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(a) Show that the union of finitely many closed sets is closed.

(b) Give an example showing that the union of infinitely many closed sets may fail to be closed.

(a)

Given that $ \bigcup_{k=1}^{n} C_k $ is closed if $ \left[ \bigcup_{k=1}^{n} C_k \right]^c $ is open.

,and $ \left[ \bigcup_{k=1}^{n} C_k \right]^c $ is open if $\bigcap_{k=1}^{n} C_{k}^c$ is open.

As It was shown previously that $\bigcap_{k=1}^{n} C_{k}^c$ is open (I did this in a previous post Intersection of open sets?), $ \bigcup_{k=1}^{n} C_k $ has to be closed.

(b) any hint for this part?

Regarding (a) Is my argumentation sufficient? is there a better alternative?

Much appreciated.

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    $\begingroup$ Hint: in many familiar topological spaces, one-point sets are closed. $\endgroup$ Jun 14, 2017 at 5:03

3 Answers 3

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Your proof of (a) is o.k.

(b): $ \bigcup_{k=1}^{\infty}[\frac{1}{k},1]$

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Your proof of (a) is OK.

In order to find a counter example (for dropping a requirement) you often can see this from the proof, but in this case you have to check the proof for the theorem that you're using, the proof where the requirement is really used.

The requirement is used in that proof by picking a point in the intersection (of open sets) and see that there's a disc around that that is within each open set and a disc with radius being the minimum of these is within the intersection (in fact you need only have the radius being a lower bound).

Here the finiteness is used to ensure that a set of radii has a minimum and that is positive.

To make that step fail you need to have an infinite set of radii that will have a lower bound of zero and that's easy to find. Just take $r_j = 1/j$. That is you pick a set of open discs with radius $1/j$ then their intersection is no longer open.

Then you just take the complement of these and get a counter example. That is $F_j = \{x: x\ge 1/j\} $ and see that $\bigcup F_j = \{x\ne 0\}$ which is not closed.

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The union of finitely many closed sets is open. The intersection of an arbitrary collection of closed sets is closed.

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