7
$\begingroup$

I want to prove that if $f,g$ are continuous functions from a topological space $(X,\tau)$ to a metric space $(Y,d)$ then the set

$$ A = \{ x \in X : f(x) = g(x) \} $$

is closed. I found a very similar question asked on this site but it referred to Hausdorff spaces, which I haven't quite met. I know I can prove this by proving that the complement of $A$ is open. If I consider $b \in X$ such that $f(b) \neq g(b)$, then I can find two open sets $U,V \subseteq Y$ such that $f(b) \in U $, $\ g(b) \in V$ and $U \cap V = \varnothing$. Then since the functions are continuous we must have that $f^{-1}(U)$ and $g^{-1}(V)$, which both contain $b$, are open. Hence so is their intersection.

Now I'm not sure how to proceed...

EDIT: Whilst there are neater methods in the answers, I've just realised how I can finish the proof above. The intersection

$$ W_b = f^{-1}(U) \cap g^{-1}(V)$$

lies entirely outside $A$ --- one can show this by supposing that there exists some $a \in A$ that is a member of $W_b$ and showing that $f(a) \in U$ and $g(a) \in V$. But $f(a) = g(a)$ by the assumption that $a \in A$, which is impossible since $U \cap V = \varnothing$.

Then the union

$$ \bigcup_{b \in X \setminus A} W_b$$

is a union of open sets and hence open, contains all $b \in X \setminus A$, and does not intersect with $A$ since $W_b \cap A = \varnothing$. Hence the union above is precisely $X \setminus A$, and hence $A$ is closed.

$\endgroup$

4 Answers 4

11
$\begingroup$

If $Y$ is a metric space, the set $\Delta =\{(y,y):y\in Y\}$ is closed in $Y\times Y$ -- this holds more generally iff $Y$ is Hausdorff (can you prove this?). Observe now that $$A=\{x\in X:f(x)=g(x)\}$$ equals the set $$B=\{x\in X:h(x)\in \Delta\}=h^{-1}(\Delta)$$ where $h:X\to Y\times Y$ is defined by $h(x)=(f(x),g(x))$. But $\Delta$ is closed, and $h$ is continuous. This proves the more general case where $Y$ is Hausdorff.

$\endgroup$
0
8
$\begingroup$

Consider the map $F:X\rightarrow \Bbb R$ given by $F(x)=d(f(x),g(x))$. $F$ is continuous (you can check this), and $F^{-1}(\{0\})=A$ so $A$ is closed since $\{0\}$ is closed in $\Bbb R$.

$\endgroup$
2
  • $\begingroup$ What is $0$ in metric space? $\endgroup$
    – Hamou
    Jul 26, 2014 at 22:38
  • $\begingroup$ @Hamou $0\in\Bbb R$! $\endgroup$
    – Pedro
    Jul 26, 2014 at 22:38
3
$\begingroup$

It's equal to

$$\bigcap \{x\in X: d(f(x),g(x))\le \epsilon\}$$

which is the intersection of closed sets since $Y$ is metric (hence Hausdorff) and $f,g$ are continuous.


We know it's closed because $Y$ is Hausdorff, so that the diagonal, $\Delta Y\subseteq Y\times Y$ is closed. This is so because the complement is open:

Proof: If $(x,y)\ne (y,y)$ then $B(x,{1\over 2}d(x,y))\times B(y,\delta)$ is an open set around $(x,y)$ not containing $(y,y)$, so the complement is open.

$\endgroup$
3
  • $\begingroup$ It doesn't matter if the intersection is finite when you're dealing with closed sets, that's only for open sets. $\endgroup$ Jul 26, 2014 at 22:42
  • $\begingroup$ @AymanHourieh I've updated my answer to address your question of "why is it closed." Really it holds more generally in all Hausdorff spaces, there you substitute the balls I used in the answer by an open set containing $x$ by not $y$ crossed with any open set around $y$. $\endgroup$ Jul 26, 2014 at 22:50
  • $\begingroup$ I think it was fine. I just got confused for a second and thought we were talking about a union. $\endgroup$ Jul 26, 2014 at 22:53
2
$\begingroup$

Let $X$ be topological space and Y be a metric space. We show that X\A is open. Let $x \in$ X\A be arbitrary.

Then $f(x)\neq g(x)$ (Otherwise x would be in A). We define $\epsilon:=d(f(x),g(x))$, so the open neighborhoods $U_{\frac{\epsilon}{2}}(f(x))$ and $V_{\frac{\epsilon}{2}}(g(x))$ are disjoint.

Hence $W:=f^{-1}(U_{\frac{\epsilon}{2}}(f(x)))\cap f^{-1}((V_{\frac{\epsilon}{2}}(g(x))) \neq \emptyset $ is an open neighborhood of x with $W \subseteq$ X\A. If this were not true then there were a $y \in W$ such that $f(y)=g(y)$ which can't be happen.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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