Analysis / compactness proof check

[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.

 

[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?

 

[eckiller]

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

 

[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)

 

[eckiller]

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

 

[fourier jr]

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.

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)

 

[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.