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## Homework Statement

This question has two parts :

(a) Let F be a finite collection of closed sets in ℝ

^{n}. Prove that UF ( The union of the sets in F ) is always a closed set in ℝ

^{n}.

(b) Let F be a finite collection of closed intervals A

_{n}= [[itex]\frac{1}{n}, 1- \frac{1}{n}[/itex]] in ℝ for n = 1,2,3... What do you notice about UF? Is it closed, open, both or niether?

## Homework Equations

I know a set S is open ⇔ it is equal to its own interior, that is S = S

^{0}.

I know a set S is closed if its compliment ℝ

^{n}-S is open.

I also know a set S is closed ⇔ B(S) [itex]\subseteq[/itex] S.

I know that the only open and closed sets are ℝ

^{n}and ∅.

I know a set S is neither open or closed if there is at least one point which has a neighborhood containing points from both S and its compliment.

## The Attempt at a Solution

(a) Suppose that F = {[itex] S_1, S_2, ..., S_p | S_i\in ℝ^n, 1 ≤ i ≤ p [/itex]}

Suppose further that :

[tex]Q = \bigcup_{i=1}^{p} S_i[/tex]

We want to show Q is always a closed set in ℝ

^{n}, so let us show that the compliment of Q, ℝ

^{n}- Q is open.

We know from F that : [itex]S_i \subseteq B(S_i), \space 1≤ i ≤ p[/itex]. That is, each individual set contains its own boundary.

This implicates to us that : [itex]ℝ^n - S_i[/itex] is open since S

_{i}is closed for 1 ≤ i ≤ p now suppose that :

[tex]Q_o = \bigcup_{i=1}^{p} (ℝ^n - S_i)[/tex]

So that Q

_{o}is composed of a finite collection of the compliments of F.

Am I on the right track here or have I missed something? Ill attempt (b) after I figure this one out.