Prove that $f^{-1} (F)$ is closed

Question

A set $F \subset \mathbb R$ is closed if for any convergent sequence $\{x_n\}$ in F converges, we have $\lim_{n \to \infty} x_n=x \in F $.
How to Prove that if $f :\mathbb R \to \mathbb R$ is continuous and $F$ is closed then

$f^{-1} (F)$ := {$x \in \mathbb R : f(x) \in F $}

is closed.

Answer 1

Hint. This works very directly. Suppose you have a sequence $(x_n)$ in $f^{-1}[F]$ converging to $x \in \mathbf R$. As $f$ is continuous (on the whole of $\mathbf R$!) we know that $f(x_n)$ converges to? As $F$ is closed and $f(x_n) \in F$, this implies $\ldots$.

Answer 2

Suppose $(x_n)$ be a sequence in $f^-$$^1$$[F]$ converging to x∈$R$

since $f$ is continuous on $R$ that means $f$ is continuous on each point of $R$

By $Sequential-Criterian$ of continuity $f(x_n)$ converges to $f(x)$.

$F$ is closed and $f(x_n)$∈$F$ implies that $f(x)$∈$F$.

Finally we have $x$∈$f^-$$^1$$[F]$, Hence $f^-$$^1$$[F]$ is closed.