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.

I'd like to know if the following proof is valid. The only thing I'm not sure about (though I can't see why it's invalid if it is) is if we can always use the Hausdorfness of $Y$ to separate an open set from $f(C)=g(C)$.

"Let $X$ be a space, $Y$ a $T_2$-space, and $f,g:X \to Y$ continuous functions. Prove that $C:=\{x \in \,|\,f(x)=g(x)\}$ is a closed subset of $X$."

Let $x \in X$ be such that $f(x) \ne g(x)$. Since $Y$ is $T_2$, there are open sets $U_\alpha \ni f(x)$ and $V_\alpha \ni g(x)$ such that $U_\alpha \cap O_\alpha=V_\alpha \cap O_\alpha=\varnothing$ for all $\alpha \in A$ where $A$ is the set indexing the points of $f(C)=g(C)$, and $O_\alpha$ is an open set containing $x_\alpha \in f(C)=g(C)$. Let $U:=\displaystyle\bigcap_{\alpha \in A} U_\alpha$ and $V:=\displaystyle\bigcap_{\alpha \in A} V_\alpha$. Then $U \cap f(C)=V \cap f(C)=\varnothing$. Therefore, since $f,g$ are continuous, $f^{-1}(U)$ and $g^{-1}(V)$ are open sets in $X$ disjoint from $C$. Since $X\setminus A$ is the union of such open sets, it is open itself and therefore $C$ is closed.

Thanks.

share|improve this question
1  
See this page for help. –  Clayton Jan 23 '13 at 23:28
    
It doesn't seem correct. In particular, how do you know that $U$ is open? –  David Mitra Jan 23 '13 at 23:29
3  
This seems in the spirit of what you're trying to do: Assume $f(x)\ne g(x)$. Choose disjoint open sets $U $ and $V $ in $Y$ with $f(x)\in U $ and $g(x)\in V $. By continuity of the functions choose open sets $N_1$ and $N_2$ in $X$ both containing $x$ such that $f(N_1)\subset U $ and $g(N_2)\subset V$. Take $O=N_1\cap N_2$. Then $O$ is open, contains $x$, and is disjoint from $C$ (since $f(N_1)\subset U $, $g(N_2)\subset V$ and $U\cap V=\emptyset)$. So $C^c$ is open. –  David Mitra Jan 23 '13 at 23:36
1  
One possible answer –  leo Jan 24 '13 at 0:21
    
@David: True, I should have realized that. That's the basic idea I was going for, yes. Thanks. –  Alex Petzke Jan 24 '13 at 2:17
add comment

1 Answer 1

up vote 4 down vote accepted

If $f,g \colon X \to Y$ are continuous, then $(f,g)\colon X\to Y\times Y$ is continuous. The set $\{x\colon f(x) = g(x) \}$ is the counter-image by means of $(f,g)$ of the diagonal $\Delta = \{ (y,y) \colon y\in Y\}$ of $Y\times Y$. So it is enough to check that the diagonal is a closed set.

Let's prove that any point $(x,y)\not \in \Delta$ has a neighborhood which does not intersect $\Delta$. In fact Hausdorff property of $Y$ states that $x$ and $y$ (being different points) have two non overlapping neighbourhoods $U$, $V$. Hence $U\times V$ is a neighbourhood of $(x,y)$ not touching $\Delta$.

share|improve this answer
    
Nice argument! $ $ –  leo Jan 23 '13 at 23:34
    
Indeed it is a great exercise to show that $\Delta_Y$ is a closed subset of $Y\times Y$ if and only if $Y$ is Hausdorff. –  Thomas Andrews Jan 23 '13 at 23:58
    
Am I correct to think that if $C=\{x \in X \,|\,f(x)=g(x)\}$ then $(f,g)(C)=\Delta_Y$, and since $\Delta_Y$ is closed, its preimage in $X$ ($C$) is as well? –  Alex Petzke Jan 24 '13 at 2:13
    
@AlexPetzke we can ensure only that $(f,g)(C)\subset \Delta_Y$ –  leo Jan 24 '13 at 4:16
    
@leo: ok, I see that. But we still have $(f,g)^{-1}(\Delta_Y)=C$ is closed, by continuity? –  Alex Petzke Jan 24 '13 at 4:40
show 1 more comment

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.