# Analysis / compactness proof check

1. May 16, 2005

### eckiller

Hi,

I think this proof is easy, but would like someone to check my work since sometimes I miss technicalities on these "easy" proofs.

Let K1, ..., Kp be compact sets in R^n. Show that union( Kj, j = 1 to p) is a compact set in R^n.

Proof.

We show that if K1 and K2 are compact then K1 union K2 is compact. Then
apply this fact finitely many times to conclude the original statement.

I have a theorem: A set E in R^n is compact IFF E is a bounded closed set.

Then if K1 and K2 are compact, they are bounded and closed.

Then the union is closed. We define a bounded set as a set contained in the
open ball B(0, r), where 0 = (0, 0, ..., 0). K1, and K2 bounded implies K1
contained in B(0, r1), and K2 countained in B(0, r2). Then pick r = max(r1,
r2). Then K1 union K2 contained in B(0, r). And hence K1 union K2 is
bounded. Thus K1 union K2 is a compact set.

Last edited: May 17, 2005
2. May 17, 2005

### quasar987

Remark: After you edited, you took out the first sentence of your proof. That is, the generalization that if the union of K1 and K2 is compact, then the union of an infinity of compact sets is also a compact.

Is your new proof based on that argument as well?

3. May 17, 2005

### eckiller

Yes, I overwrote it by accident when I copied my new version of the proof in.

4. May 17, 2005

### quasar987

Ok, well I'm no analysis guru but I believe this argument is not valid. For exemple, in R, the union of two closed sets is closed but the union of an infinity of closed is generally not closed.

And here's a disturbing example of an infinite union of compact set of R that is not closed, and hence not compact:

$$F_n = \left[\frac{1}{n},1\right], \ \ n \in \mathbb{N}$$

F_n corresponding to each integer is closed and bounded by 1, and hence compact (according to the theorem you state). But

$$\bigcup_{n=1}^{\infty}\left[\frac{1}{n},1\right]=(0,1]$$

is not closed. (Example taken from my real analysis textbook)

5. May 17, 2005

### eckiller

Thanks for the reply. But note that mu union is finite. j = 1 to finite number p.

6. May 17, 2005

### fourier jr

no need to use that; it's not that complicated. since each one of your sets has a finite subcover, their union will also have a finite subcover. it's just a matter of writing it down in a rigourous way now & fiddling with the notation. (& yes the theorem is only true for a finite # of sets)

7. May 17, 2005

### quasar987

Aw man, I hadn't paid attention to "from j=1 to p" and i misread "finitely" to "infinitely" :grumpy:.. time to go to sleep I think.