Any collection of neighborhoods of a dense set cover the space?

Question

Trying to solve an exercise, I'm trying to see if this is true:

Let $X$ a Hausdorff space and $D\subset X$ dense. Also let a function $$f:D\to\wp(X),\quad x\mapsto U_x$$ where $U_x$ is a neighborhood of $x$. Then we can say that $X=\bigcup f(D)$?

I'm unable to prove it so probably it's false. My first idea was taking each $U_x$ open, then $U:=\bigcup f(D)$ is also open and $U^\complement$ closed. Then if my assertion would be true, it must be the case that $U^\complement=\emptyset$, but I'm unable to show this.

Can someone confirm if there exists some Hausdorff space where my hypothesis doesn't work? Or, by the contrary, if my hypothesis is true?

Answer 1

This is false even in the real line. Let $\{r_1,r_2,...\}$ be an ennumeration of the rationals and consider the neighborhoods $(r_n-\frac 1 {2^{n}},r_n+\frac 1 {2^{n}})$. Since the total length of these intervals is $\sum_{n=1}^{\infty} \frac 1 {2^{n-1}}=2$, they cannot cover the entire line. [ If you know measure theory you can see the reason immediately. Otherwise you can see that the intervals don't even cover $[0,3]$ using a compactness argument.]

Answer 2

If $y\notin D$ then you can choose for $U_x=X-\{y\}$ for every $x\in D$ - which is evidently an open neighborhood of $x$ in Hausdorff space $X$.

Then the union of these sets is again $X-\{y\}\neq X$.

Apparantly that statement is false already if $X$ is a $T_1$-space.