Proofs about continuity and convergence in topological spaces

Question

I'm working on the following exercise:

Let $f:(X,T)\to(Y,S)$ and $x\in X$. Prove that if $f$ is continuous at $x$ then if a sequence $\{x_n\}$ converge to $x$ we have $f(\{x_n\})\to f(x)$, show that the reciprocal is true if $X$ is a first-countable space. Give an example to show that the former propositions are not always equivalent.

I will divide my work in different parts:

$(a)$ Prove that if $f$ is continuous at $x$ and $\{x_n\}\to x$ then $\{f(x_n)\}\to f(x)$

We have the following definitions: $f$ continous at $x$ if for every $V$ such that $f(x)\in V$ is $f^{-1}(V)$ open; and $\{x_n\}\to x$ if given a set $W$ such that $x\in W$ then there is a $n_0$ such that $x_n\in W\;\; \forall n\geq n_0$.

So it looks that this is about to prove that if an open set $U$ contains $f(x)$, then exists $n_0$ such that $f(x_n)\in U$ for every $n\geq n_0$. For what I know because of the continuity of $f$, for the open set $U$ it must be true that $f^{-1}(U)$ is open; also $x\in f^{-1}(U)$, due to the convergence of $\{x_n\}$ if we take $V\subset f^{-1}(U)$ then exists $n_0 $ such that $x_n\in V\subset f^{-1}(U)$ for every $n\geq n_0$, then $f(x_n)\in f(V)\subset U$ for every $n\geq n_0$, which menas that $f(x_n)\to f(x)$.

(b) Prove that the reciprocal is true if $X$ is a first countable space.

Here I suppose that if $\{x_n\}\to x$ then $\{f(x_n)\}\to f(x)$. Let $V$ be a closed set in $Y$ and take preimage of it. Now for $f^{-1}(V)\subset X$ consider any sequence $\{{x_n}\}$ that converges to some point $x$, by the hypothesis looks like $f(x_n)\to f(x)\in V$, this means that $x\in f^{-1}(V)$ and therefore, $f^{-1}(V)$ is closed (here I'm using the fact that if every convergent sequence in a set converge to a point inside such set then the set is closed). Follows that $f$ is continuous by definition.

(c) Finding a counterexample for the reciprocal.

I couldn't find an example to do this part, my textbook suggest that I should take $f: (X,T)\to (X,D)$ being $X$ an uncountable set, $T$ the cocountable topology and $D$ the discrete topology, but I couldn't define anything that works properly (for some reason I'm always struggling to construct counterexamples)

Also, please verify if what I did in the former proofs was ok, because in particular I wonder if taking $V\subset f^{-1}(U)$ was ok since looking back to it looks like an unnecessary step.

Answer 1

Have you looked at steen and Seebach's Counterexamples in Topology? It has lots of great examples like this.

I can't think of any good counterexamples this late at night so can't give you a specific hint, but here's a general one: if you know that first countable is necessary, go back through your proof and see which step used it. Did you count something somewhere? Did you need something to have a countable neighborhood basis? Once you've answered that, you'll have a pretty good idea how to break the proof.

You have good intuition that V was unnecessary. (Always double check if you find yourself taking an open set inside an open set - what is it going to gain you? We take open sets inside closed sets, or intersections of open sets, etc, but, in the absence of further information, V is no more specific than U.)

"By the hypothesis looks like..." The rest of that paragraph should be rephrased. You got a little vague.

Note: for continuity, if U lives in Y and V in X in part (a), then keep it that way in parts (b) and (c), too. Similarly, if V is the name of an open set, pick a new letter for a closed one.