Dense and continous function in topological space

Question

Let $X, Y$ be two metric spaces and $D$ is a dense subset of $X$. Let $f, g$ be Two continous function from $X$ to $Y$ S. t
$f(x) =g(x)$ for all $x$ in $D$. Then $f=g$ on $X$.

Is this result true in arbitrary topological space??

I know this is not true in topological space. But I can't find any example. Please give me an example.

Again this result will true in topological space provided that $Y$ is Hausdorff. But how to prove it. Please help me. Thank you

Answer 1

In an Hausdorff space $Y$ the diagonal $\Delta\subset Y\times Y$ is closed. The function

$H: X\to Y\times Y$

that maps every $x\in X$ to $H(x):=(f(x),g(x))$ is continuos and so the inverse image of $\Delta$ twith respect the function $H$ is closed, but this set is

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

When this set is dense you have that $D=D^{cl}=X$ and so $f=g$

If you consider a $T_1 $ connected topological space $X$ in which $S:=X/ \{p\}$ is dense in $X$ (for example $\mathbb{R}$ with the standard topology) you have that

1. $X/S$ is not T_1 (and so it is not T_2) because if you consider $x\in S$ you have that $\pi^{-1}(x^\sim)=X/ \{p\}$ that it is open, so $\{x^\sim\}$ is open in $X/S$ but it can not be closed because $\{x^\sim\}$ would be a proper closed and open set of the connected space $X/S$ ;

2. $\pi: X\to X/S$ and $c: X\to X/S$ are continuos function, where $c$ is the constant function that maps every $z\in X$ to $\pi(S)$ ;

3. $D:=\{x\in X: \pi(x)=c(x)\}=X/ \{p\}$ and D is dense;

   $\pi\neq c$

Answer 2

A simple counterexample: let $Y = \mathbb{R}$ in the indiscrete (trivial) topology. Let $X$ be the $\mathbb{R}$ in the usual topology. $f(x) = 0$ for all $x$ and $g(x) = 0$ for $x \in \mathbb{Q}$ and $g(x) =1 $ otherwise. Then both are continuous (any map with codomain the indiscrete topology is) and they agree on the dense set $\mathbb{Q}$ but $f \neq g$.

We could also take the cofinite topology on $\mathbb{R}$ as $Y$, and use $f(x) = x, x \in \mathbb{R}$ and $g(x) = x$, for $x \in \mathbb{Q}$, and $g(x) = x+1$, $x \in \mathbb{R}\setminus \mathbb{Q}$ to get a $T_1$ counterexample. These maps are bijections and bijections between $T_1$ spaces are always continuous. Again they agree on $\mathbb{Q}$ and are different.

The proof by Federico Falluca is fine. You could also use nets if you preferred those.