Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given that $X$ is an arbitrary topological space, $Y$ is a totally ordered set in the order topology, and $f$, $g \colon X \to Y$ are continuous, how to show that the subset $A$ of $X$ given by $$ A := \{ x \in X \colon f(x) \leq g(x) \} $$ is closed in $X$?

share|improve this question

2 Answers 2

You can prove it directly from the definitions. Let $U=X\setminus A=\{x\in X:f(x)>g(x)\}$; we need only prove that $U$ is open. Let $x\in U$ be arbitrary; we’ll find an open nbhd of $x$ contained in $U$. Since $x\in U$, we know that $g(x)<f(x)$. There are now two cases.

  1. There is some $a\in Y$ such that $g(x)<a<f(x)$. Let $$V=g^{-1}\big[(\leftarrow,a)\big]\quad\text{and}\quad W=f^{-1}\big[(a,\to)\big]\;.$$ Then $V\cap W$ is an open nbhd of $x$ (why?), and for each $y\in V\cap W$ we have $g(y)<a<f(y)$, so $V\cap W\subseteq U$.

  2. The interval $\big(g(x),f(x)\big)$ in $Y$ is empty. Then let $$V=g^{-1}\big[(\leftarrow,f(x))\big]\quad\text{and}\quad W=f^{-1}\big[(g(x),\to)\big]$$ and argue almost exactly as in the first case.

share|improve this answer

The set $\{(x,y)\in Y^2\mid x\le y\}$ is closed in $Y^2$ and $X\to Y^2, x\mapsto (f(x),g(x))$ is continuous.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.