Check a topology Hello I am starting my topology studies, and I am having difficulties with this exercise
Let $\infty \in X$ be a given element in a set. Check that $\tau = \{ A \subset{X}|  \infty \not\in A  $ or $X-A$ is finite $\}$
defines a topology on $X$.
I have to check that:

*

*$\varnothing$ and $X$ are open sets

*The union of any number of open sets is an open set

*The intersection of two open sets is an open set.

I don't know how to check it
Thanks
 A: We have a space $X$ with a distinguished point $\infty$.
Let $\tau$ be a topology (using the set of open sets definition) is defined as follows:
$$ \tau = \{ A \mathop| A \subset X \land (\infty \notin A \lor |X-A| \;\text{is finite}) \} $$
Let's check that condition 1 holds.
$\varnothing$ is an open set because $\infty$ is not in it.
$X$ is an open set because $|X-X|$ is empty and thus finite.
Let's check that condition 2 holds.
Let $F$ be a family of open sets. Suppose none of them contain $\infty$, then $\bigcup F$ also doesn't contain $\infty$ and is therefore open.
Suppose at least one element of $F$ contains $\infty$, let's pick one of them and call it $F_0$. Since $F_0$ is open and contains $\infty$ that means that $X-F_0$ is finite. $X-\bigcup F $ is necessarily a subset of $X-F_0$ and is therefore also finite. Therefore $\bigcup F$ is open.
Let's check that condition 3 holds.
Suppose we have two open sets $A$ and $B$.
If $A$ or $B$ or both fail to contain $\infty$, then $A \cap B$ also fails to contain $\infty$ and is therefore open.
If $|X-A|$ is finite and $|X-B|$ is finite then $|(X-A) \cup (X-B)|$ is finite and thus $|X - (A \cap B)|$ is finite. Therefore $A \cap B$ is open.
