If $f,g$ are continuous from $X \rightarrow \Bbb R$ where $(X,J)$ is a topology, the set of $x$ where $f(x) \geq g(x)$ is closed.

Question

If $f,g$ are continuous maps from $X \rightarrow \Bbb R$ where $(X,J)$ is a topology, then I wish to show that the set $A$ where $f(x) \geq g(x)$ is closed.

My first thought is to show that $A$ contains its limit points. So if $x$ is a limit point such that every punctured neighborhood of $x$ hits a point in $A$, I don't see how that helps (there would be some $y$ nearby where $f(y) \geq g(y)$ holds).

I need a hint to understand this problem.

Answer 1

Suppose that $p$ is not in $A = \{x \in X: f(x) \ge g(x)\}$. We will show that $p$ is not a limit point of $A$:

We know that $f(p) < g(p)$ and as these are real numbers, we can find some $r \in \mathbb{R}$ such that $f(p) < r < g(p)$. But then note that $U=g^{-1}[(r, +\infty) \cap f^{-1}[(-\infty, r)]$ is an open set of $X$ (here we use continuity of $f$ and $g$ and the fact that open segments are open in the reals) that contains $p$ and such that any $x \in U$ obeys $f(x) < r$ and $g(x) > r$, which ensures that $x \notin A$.

So $U$ is an open neighbourhood of $p$ that does not intersect $A$. So any $p \notin A$ cannot be a limit point of $A$ (or logically equivalent: $A$ contains all its limit points). $A$ is thus closed.

Answer 2

Note $f−g$ is continuous, and the inverse image of a $[0, \infty]$ (a closed set) is closed, which is equivalent to the set of $x$ where $f(x)≥g(x)$, therefore show that the set $A=\{ x| f(x)≥g(x) \}$ is closed.