Reading a Proof About Showing A Set is Closed

Question

I am reading this, specifically Proposition 0.2, and there are a few things that I am not understanding. Here is the problem statement, and you can click the link for more details:

Let $Y$ be an ordered set in the order topology. Let $f,g : X \rightarrow Y$ be continuous. Show that the set $\{x ~|~ f(x) \le g(x) \}$. is closed in $X$.

First, the author that claims because $f^{-1}(U) \cap g^{-1}(V)$ is open in $X$, there exists a "smaller" open neighbhorhood $W$ of $x$ such that $x \in W \subseteq f^{-1}(U) \cap g^{-1}(V)$. What guarantees that $W$ exits? If $X$ had a basis, then I believe this would be the case; but it isn't immediately obvious.

Next, why does $f(W) \cap g(W) = \emptyset$ imply that $g(w) < f(w)$ holds for all $w \in W$? Here is my attempt at showing this.

Because the two sets are disjoint, given an arbitrary $w \in W$, $f(w) \notin g(W)$ which implies $f(w) \neq g(z)$ for all $z \in W$. In particular, when $z = w$, this gives us $f(w) \neq g(w)$ which implies either $f(w) < g(w)$ or $f(w) > g(w)$.

I have tried, but I cannot see anyway of ruling out the possibility of $f(w) < g(w)$. Would someone mind help me make sense of this proof?

Answer 1

First, the author that claims because $f^{-1}(U) \cap g^{-1}(V)$ is open in $X$, there exists a "smaller" open neighbhorhood $W$ of $x$ such that $x \in W \subseteq f^{-1}(U) \cap g^{-1}(V)$. What guarantees that $W$ exits? If $X$ had a basis, then I believe this would be the case; but it isn't immediately obvious.

This is always true. For any $x$ in any open set $U$, there always exists an open set $W$ such that $x \in W \subseteq U$. In particular, you can just take $W=U$. So, in your example, you can just take $W = f^{-1}(U) \cap g^{-1}(V)$.

Next, why does $f(W) \cap g(W) = \emptyset$ imply that $g(w) < f(w)$ holds for all $w \in W$? Here is my attempt at showing this.

Because the two sets are disjoint, given an arbitrary $w \in W$, $f(w) \notin g(W)$ which implies $f(w) \neq g(z)$ for all $z \in W$. In particular, when $z = w$, this gives us $f(w) \neq g(w)$ which implies either $f(w) < g(w)$ or $f(w) > g(w)$.

I have tried, but I cannot see anyway of ruling out the possibility of $f(w) < g(w)$. Would someone mind help me make sense of this proof?

They failed to mention that $U$ and $V$ are constructed as in the proof of Lemma 0.1. So in particular, each element of $f(U)$ is larger than each element of $g(V)$.

Answer 2

You can simply prove it as follow. Since $f,g$ are continuous, $h=f-g$ is continuous. So $\{x ~|~ f(x) \leqslant g(x) \}=\{x ~|~ f(x)-g(x) \leqslant 0\}=h^{-1}(-\infty,0]$. $h^{-1}(-\infty,0]$ is closed because $(-\infty,0]$ is closed.

Answer 3

So we have $x_0 \in A $ where $A$ is the complement of the set. So $f(x_0) > g(x_0)$, and $U$ and $V$ are dijsoint neighbourhoods of $f(x_0)$ and $g(x_0)$ in $Y$, by the lemma shown before it (Hausdorffness). So we have that $x_0 \in f^{-1}[U] \cap g^{-1}[V]$ by definition, and this intersection is itself open, as $f$ and $g$ are continuous. We can just take $W$ to be this intersection. If $w \in W$, then $f(w) \in U$ and $g(w) \in V$. These sets are disjoint so either $f(w) < g(w)$, or $g(w) < f(w)$. Because $U$ and $V$ are constructed to be open intervals (or half-open ones) we know that $U > V$ (every element of $U$ lies above all elements of $V$, see the proof of 0.1), as $f(w_0) > g(w_0)$. This an be easily checked from the proof of 0.1.

Alternatively, the proof of 0.1 before it shows that $L = \{(x,y) \in Y \times Y\}$ is closed in $Y \times Y$. And your set is just $(f \times g )^{-1}[L]$, so also closed, where $(f \times g)(x) = (f(x), g(x))\in Y \times Y$ is continuous as $f$ and $g$ are.