Proof Verification: Continuity and Open sets

Question

Show that if $f : \mathbb{R} \to \mathbb{R}$ is continuous, then the set $A =\{x \in \mathbb{R}:f(x)<\alpha\}$ is open in $\mathbb{R}$ for each $\alpha \in \mathbb{R}.$

My attempt:

$f(x)< \alpha$ implies that if we have the open set $(-\infty, \alpha)$ in $\mathbb{R}$, then by the global continuity theorem, we have that there exists an open set $H$ in $\mathbb{R}$ such that $H \cap \mathbb{R}= f^{-1}(G).$

This implies that $H= f^{-1}(G)$ which by definition is $\{x \in \mathbb{R}:f(x)<\alpha\}$ and is open.

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

Thank you!!

Answer 1

It is correct, but:

1. Since $H\subset\mathbb R$, why do you write aboute $H\cap\mathbb R$? Why not just $H$?
2. Are you sure that you can use the “global continuity theorem”? It seems more likely that you are suppposed to prove it in this particular case.