real analysis - Show the inverse of a closed and continuous function is also closed - please help

I want to show that if I have some closed set $F \subset \mathbb{R}$, such that, for any sequence $(x_n) \in F$ that converges, we have $\lim_{n \to \infty} x_n \in F$

then if $f: \mathbb{R} \to \mathbb{R}$ is continuous and $F$ is closed, means $f^{-1}(F):= \{x \in \mathbb{R} : f(x) \in F \}$ is also closed.

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How can I approach this problem? My understanding is that I need to prove that

$\lim_{n \to \infty} f(x_n) \in F$ (or is it perhaps $\lim_{n \to \infty}(x_n) \in F$ ?)

Starting ideas so far: -use sequential characterization of continuity.

Assume $F$ is closed and $f: \mathbb{R} \to \mathbb{R}$ is continuous.

Then, $x \in f^{-1}(F) \implies f(x) \in F$

Since $f$ is continuous, by a sequential characterisation of continuity, for any $(x_n) \in \mathbb{R}$ which converges to a point $c \in \mathbb{R}$, we have, by definition $f(x_n) \to f(c)$.

I feel like I am on the right track, but I am unsure what needs polishing and how to proceed. Any help would be appreciated.

Answer

Overall, if you want to show that $f^{-1}(F) = \{x \in \Bbb R : f(x) \in F\}$ is closed for closed $F \subset \Bbb R$, you can pick any convergent sequence $(x_n)$ of points in $f^{-1}(F)$, and show that it converges to something in $f^{-1}(F)$.

So you have all the ideas in place, but you're right, it's just a bit rough. I would say a good "flow" would be to:

• Let $(x_n)$ be a sequence of points in $f^{-1}(F),$ with $x_n \to x^*$. Your goal is to show that $x^* \in f^{-1}(F)$, that is, $f(x^*) \in F$.




• Let $y_n = f(x_n)$, so that $(y_n)$ is a sequence of points in $F$.




• Now I would use the sequential characterization of continuity, and the fact that $F$ is closed.

  
  
  
  
  
  • Briefly, $x_n \to x^*$ means $y_n \to f(x^*) \overset{\text{def}}{=}y^*$ (why?). What can we say about where $y^*$ lives, and why can we say that? What does that imply about where $x^*$ lives?