$X$ open $\implies A = \{x\in X; f(x)\neq g(x)\}$ is open, and $X$ closed $\implies F = \{x\in X; f(x)= g(x)\}$ is open

Question

I need to prove:

I've found this question that basically proves it for the open case, but there's no assumption on $X$ being open... Also, I do know the property that $f$ continuous then $f^{-1}$ takes open sets to open sets.

I need, somehow, to use the definition of continuity that says $\lim_{x\to a} f(x) = f(a)$. For an open set, I think I need to use the definition of open set that talks about open balls. Is there an elementar way to do it?

What is X here? What are f and g and what are their domains? — Prithviraj Chowdhury, Jul 02 '16 at 19:47
Then use the fact that $h=f-g$ is continuous if $f$ and $g$ are continuous. Set $A$ is preimage under $h$ of an open set, hence is open. — drhab, Jul 02 '16 at 19:51
The second is false. Consider $X={x}$ a singleton. The set you talk about is closed. — , Jul 02 '16 at 20:01
@G.Sassatelli Seems to me that OP should quantify; I'm reading $F$ as open in $X$, not necessarily in $\mathbb{R}$. — Neal, Jul 02 '16 at 20:12
@neal which is still false. It's closed nevertheless and almost never open in $ X $. — , Jul 02 '16 at 20:57

drhab · Accepted Answer · 2016-07-02T20:35:01.917

1

If functions $f,g:X\to\mathbb R$ are continuous then function $k:X\to\mathbb R^2$ prescribed by $x\mapsto\langle f(x),g(x)\rangle$ is continuous.

Also function $s:\mathbb R^2\to\mathbb R$ prescribed by $\langle u,v\rangle\mapsto u-v$ is continuous.

Then composition $h:s\circ k:X\to\mathbb R$ prescribed by $x\mapsto f(x)-g(x)$ is continuous.

$A=h^{-1}(\mathbb R-\{0\})$ and $\mathbb R-\{0\}$ is an open set.

The continuity of $h$ tells us now that $A$ is open in $X$.

This means that $A=X\cap U$ where $U$ is an open set in $\mathbb R$.

If moreover $X$ is open in $\mathbb R$ then we can conclude that $A$ (as intersection of open sets) is also open in $\mathbb R$.

If $X=\mathbb R$ then $X$ is closed in $\mathbb R$. However it cannot be shown that $F=\{x\in\mathbb R\mid f(x)=g(x)\}$ is an open set. E.g. take $f$ prescribed by $x\mapsto x$ and $g$ prescribed by $x\mapsto-x$. Then $F=\{0\}$ wich is not an open set.

edited Jul 02 '16 at 20:35

answered Jul 02 '16 at 20:01

drhab

151,093

$A = {x\in\mathbb{R}; f(x)-g(x)\neq 0}$, right? So $A = h^{-1}(\mathbb{R}-{0})$ because it's the set of $x$ that maps to $f(x)-g(x)$ when this difference is not $0$? Since $\mathbb{R}-{0}$ is open, it's inverse by $h$ is open in $X$, where $X$ is the set of such $x$ that maps to $f(x)-g(x)$ when this differente is not $0$. But we want to prove that it's open in $\mathbb{R}$, not just $X$. So since $A$ is open in $X$, we can write it as $A = X\cap U$ where $U$ is open in $\mathbb{R}$. If $X$ is open in $\mathbb{R}$ then we can finally conclude that $A$ is open in $\mathbb{R}$. Am I right? – Guerlando OCs Jul 02 '16 at 23:16
Yes. That is exactly what I am saying. – drhab Jul 03 '16 at 06:34

$X$ open $\implies A = \{x\in X; f(x)\neq g(x)\}$ is open, and $X$ closed $\implies F = \{x\in X; f(x)= g(x)\}$ is open

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