Can I apply the sequential criterion of continuity here? Suppose that $f:D\subseteq\mathbb{R}\to\mathbb{R}$ is continuous, and $A\subseteq\mathbb{R}$ is a closed set in the reals. I need to prove or disprove: $f^{-1}(A)$ is closed.
I think since $f$ is continuous, we have $f^{-1}(A)$ is closed in $D\,$?
$$f^{-1}(A)=\{x\in D: f(x)\in A\}.$$
I was thinking how can I use the sequential criterion here. Should I take a sequence $f(x_n)$ in $A$ such that $f(x_n)\to y\in A$, and hence $x_n\to f^{-1}(y)\in f^{-1}(A)\in D\,$?
Thanks for helping.
 A: We have $f:D\xrightarrow{\text{cts}}\mathbb{R}$, and $A$ closed with respect to $\mathbb{R}$. We show that
$$f^{-1}(A)=\{x\in D: f(x)\in A\}$$ is closed with respect to $D\subset\mathbb{R}$.
The sequential criterion (or definition) says $f$ is continuous at $a\in D$ iff:

$\lim_{n\to\infty} x_n=a\quad$ implies $\quad \lim_{n\to\infty} f(x_n)=f(a),\;\;$ for each sequence $x_1,x_2,\dots$ in $D$.

We need to show that every closure point (with respect to $D$) for $f^{-1}(A)$ belongs to $f^{-1}(A)$. By definition, if $t$ is such a $D$-closure point, then $t\in D$ and there is $t_n\to t$ (a convergent sequence) with each $t_n\in f^{-1}(A)$, so that each $f(t_n)\in A$.
Let $t_n$ be any such sequence. Since $f$ is continuous at $t$, $\;\; \lim_{n\to\infty} f(t_n)=f(t).$
But this shows that $f(t)$ is a closure point (with respect to $\mathbb{R}$) of $A$, which we can denote $f(t)\in \overline A$, where $\overline A$ is the set of all closure points, called the closure of $A$.
But $A$ is closed, which means $A=\overline A$. So finally $f(t)\in A$, and $t \in f^{-1}(A)$.
A: If you want to show a set is closed using sequences you need to start out with an (arbritrary) convergent sequence in the set in question, i.e. you have to look at $b_n\in f^{-1}(A)=:B$ such that $b_n\rightarrow b$ You then have to conclude that $b\in B$ in order to show $B$ is closed.
(Note: for relative closure (i.e. for the case where $D$ is open) you may assume $b\in D$)
