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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 functions, how to show that the subset $A$ of $X$ given by $$ A := \left\{ \ x \in X \ \colon \ f(x) \leq g(x) \ \right\} $$ is closed in $X$?

Edit based on the answer by Heigen von Eitzen:

Let $U \colon= \{ u \times v \in Y \times Y \ \colon \ u < v \ \}$. We show that $U$ is open in $Y \times Y$.

Let $u \times v \in U$. Then $u < v$.

Case I. If there is some $y \in Y$ such that $u < y < v$, then $u \times v \in (-\infty, y) \times (y, +\infty) \subset U$.

Case 2. If $(u,v)$ is empty, then $u \in (-\infty, v)$ and $v \in (u, +\infty)$, and so $u \times v \in (-\infty, v) \times (u, +\infty)$.

Moreover, if $a \times b \in (-\infty, v) \times (u, +\infty)$, then we must have $a < v$ and $b> u$. So $a \leq u < v \leq b$, which implies that $a < b$ and so $a \times b \in U$.

Thus, $u \times v \in (-\infty, v) \times (u, +\infty) \subset U$.

So $U$ is open in $Y \times Y$.

Similarly, we can show that the set $$V \colon= \left\{ \ u \times v \in Y \times Y \ \colon \ u > v \ \right\}$$ is open in $Y \times Y$.

Thus, it follows that the set $A \colon= \{ \ u \times v \in Y \times Y \ \colon \ u \leq v \ \}$ is closed in $Y \times Y$.

Now since the maps $f \colon X \to Y$ and $g \colon X \to Y$ are continuous, so is the map $f \times g \colon X \to Y \times Y$ defined as $$(f \times g)(x) \colon= f(x) \times g(x) \ \mbox{ for all } \ x \in X.$$

Thus, the inverse image under $f \times g$ of the set $A$ is closed in $X$.

But $$(f \times g)^{-1} (A) = \left\{ \ x \in X \ \colon \ (f \times g)(x) \in A \ \right\} = \left\{ \ x \in X \ \colon \ f(x) \leq g(x) \ \right\}.$$

share|cite|improve this question
up vote 5 down vote accepted

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|cite|improve this answer
    
your answers are so instructive and illuminating!! How much I've benefitted from your presence on this forum!! God bless you!! – Saaqib Mahmuud Mar 24 '15 at 6:36
    
@Saaqib: Thank you! – Brian M. Scott Mar 24 '15 at 6:38

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|cite|improve this answer
    
how is the set $\{ (x,y) \in Y^2 \ \vert \ x \leq y \ \}$ is closed in $Y^2$? – Saaqib Mahmuud Mar 24 '15 at 6:41
    
how is the set $$\left\{ \ u \times v \in Y \times Y \ \colon \ u \leq v \ \right\}$$ closed in $Y \times Y$? Can you please elaborate? – Saaqib Mahmuud Jan 27 at 14:25

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.