Sublevel sets of continuous functions are closed

Question

Consider a continuous function $f:\mathbb R^n\to\mathbb R$. The set $$M=\{x\in\mathbb R^n ; f(x)\leq c\}$$ where $c$ is a real number. It follows from continuity that $M$ is closed.

I have seen a proof of the above statement using the fact that we can define the set of $M$ as the preimage $f^{-1}((-\infty,c])$ and since $(-\infty,c]$ is closed we use continuity to conclude that $M$ is closed.

I am not sure why this argument works. Which definition of continuity do we use here?

Also, is there a different way to prove this statement? Maybe by using sequences?

I'd appreciate any help/hint. Thank you.

Answer 1

Suppose $x_k$ is a sequence in $M$ and $x_k\to x_0.$ By continuity, $f(x_k) \to f(x_0).$ The limit of a sequence whose terms are all $\le c$ must itself be $\le c.$ Thus $f(x_0)\le c,$ which implies $x_0\in M.$ It follows tha $M$ is closed.

Answer 2

Are you aware of the typical topological characterization of a continuous function $f$? In the topological definition of continuity, the function $f : \mathbb R^n \to \mathbb R$ is defined to be continuous if for every open set $U \subset \mathbb R,$ the preimage $f^{-1}(U)$ is open in $\mathbb R^n$.

Suppose $f:X \to Y$ is continuous, and let $N \subseteq Y$ be a closed set. So $Y\setminus N$ is an open set. Hence $f^{-1}(Y \setminus N)=X\setminus f^{-1}(N)$ is open. Therefore $f^{-1}(N) \subseteq X$ is a closed set.

In particular we know that $f: \mathbb R^n \to \mathbb R$ is continuous, $N=(-\infty, c]$ is closed, and so $M=f^{-1}(N)$ is closed.

There is also the sequential argument which zhw. has correctly provided.