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Show that a finite intersection of open sets in $\mathbb{C}$ is an open set in $\mathbb{C}$.

Attempt: I want to show $\bigcap_{i=0}^{n}A_i$ is open. Let $z\in\bigcap^{n}A_i$ for open $A_i$ in $\mathbb{C}$. Then, for some $i$, $z\in A_n$. Since $A_i$ is open, $\exists\epsilon>0$ s.t. $B_{\epsilon}(z)\subseteq A_i$. But that means $B_{\epsilon}(z)\subseteq A_i\subseteq \bigcap^{n}A_i$ thus $\bigcap^{n}A_i$ is open.

A finite union of closed sets in $\mathbb{C}$ is a closed set in $\mathbb{C}$.

Attempt: I want to show that $\bigcup_{i=0}^{n}A_i$ for closed $A_i$ is closed, so I'll show ($\bigcup_{i=0}^{n}A_i )^{c}$ is open. Let $z\in(\bigcup_{i=0}^{n}A_i )^{c}$. Then $z\in A_{i}^{c}$ for some $i$. Since all $A_{i}$ are closed, $A_{i}^{c}$ is open. Then I'm guessing you show that since you can find an open ball around $z$ in $A_{i}^{c}$, you can find an open ball around the point in ($\bigcup_{i=0}^{n}A_i )^{c}$, thus the complement is open?

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    After the edit: The proof is still incorrect. The problem in the proof is not using $n$ instead of $i$. The intersection is a subset of each $A_i$, and not a superset (generally speaking). The correct proof is explained informally in Patrick's answer.2012-01-25

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Your proof for the intersection is actually the proof for the union if you replace $\bigcap$ by $\bigcup$ and it works for a union of arbitrarily many sets (just use the same proof). For the intersection, you need to take the minimum of the radii of each ball included in $A_i$ centered at $z$ (you seem to know how to write these things so I'll let you do that). For the finite union of closed sets, you just use De Morgan's laws, because its equivalent to having finite intersections of open sets being open under these laws.

Hope that helps,