \mathbb{R}$ whose points of discontinuity form an infinite uncountable set.
Not sure.
(h) A function $v : \mathbb{R} → \mathbb{R}$ which is nowhere continuous but differentiable at one point.
Not sure
(i) A series $\sum a_n$ an that converges such that $\sum a_n^2$ diverges.
Will attempt again later, but if you could leave an answer so I could check, I would be grateful!
(j) A sequence of discontinuous functions on $[0, 1]$ converging uniformly to a continuous function.
Same as above.
 A: For $(d)$, that's not right - the empty set is open! (And no, there are sets which are neither open nor closed; it's not the best terminology. :P)
For $(e)$, you've designed a function with no minimum. But that's easily fixed.
For $(f)$, that's not quite correct: for example, the function $f(x)={1\over x}$ is a continuous function from $(0, 1)$ to $\mathbb{R}$ but the Cauchy sequence $\{{1\over n}: n\in\mathbb{N}\}$ doesn't get mapped to a Cauchy sequence by $f$. We need more: something special about $[0, 1]$ versus $(0, 1)$ . . .
For $(g)$, it might be easier to find an everywhere discontinuous function. (Hint: can you make it $0$ "some of the time," and $1$ "the rest of the time," in such a way that the $0$-valued points and the $1$-valued points "interlace" everywhere?)
For $(h)$, check the definition of differentiability . . .
I'll leave $(i)$ and $(j)$ off until you've had time to take a crack at them.
A: (i) $\displaystyle a_n = \frac{(-1)^n}{\sqrt{n}}$.
$\sum_n a_n$ converges but $\displaystyle \sum_n a_n^2 = \sum_n\frac{1}{n}$ does not.
