# Analysis / compactness proof check

• eckiller
In summary, a set is compact if it has a finite subcover and the union of any finite number of compact sets is also compact.

#### 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.

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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?

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

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)

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

eckiller said:
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)

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.

## 1. What is a compactness proof?

A compactness proof is a mathematical technique used to show that a set of assumptions or axioms can lead to a conclusion by demonstrating that every possible counterexample can be reduced to a finite subset of the original assumptions.

## 2. How is compactness used in mathematical analysis?

In mathematical analysis, compactness is often used to prove the existence of solutions to equations or inequalities. It can also be used to show that a function or set is bounded or has a maximum or minimum value.

## 3. What is the difference between compactness and completeness?

Compactness refers to the property of a set to contain all of its limit points, while completeness refers to the property of a set to contain all of its Cauchy sequences. In other words, a set is compact if it is closed and bounded, while a set is complete if it contains all of its possible limits.

## 4. What are some common techniques used in compactness proofs?

Some common techniques used in compactness proofs include the Bolzano-Weierstrass theorem, the Arzelà-Ascoli theorem, and the Heine-Borel theorem. These theorems provide useful criteria for determining whether a set is compact or not, and can be used to construct compact sets or identify counterexamples.

## 5. Can compactness be generalized to higher dimensions?

Yes, compactness can be generalized to higher dimensions. In fact, compactness is an important concept in topology, which studies the properties of spaces and sets that remain unchanged under continuous transformations. Compactness can also be extended to infinite-dimensional spaces, such as functional spaces, and is an essential tool in many areas of mathematics and science.