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Let $f , g : X \rightarrow Y$ be continuous where $Y$ is Hausdorff. Prove that $A = \{x : f(x) = g(x)\}$ is closed in $X$. I have done the followings.

$f(X)$ and $g(X)$ are two subspaces of $Y$.

As Y is Hausdorff, $f(X), g(X)$ and $f(X) \times g(X)$ are also.

$L = \{(f(X),g(X)) : f(X) = g(X)\}$ is closed in $f(X) \times g(X)$.

Inverse image of a closed set is closed under continuous mapping.

Thus $\{x : f(X) = G(X)\} \subset X$ is closed.

Is my approach correct? Last steps (closedness of $L$ and closedness of $L \Rightarrow$ closedness of $A$) is not clear to me. Please explain. if this process is not correct or any easier method is available, please give it.

Thank you.

Brian M. Scott
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Supriyo
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3 Answers3

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I don't completely understand what you are doing, but this is how I would do this problem : To show that $C = \{x\in X : f(x) = g(x)\}$ is closed, you can show that $C^c$ is open. Choose $x \in C^c$, then $f(x) \neq g(x)$. Since $Y$ is Hausdorff, there exist open sets $U, V \subset Y$ such that $$ f(x) \in U, g(x) \in V \text{ and } U\cap V = \emptyset $$ Take $W = f^{-1}(U)\cap g^{-1}(V)$, then $W$ is open (since $f$ and $g$ are continuous), and for any $z \in W$, $$ f(z) \in U, g(z) \in V \Rightarrow f(z) \neq g(z) $$ Hence, $W$ is a neighbourhood of $x$ that is contained in $C^c$. Hence, $C^c$ is open.

Prahlad Vaidyanathan
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  • It is clear almost. You have written "for any $z \in W, f(z) \in U, g(z) \in V \Rightarrow f(z) \neq g(z)$". Is this step very much necessary to show $W$ is a nbd of $x$?. Why? $W$ is open in $X$, containing $x$, should be enough. – Supriyo Aug 12 '13 at 06:45
  • You need $W$ to be a subset of $C^c$, otherwise it is no good. – Prahlad Vaidyanathan Aug 12 '13 at 06:47
  • If $W$ is not a proper subset of $C^c$ and open in $X$, $W \intersection C^c$ is also open in $C^c$ and so ? – Supriyo Aug 12 '13 at 06:51
  • No No. You want $C^c$ to be open in $X$, not itself! So you want to show that for any $x \in C^c$, there is an open set $W$ of $X$ such that $x \in W$ and $W \subset C^c$. (Think about why $[0,1]$ is not open in $\mathbb{R}$ - a half-open interval may be open in $[0,1]$, but not in $\mathbb{R}$) – Prahlad Vaidyanathan Aug 12 '13 at 07:01
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$Y$ is Hausdorff iff the diagonal $\Delta\subseteq Y\times Y$ is closed. And $A=(f,g)^{-1}(\Delta)$.

Hagen von Eitzen
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Your $L$ is empty if $f[X]\ne g[X]$, and it contains the single pair $\langle f[X],g[X]\rangle$ if $f[X]=g[X]$. This is a pair of subsets of $Y\times Y$ and is definitely not what you want. I'll get you started on a correct version of this approach.

Let $Y_f=f[X]$ and $Y_g=g[X]$, and define a function

$$\varphi:X\to Y_f\times Y_g:x\mapsto\langle f(x),g(x)\rangle\;.$$

Let $\Delta=\{\langle y,y\rangle:y\in Y\}$, and note that $Y_f\times Y_g$ and $\Delta$ are both subsets of $Y\times Y$, so it makes sense to let $L=(Y_f\times Y_g)\cap\Delta$.

  • Use Hausdorffness of $Y\times Y$ to show that $\Delta$ is closed in $Y\times Y$.
  • Use this to show that $L$ is closed in $Y_f\times Y_g$.
  • Show that $\varphi$ is continuous.
  • Conclude that $\varphi^{-1}[L]$ is closed in $X$.
Brian M. Scott
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  • Thank you sir. At last I got it. In the morning I could not construct the mapping $\phi : X \rightarrow Y_f \times Y_g$, but I understood that such a mapping is working. It stacked me in two points. – Supriyo Aug 12 '13 at 17:14
  • @Samprity: You're welcome. – Brian M. Scott Aug 12 '13 at 17:24