$\{x\in X: f(x)=g(x)\}$ is closed in $X$ if $f$ and $g$ are continuous on $X$.

Question

could you help me demonstrate the following please

Let $f,g:(X,d_X)\rightarrow\mathbb{R}$ be continuous functions. Prove that $\{x\in X: f(x)=g(x)\}$ is closed in $X$.

It is a result that is very interesting to me, but I really do not know how to start the demonstration, it is like relative closed, but I do not know how to use the continuity hypotheses to arrive at the result. I have no attempts, because I have no idea how to attack the problem

Closed under what binary operation? Function composition? – Taylor Rendon Jul 01 '20 at 17:48 — Taylor Rendon, Jul 01 '20 at 17:48
Closed set in the topological meaning. – Mark Jul 01 '20 at 17:51 — Mark, Jul 01 '20 at 17:51

score 6 · Accepted Answer · answered Jul 01 '20 at 17:44

6

The inverse image of closed sets through continuous functions are closed sets. Take the function $f-g$, it is continuous as a difference of continuous functions. Take the closed subset $\{0\}$ of $\mathbb{R}$. Do you see why your set is equal to $(f-g)^{-1}(\{0\})$?

answered Jul 01 '20 at 17:44

Just dropped in

15,513

the truth is that I don't see why my outfit is the same, could you explain why? – Haus Jul 02 '20 at 23:40
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@Haus what do you mean outfit? – Just dropped in Jul 03 '20 at 01:17
Is correct that ${x\in X: f(x)=g(x)}={x\in X: (f-g)(x)=0 }= (f-g)^{-1}(0)$? – Haus Jul 03 '20 at 01:39
If the above is correct, since {0} is closed and the inverse image of closed is closed, we conclude – Haus Jul 03 '20 at 04:05
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Yes, that is correct @Haus – Nicholas Roberts Jul 03 '20 at 04:54

score 4 · Answer 2 · answered Jul 01 '20 at 17:44

4

Consider the continuous function $f-g$. Then your set is the inverse image $(f-g)^{-1}(\{0\})$ and thus closed.

Alternatively, take a sequence $(x_n)_n$ in $\{x \in X: f(x) = g(x)\}$ with $x_n \to x$. Then by continuity $f(x_n) \to f(x)$ and $g(x_n) \to g(x)$. But since $f(x_n) = g(x_n)$ for all $n$ it follows that $f(x) = g(x)$. Thus your set must be closed.

answered Jul 01 '20 at 17:44

J. De Ro

21,438

This is using uniqueness of limits in a Hausdorff space right? – Nicholas Roberts Jul 01 '20 at 17:48
Yes, that was implicitely used. Also in the first approach some separation axiom was used: singeltons are closed in metric spaces – J. De Ro Jul 01 '20 at 17:50
Could you explain why my set is the reverse image? – Haus Jul 03 '20 at 00:28
@Haus Simply by definition of reverse image – J. De Ro Jul 04 '20 at 11:41

score 2 · Answer 3 · answered Jul 01 '20 at 17:46

Since it is a metric space a set is closed if and only if it is sequentially closed. So suppose $x_n$ is a sequence from your set which converges to $x$. Then by continuity $f(x_n)\to f(x)$ and $g(x_n)\to g(x)$. But for each $n\in\mathbb{N}$ we have $f(x_n)=g(x_n)$, hence by passing to the limit $f(x)=g(x)$.

score 2 · Answer 4 · answered Jul 01 '20 at 18:15

The key property is that ${\Bbb R}$ is Hausdorff. More generally, consider two continuous functions $f, g$ from a topological space $X$ to a Hausdorff topological space $E$. Then the map $f \times g: X \to E \times E$ is continuous. Moreover, according to this result, the diagonal of $E \times E$, that is the set $D = \{(x, x) \mid x \in E\}$, is closed. It follows that $(f \times g)^{-1}(D)$ is also closed. But $$ (f \times g)^{-1}(D) = \{x \in X \mid (f(x), g(x)) \in D\} = \{x \in X \mid f(x) = g(x) $$

score 1 · Answer 5 · answered Jul 01 '20 at 17:45

1

consider a sequence in the set that converges to an $x\in X$. Then prove the statement by using sequential continuity

answered Jul 01 '20 at 17:45

alphaomega

2,216

$\{x\in X: f(x)=g(x)\}$ is closed in $X$ if $f$ and $g$ are continuous on $X$.

5 Answers5