Dense and continous function in topological space 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 
 A: 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

*

*$X/S$ 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  $X/S$ ;


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


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