Proof Verification: Continuity and Closed sets

Question

Show that if $f:R→R$ is continuous, then the set $A=\{x∈R:f(x)\le α\}$ is closed in $\mathbb{R}$ for each $α∈\mathbb{R}$ My attempt:

$f(x) \le α$ $\implies$ we have $(-\infty, \alpha] \subset \mathbb{R}$ which is closed as its complement $(\alpha, \infty)$ is open, then by the global continuity theorem, we have that there exists an open set $H \subset \mathbb{R}$ such that $H=f^{-1}((\alpha, \infty))$ This implies that $H = \{x∈\mathbb{R}:f(x)>α\}$ is open which implies that $H^c = \{x∈R:f(x)\ge α\}$ is closed.

Can anyone please verify this and check if my reasoning is ok?

Thank you.

Answer 1

The idea is perfect, though your answer seems to have some confusion between $f(x)\leq\alpha$s and $f(x)\geq\alpha$s.

Answer 2

You are being inconsistent: areo intereisted in $f(x)\geqslant\alpha$ or in $f(x)\leqslant\alpha$?

Although correct from the logica point of view, the expression “there exists an open set $H \subset \mathbb{R}$ such that $H=f^{-1}((\alpha, \infty))$” is strange. It would be more natural to write that the set $H=f^{-1}((\alpha,\infty))$ is open.

Otherwise, it is correct.