Unspecified topology in an exercise from Munkres

Question

Below is an exercise from Munkres. He never mentions what topology $X$ is endowed with. How can I solve the exercise if I don't know the topology of $X$? What is it supposed to be?

Answer 1

First show that $h:X \to Y \times Y, x \to (f(x),g(x))$ is continuous.

This only uses the universal mapping property for product spaces.

Then that $A = \{ (y_1,y_2) : y_1 \le y_2 \}$ is a closed subset of $Y \times Y$.
This is a nice exercise in the order topology.

Since

$$\{x: f(x) \le g(x)\} = h^{-1}[A]$$ it is closed as the inverse image of a closed set under a continuous map.
Details are left for the enjoyment of the reader.

Answer 2

$X$ can be any topological space. All you need is that $f$ and $g$ are continuous as maps from $X$ to $Y$, which has the order topology.

Hint elaboration for (b): $X_1 = \{x: f(x) \le g(x) \}$ is closed by a). Ditto for $X_2 = \{x: g(x) \le f(x) \}$.

Clearly $X = X_1 \cup X_2$. And $h|X_1 = f$ and $h|X_2 = g$.

Answer 3

For part (a). For $y\in Y$ let $(\leftarrow,y)=\{y'\in Y: y'<y\}$ and $(y,\rightarrow)=\{y'\in Y:y'>y\}$.

If $g(x)<f(x)$ then there exist disjoint open $G,F \subset Y$ with $g(x)\in G$ and $f(x)\in F$ such that $\forall y\in G \;\forall y'\in F\;(y<y').$

Because if the interval $(g(x),f(x))$ is empty we can let $G =(\leftarrow,f(x))$ and $F=(g(x),\rightarrow),$ but if $g(x)<z<f(x)$ for some (any) $z\in Y ,$ we can let $G=(\leftarrow,z)$ and $F=(z,\rightarrow).$

So if $g(x)<f(x)$ then $U=(g^{-1}G)\cap (f^{-1}F)$ is an open subset of $X,$ containing $x,$ and such that $\forall x'\in U\;(\;g(x')<f(x')\;).$ Therefore $\{x: g(x)<f(x)\}$ is open in $X.$ So its complement $\{x:f(x)\leq g(x)\}$ is closed in $X.$