A topological function with only removable discontinuities

Solution 1:

Note that property $(1)$ is just equivalent to regularity, so we already know many examples of spaces for which the proposition holds.

Solution 2:

If you require the limit to be unique (as you said, like if we assume $Y$ to be Hausdorff), then $g$ must be continuous.

First, the function $g$ is defined everywhere, since the limit of $f$ exists and is unique everywhere. Now you have to show that the limit of $g$ exitst (and is unique, but that's implied by the Hausdorff assumption) and that it matches the value of the function. Well, assume the limit does not exist at a point $x_0$. Then,

$$\exists V\subset Y\ \text{ s.t. }\ g(x_0)\in V, \text{ and }\, \forall U\subset X \,\, \exists y_U\in U \,\, \text{s.t. } g(y_U)\notin V\tag{1}$$

To fix ideas, if $X,Y$ were also metric spaces, this would mean

$$ \exists \varepsilon>0 \text{ s.t. } \forall \delta>0\ \exists x_\varepsilon \,\text{ s.t. }\, |x_\varepsilon-x_0|<\delta, |g(x_\varepsilon)-g(x_0)|>\varepsilon. $$

Given this set $V$, by definition of limit (and the definition of $g$), there is $U\subset X$ s.t. $\forall x\in U$, $f(x)\in V$. However, by (1), there is $y_U\in U$ s.t. $g(y_U)\notin V$. There are two scenarios: $g(y_U)\notin \partial V$ and $g(y_U)\in \partial V$.

1) If $g(y_U)\notin \partial V$, then $g(y_U)\in (\overline{V})^c$ and, by definition of $g$ as limit of $f$, $\exists U_y\subset U$ such that $y_U\in U_y$ and $f(x)\notin V\ \forall x\in U_y$, which contradicts the fact that $\lim_{x\to x_0} f(x)$ exists.

2) If $g(y_U)\in\partial V$, then consider $W\subset V$ such that $g(y_U)\notin \overline{W}$ (I believe this is possible, because $Y$ is Hausdorff, but you may want to double check this assumption). Then, by definition of limit at $x_0$, we have $Z\subset U$ such that $\forall x\in Z, f(x)\in W$. On the other hand, because of (1), we have that there is $y_W\in Z$ such that $g(y_W)\notin V$, and therefore $g(y_W)\notin \overline{W}$. Hence, we can use part 1) to conclude.

By the same argument I think you should be able to prove that the limit as $x\to x_0$ of $g$ has to be $g(x_0)$.

Edit: the proof was correct only if $g(y_U)\notin \bar{V}$. I added a proof for the case $g(y_U)\in\partial V$.