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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?

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  • $\begingroup$ 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. $\endgroup$ – Brian M. Scott Mar 9 at 23:16
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  1. It is enough (I would change the phrasing "is absurd" - but I am not a native english speaker)

  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.

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  • $\begingroup$ 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. $\endgroup$ – ask 1234 Mar 9 at 18:40
  • $\begingroup$ $Y$ 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) $\endgroup$ – shortmanikos Mar 9 at 19:16
  • $\begingroup$ @shortmanikos: The phrasing is absurd is fine and not uncommon in this context. $\endgroup$ – Brian M. Scott Mar 9 at 23:17

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