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.

Solution 1:

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 $t \in f^{-1}(A)$.