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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    $\begingroup$ Is this not true in any metric space? $\endgroup$ – user21436 Jan 25 '12 at 2:51
  • $\begingroup$ There's a (serious, in my view) mistake in the proof: It is not true that $A_n$ is a subset of the intersection $\bigcap_{i=1}^n A_i$; in fact, quite the opposite is true. $\endgroup$ – Srivatsan Jan 25 '12 at 2:52
  • $\begingroup$ @Kannappan : Indeed. It is true in any topology (because it is in the definition of topologies), and the fact about closed sets is also true in any topology by using De Morgan's laws. It's not restricted to metric spaces. Although this proof is needed to show that metric spaces can be endowed with the topology generated by the balls. $\endgroup$ – Patrick Da Silva Jan 25 '12 at 2:56
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    $\begingroup$ @Emir : +1 for showing what you've tried. $\endgroup$ – Patrick Da Silva Jan 25 '12 at 2:59
  • $\begingroup$ @Srivatsan it was a typo -- I began with "open sets $A_n$ but then I changed the index set to $I$ midway for some reason. $\endgroup$ – Emir Jan 25 '12 at 3:32

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,


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