Let $X,Y$ be Hausdorff and $f,g : X \to Y$ continuous. Let $A \subset X$ be dense and suppose that $f\mid_A = g\mid_A$. Show that $f=g$.

Question

Let $X,Y$ be Hausdorff and $f,g : X \to Y$ continuous. Let $A \subset X$ be dense and suppose that $f\mid_A = g\mid_A$. Show that $f=g$.

Suppose the contrary that $f \ne g$. This implies that there exists $x \in X$ such that $f(x)\ne g(x)$. As $Y$ is Hausdorff we can consider neighborhoods $U_{f(x)}$ and $U_{g(x)}$ for which we have that $U_{f(x)} \cap U_{g(x)} = \emptyset$. Now $$f^{-1}(U_{f(x)} \cap U_{g(x)})=f^{-1}(U_{f(x)}) \cap f^{-1}(U_{g(x)})=\emptyset$$ but this is a contradiction since both of them contain at least $x$?

I'm not sure if the proof is correct since I'm not using the density of $A$ at all. Isn't it true that $x \in f^{-1}(U_{f(x)}) \cap f^{-1}(U_{g(x)})$?

You do not use the set $A$ at all. That $x$ is in this intersection is not true. We always have $x\notin f^{-1}(U_{g(x)})$ since $f(x)\notin U_{g(x)}$ by how you chose these two neighborhoods. — J. De Ro, Jan 08 '22 at 11:52
If you know about filters or nets there is an easy proof using the fact that limits in Hausdorff spaces are unique. — J. De Ro, Jan 08 '22 at 11:53
For fixed $x\in X$ take a net converging to $x$ and use continuity of $f$. — AlvinL, Jan 08 '22 at 12:35
I understand that you're mainly interesting in analysis of your attempt (after all, the post is tagged solution-verificaion). Still, I'll add links to some other posts: $f,g$ continuous from $X$ to $Y$. if they are agree on a dense set $A$ of $X$ then they agree on $X$ How to prove the uniqueness of a continuous extension of a densely defined function?, If two continuous maps into a Hausdorff space agree on a dense subset, they are identically equal — Martin Sleziak, Jan 09 '22 at 08:23

score 1 · Answer 1 · answered Jan 08 '22 at 11:51

1

Why would $f^{-1}[U_{g(x)}]$ contain $x$?

Without density of $A$, the claim is obviously false. For example, consider $A=\emptyset$.

Notice that the density of $A$ means that there is a net in $A$ convergent to $x$.

(You also only need $Y$ to be Hausdorff, $X$ can be arbitrary, as far as I can tell.)

answered Jan 08 '22 at 11:51

tomasz

35,474

Silly mistake. I considered the preimage under $f$ for both sets $U_f$ and $U_g$. It's true however that $f^{-1}[U_{f(x)}]$ contains $x$? – Locomot Ragsar Jan 08 '22 at 11:55
@LocomotRagsar: of course, it contains the preimage of $f(x)$, after all. – tomasz Jan 08 '22 at 12:05
I think you can use the fact that $X$ is Hausdorff to show that $f^{-1}(U_{f(x)}) \cap g^{-1}(U_{g(x)}) = \emptyset$ by using continuity of the maps. – Locomot Ragsar Jan 08 '22 at 12:07
As $f$ is continuous we have that $f^{-1}(U_{f(x)})$ is open neighborhood of $x$ in $X$ and also as $g$ is continuous we have that $g^{-1}(U_{g(x)})$ is open neighborhood of $x$ in $X$. Since $X$ is Hausdorff the sets $f^{-1}(U_{f(x)})$ and $g^{-1}(U_{g(x)})$ don't intersect, but they both contain $x$ again? – Locomot Ragsar Jan 08 '22 at 12:11
No, the preimages $f^{-1}[U_{f(x)}]$ and $g^{-1}[U_{g(x)}]$ always intersect, because $x$ is in both of them. If you want to do this via neighbourhoods instead of nets, you should proceed as before, by showing that $f^{-1}[U_{f(x)}]$ and $f^{-1}[U_{g(x)}]$ intersect. It will probably help if you first observe that $f[A]=g[A]$ is dense in the range of $g$. – tomasz Jan 08 '22 at 16:11

score 1 · Answer 2 · answered Jan 08 '22 at 22:47

Consider $f^{-1}[U_{f(x)}] \cap g^{-1}[U_{g(x)}]$ which is open in $X$ by continuity of $f$ and $g$, and non-empty as $x$ is in it.

So it intersects the dense set $A$ in some $a$. For $a\in A$ we know $f(a)=g(a)$ and $f(a) \in U_{f(x)}$ and $g(a) \in U_{g(x)}$ which together contradicts the disjointness of the chosen neighbourhoods in $Y$. $X$ being Hausdorff is superfluous.

Let $X,Y$ be Hausdorff and $f,g : X \to Y$ continuous. Let $A \subset X$ be dense and suppose that $f\mid_A = g\mid_A$. Show that $f=g$.

2 Answers2