1
$\begingroup$

Prove a finite union of closed sets is closed

Proof: Suppose $A_{\alpha_i}$, $i = 1,2,\ldots,n$ are closed and we want to prove $\bigcup_{i = 1}^n A_{\alpha_i}$ is closed, we know from De Morgans Law that $(\bigcup_{i = 1}^n A_{\alpha_i})^c = \bigcap_{i = 1}^n A_{\alpha_i}^c$ also from the previous problem we have proved that a finite intersection of open sets is open. Therefore, $\bigcup_{i = 1}^n A_{\alpha_i}$ is closed.

I am not sure if am right, any suggestions would be greatly appreciated. For brevity's sake just assume that I proved that a finite intersection of open sets is open.

$\endgroup$
  • $\begingroup$ It is standard to use "$\cap$" in things like $A\cap B$ and $A_1\cap\cdots\cap A_n$, and "$\bigcap$" in things like $\bigcap_{i=1}^n A_i$. I have edited accordingly. ${}\qquad{}$ $\endgroup$ – Michael Hardy Feb 22 '15 at 14:45
1
$\begingroup$

Yes, this is quite OK. In my preferred set-up of the theory, the intersection of a finite number of open sets is open by definition of a topology. Probably you have a different approach here, but the de Morgan approach is fine regardless.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.