# Infinite set is dense set in cofinite topology

Define $$\tau = \{U \subset X; U^c \ \text{is finite or} \ U=\emptyset\}.$$ Let $$X$$ be an infinite set and $$Y \subset X$$ be an infinite subset of $$X$$. Prove that $$Y$$ is dense in the cofinite topology and that $$(X,\tau)$$ is separable.

My attempt:

Let $$x \in X$$ and $$O$$ be an open set that contains $$x$$. Using the definition we have that $$X\setminus O$$ is finite. If $$Y \cap X=\emptyset$$, then $$Y \subset X\setminus O$$ which is absurd. So $$Y$$ intersects $$O$$ and, as $$x$$ is a neighbourhood of $$O$$, $$x \in \overline{Y}$$ and we have $$\overline {Y}=X$$.

Is this enought or we have some mistake?

For the second part, i don't know to prove that $$(X,\tau)$$ is separable, any hint?

• Where you write as $x$ is a neighbourhood of $O$ you actually mean as $O$ is a neighbourhood of $x$. I would say as $O$ is an arbitrary neighbourhood of $x$: the fact that this applies to every nbhd of $x$ is crucial. Otherwise it’s fine. Mar 9, 2021 at 23:16

2. "For the second part, i don't know to prove that $$(X,\tau)$$ is separable", you just did prove it - think about the definition of separable space.
• My definition for separable spaces is: If $Y\subset X$ is dense and enumerable, then $(X,\tau)$ is separable. I don't really get when i proved that $Y$ is enumerable. Mar 9, 2021 at 18:40
• $Y$ isn't necessarily enumerable but what you have proved is that every infinite set is dense. Since $X$ is infinite it follows that... (basic set theory result) Mar 9, 2021 at 19:16