1. ## functions of topologies

Let X be a topological space and let Y be a simply ordered set in the order topology. Let $f,g:X\to Y$ be two continuous functions.

Show that the set $\{x\in X:f(x)\leq g(x)\}$ is closed in X.

Since f is continuous, for every open set $V\subset Y$, $f^{-1}(V)=\{x\in X: f(x)\in V\}$ is an open subset of X. Similar, for every open set $U\subset Y$, $g^{-1}(U)=\{x\in X: g(x)\in U\}$.

How do I go from this to showing what needs to be shown?

2. ## Re: functions of topologies

Originally Posted by dwsmith
Let X be a topological space and let Y be a simply ordered set in the order topology. Let $f,g:X\to Y$ be two continuous functions.

Show that the set $\{x\in X:f(x)\leq g(x)\}$ is closed in X.

Since f is continuous, for every open set $V\subset Y$, $f^{-1}(V)=\{x\in X: f(x)\in V\}$ is an open subset of X. Similar, for every open set $U\subset Y$, $g^{-1}(U)=\{x\in X: g(x)\in U\}$.

How do I go from this to showing what needs to be shown?
Let $\Sigma=\left\{(x,y)\in Y^2:x\leqslant y\right\}$ then our set is the preimage of $\Sigma$ under the map the continuous map $X\to Y^2$ given by $x\to (f(x),g(x))$ and so it suffices to prove that $\Sigma$ is closed in $Y^2$. To see this we merely note that $Y^2-\Sigma=\left\{(x,y)\in Y^2:x>y\right\}$ and let $(x,y)\in Y^2-\Sigma$. We have two choices, if there exists $x then set $U=(z,\infty)\times (-\infty,z)$. If there is no point $z$ such that $x then set $U=(y,\infty)\times(-\infty,x)$. Either way this is an open set in $Y^2$ with the order topology which contains $(x,y)$ and is disjoint from $\Sigma$. It thus follows that $Y^2-\Sigma$ is open and so $\Sigma$ is closed, so that the conclusion follows from previous comment.