# Showing that the Cofinite topology on $X$ indeed is a topology on $X$.

Show that the cofinite topology on $$X$$ indeed is a topology on $$X$$.

The cofinite topology seems to be defined as such $$\tau = \{U \subset X \mid U = \emptyset \text{ or U^c is finite.}\}$$

Now clearly $$\emptyset \in \tau$$ and $$X \in \tau$$ since $$X^c = \emptyset$$.

To show that the arbitary union is in $$\tau$$ I've tried the following. Let $$\{U_i\}_{i \in I}$$ be a collection of subsets of $$\tau$$. We have that \begin{align} \quad \left ( \bigcup_{i \in I} U_i \right )^c = \bigcap_{i \in I} U_i^c \end{align} but I'm not sure how I should proceed here. Do I need that $$(\bigcup_{i \in I} U_i)^c$$ to be finite or what?

• Yes, and what do you know about the arbitrary intersection of finite sets? Jun 2, 2021 at 9:08

If each $$U_i$$ is empty, then $$\bigcap_{i\in I}U_i^{\,\complement}=X$$, and $$X\in\tau$$.
Otherwise, $$U_j^{\,\complement}$$ is finite, for some $$j\in I$$, and then, since$$\bigcap_{i\in I}U_i^{\,\complement}\subset U_j^{\,\complement},$$$$\bigcap_{i\in I}U_i^{\,\complement}$$ is finite. So, $$\bigcup_{i\in I}U_i\in\tau$$ in this case too.
• What if $I$ is uncountable? Would this create any problems? Jun 2, 2021 at 9:11
• Why would that be the case? Are asking whether the fact that $\bigcap_{i\in I}U_i^{\,\complement}\subset U_j^{\,\complement}$ may be false if $I$ is uncountable? Jun 2, 2021 at 9:13