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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?

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  • $\begingroup$ Yes, and what do you know about the arbitrary intersection of finite sets? $\endgroup$
    – Dan Rust
    Jun 2, 2021 at 9:08

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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.

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  • $\begingroup$ What if $I$ is uncountable? Would this create any problems? $\endgroup$
    – Danlo
    Jun 2, 2021 at 9:11
  • $\begingroup$ 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? $\endgroup$ Jun 2, 2021 at 9:13

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