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The open sets of the cofinite topology on $\mathbb{R}$ are defined as all sets in $\mathbb{R}$ whose complement is finite. Can someone point out the error in my logic with regards to the union of any open sets also being open?

Let: $$U = \bigcup_{i=1}^{\infty}\{U_i \mid \mathbb{R} - i\}$$ So for example, $U_1 = \mathbb{R} - 1$, $U_2 = \mathbb{R} - 2$, etc. Thus each $U_i \in U$ is cofinite and in the set of open sets.

However, isn't the set $U$ not cofinite since $U = \mathbb{R} - \mathbb{N}$ meaning the complement of $U$ is $\mathbb{N}$ which is not finite?

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You're mixing up unions and intersections. What you've shown is that the countable intersection of cofinite sets needn't be cofinite. – goblin Feb 28 '14 at 23:58

3 Answers 3

up vote 2 down vote accepted

The more sets you throw into a union, the smaller the complement becomes. In your example, $U_1 \cup U_2 = \mathbb{R}$.

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$(\Bbb R\setminus\{1\}) \cup (\Bbb R\setminus\{2\})=\Bbb R$

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Let $(U_i)_{i\in I}$ be a family of open sets, which means their complements are either finite or $\mathbb{R}$. Proving that $$ U=\bigcup_{i\in I}U_i $$ is open is the same as showing that the complement is closed (that is, either finite or $\mathbb{R}$); now, by De Morgan, $$ \mathbb{R}\setminus U=\mathbb{R}\setminus\biggl(\bigcup_{i\in I}U_i\biggr)= \bigcap_{i\in I}(\mathbb{R}\setminus U_i) $$ If one of the $U_i$, say $U_{i_0}$ is non empty, we have $$ \bigcap_{i\in I}(\mathbb{R}\setminus U_i)\subseteq \mathbb{R}\setminus U_{i_0} $$ which is finite. Otherwise every $U_i$ is empty and the union is empty, thus open.

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