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If $Y_1,...,Y_n$ are a finite collection of compact subsets of $X$, then their union $Y_1\cup....\cup Y_n$ are also compact.

Proof:

Assume that $Y_1,...Y_n$ are a finite collection of compact subsets of $X$

WTS that $Y_1\cup....\cup Y_n$ are compact

In order for $Y_1 \cup...\cup Y_n$ to be compact, their union must be closed and bounded

Lets first prove that the union of $Y_1\cup...\cup Y_n$ is closed

I will do this using a previous proposition which I learned

Prop: If $Y_1,...Y_n$ are are finite collection of of closed sets in X, then $Y_1\cup...\cup Y_n$ is also closed

(I have the proof of this proposition)

By our assumption $Y_1,...,Y_n$ are a finite collection of compact subsets of $X$. And since they are compact, then they must be closed

Now I have to prove that the union of $Y_1\cup...\cup Y_n$ is bounded

By our assumption $Y_1,...,Y_n$ are a finite collection of compact subsets of $X$. And since they are compact, then they must be bounded

Since $Y_1,...,Y_n$ are bounded subsets of X then by definition $\exists B(x,r)$ in X which contains $Y_1,...,Y_n$

If $\exists B(x,r)$ in X which contains $Y_1,...,Y_n$ then this means that this ball $B(x,r)$ will also contain $Y_1\cup...\cup Y_n$ (Intuitively, this seems true to me, but I am not sure if it is and how to show this)

Can I please get some help in proving this?

thisisme
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  • For each $1 \le k \le n,$ there is a ball $B(x_k, r_k)$ that contains $Y_k$. Your problem is to prove that there is one ball that contains them all. Yes, it's obvious that if $B(x,r)$ contains each of the, it contains their union. That's the definition of union. But it's not justified to simply assert the existence of $B(x,r).$ – saulspatz Jan 31 '18 at 06:25
  • This proposition is easier to prove in general. Recall that if $X$ is a topological space and $U\subset X$, then $U$ is called compact if every open cover of $U$ admits a finite subcover. This definition makes this property almost trivial. – Mathematician 42 Jan 31 '18 at 06:28
  • @Mathematician42 I started to write the very same thing, but then I noticed that the tag was real-analysis, not topology. – saulspatz Jan 31 '18 at 06:32

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