Let be $X$ a space with dense subset $D$. If $Y$ is a Hausdorff space and $f,g:X\to Y$ are continuous maps so that $f=g$ on $D$, then $f=g$ on $X$.

Question

There are a few proofs out there that involve open sets, but I was wondering if my proof using limits was correct. Here it is:

Proof. Let $x\in X$. Since $D$ is dense in $X$, there is a sequence $\{x_n\}_{n=1}^\infty$ in $D$ so that $x_n\to x$. Hence $f(x_n)=g(x_n)$. Since limits in a Hausdorff space are unique, and $f$ and $g$ are continuous, this implies that $$ f(x)=\lim_{x_n\to x} f(x_n)=\lim_{x_n\to x} g(x_n)=g(x). ~~~\square$$

Is this correct? I believe this works in $\mathbb{R}$, but my main concern was where to use Hausdorfness.

Answer 1

If you aren't familiar with nets, here's how you can approach the general case: suppose for a contradiction there were $x\in X$ with $f(x)\neq g(x)$; since $Y$ is Hausdorff you can take disjoint open neighborhoods $V$ and $W$ of $f(x)$ and $g(x)$, and then $f^{-1}(V)$ and $g^{-1}(W)$ are both open neighborhoods of $x$, as then is $f^{-1}(V)\cap g^{-1}(W)$. Since $D$ is dense in $X$ you can find some $d\in f^{-1}(V)\cap g^{-1}(W)\cap D$, but then the element $y:=f(d)=g(d)$ is inside $V\cap W=\emptyset$ giving a contradiction.

Answer 2

azif00 already mentioned in the comments that only $X$ being first-countable guarantees, that for any point $x\in X$, you find a sequence $(x_n)_{n\in\mathbb{N}}$ with $x_n\in D$ for all $n\in\mathbb{N}$ and $x_n\rightarrow x$ (See also here), so your proof only works with this additional condition. A relation of first-countable to other properties can be found here.

azif00 also already mentioned, that this can be bypassed by using nets, where we have, that for any point $x\in X$ there exists a net $(x_i)_{i\in I}$ with $x_i\in D$ for all $i\in I$ and $x_i\rightarrow x$ (See also here under Closed sets and closure). For a net like this, we have $f(x_i)\rightarrow f(x)$ and $g(x_i)\rightarrow g(x)$ since $f$ and $g$ are continuous (See also here under Continuity). In a Hausdorff space like $Y$, the limit point of nets is unique (See Theorem 3 in "General Topology" by John Kelley here on page 67), so $f(x)=g(x)$.

I wouldn't use nets though, as there are a lot simpler and more elementary proofs as you said. For example is it a direct corollary (using $X=\overline{D}\subseteq X'\subseteq X$, so $X'=X$) of the following more general lemma:

Lemma: For $f,g\colon X\rightarrow Y$ continuous with $Y$ Hausdorff, the subset $X':=\{x\in X|f(x)=g(x)\}\subseteq X$ is closed.

Proof: $(f,g)\colon X\rightarrow Y\times Y$ is continuous. Since $Y$ is Hausdorff, the diagonal $\Delta_Y\subset Y$ is closed and therefore $X'=(f,g)^{-1}(\Delta_Y)$ is as well.