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Let $f, g : X \to Y$ be continuous functions. Assume that $Y$ is Hausdorff and that there exists a dense subset $D$ of $X$ such that $f(x) = g(x)$ for all $x \in D$. Prove that $f(x) = g(x)$ for all $x \in X$.

Here is what I have so far,

Proof:
Let $f : X \to Y$ and $g : X \to Y$ be continuous and suppose that $f(x)=g(x)$ for some dense $D\subset X$. Let $x \in X$, since $D$ is a dense subset of $X$, $x \in \mathrm{Cl}(D)$. I'm unsure how to proceed from here.

Null
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2 Answers2

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$Y$ being Hausdorff is equivalent to the diagonal $Δ = \{(y,y);~y ∈ Y \} ⊂ Y × Y$ being closed.

The map $〈f,g〉\colon X → Y × Y$ is continuous and has $〈f,g〉(D) ⊂ Δ$. By continuity, you get $$〈f,g〉(\overline{D}) ⊂ \overline{Δ}.$$

Martin Sleziak
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k.stm
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    So, is your argument basically saying that the closure of the composition is a subset of the closure of the diagonal ->f(x)= g(x):∀x∈X – Null Dec 28 '14 at 23:57
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    Good answer, but maybe not the elementary answer that Cody Silvers is looking for. I think your $f \times g$ should read $\langle f, g \rangle$ (i.e., you want the function from $X$ to $Y \times Y$ that exists by the universal property of products rather than the function from $X \times X$ to $Y \times Y$ that exists because products are a bifunctor). – Rob Arthan Dec 28 '14 at 23:57
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    @RobArthan Hm, yeah. I’m still not sold on this convention, but it seems to be more common, so I will adapt. – k.stm Dec 28 '14 at 23:58
  • @CodySilvers What do you mean by “composition”? – k.stm Dec 29 '14 at 00:01
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    I'm really confused... We haven't covered 'diagonals' in class, so I'm having difficulties following your proof. I understand what a hausdorff/T2 space is but I don't understand how it relates to, or equates to the diagonal. – Null Dec 29 '14 at 00:05
  • @CodySilvers Maybe you haven’t defined the product topology yet or seen the used characterization for continuity? In that case you could stick to the hint given by Rob in the comments. The diagonal $Δ$ is really just the set of pairs $(y,y)$ with $y ∈ Y$ – that’s it, nothing more, nothing scary. The result that $Y$ is Hausdorff if and only if $Δ ⊂ Y × Y$ is closed (with respect to the product topology) is very fundamental and can be used for all sorts of arguments involving the Hausdorff property (you should learn it!). – k.stm Dec 29 '14 at 00:08
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When we have that $f(x)\neq g(x)$ then there exists opens $U_f$ and $U_g$ such that $U_f\cap U_g=\emptyset$, here we use the Hausdorff condition. Now $A:=f^{-1}(U_f)\cap g^{-1}(U_g)$ is an open since $f,g$ are continuous. Then we have that on $A$ the functions do not coincide. This contradicts the fact that they coincide on a dense subspace.

Edit: Here we have that $U_f$ is an open containing $f(x)$ and similarly $U_g$ contains $g(x)$.

Kaladin
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    What are $U_f$ and $U_g$? – Rob Arthan Dec 29 '14 at 00:18
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    $U_f$ is an open containing $f(x)$ and $U_g$ is an open containing $g(x)$. – Kaladin Dec 29 '14 at 00:20
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    well you should state that in your answer! – Rob Arthan Dec 29 '14 at 00:20
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    The formulation could be more precise. That the functions do not coincide on $A$ is too weak for a conclusion. You need that $A$ is a non-empty (since it contains $x$) open subset such that $f$ and $g$ differ at every point of $A$. Then $A\cap D$ contains at least one point$~p$, with $p\in D$ and $f(p)\neq g(p)$, for a contradiction. – Marc van Leeuwen Dec 29 '14 at 08:11
  • Marc and Kaladin y'all make a good proof! – TeeJ Lockwood Dec 14 '15 at 07:34
  • We can see the another contradiction for the disjoint sets $U_f$ and $U_g$ if we assume that $f(p) = g(p)$. We have that $A$ is nonempty because it contains $x$. We then have that there exist some $p \in A\cap D$, since $D$ is a dense subset and for any open set of $X$ there contains elements of $D$. However this implies that $f(p) = g(p)$, and further more $f(p) \subset U_g$ and $g(p) \subset U_f$, also written as $f(p)\in U_f \cap U_g$ a contradiction since these sets are disjoint. – TeeJ Lockwood Dec 14 '15 at 07:43