# Compact Set

1. Apr 8, 2006

Suppose $$K\subset\mathbb{R}$$. Prove that K is compact $$\Leftrightarrow$$ (whatever be the indexing set I, $$i\in I, F{i}\subset\mathbb{R}$$, $$F_i$$ are closed such that for all finite $$J\subset I$$, with $$\bigcap_{i\in J}F_i\bigcap K\neq\emptyset\Rightarrow\bigcap_{i\in I}F_i\bigcap K\neq\emptyset$$).

I can't make sense of this question.

It's in the form $$A\Leftrightarrow(B\Rightarrow C)$$

Last edited: Apr 8, 2006
2. Apr 8, 2006

### AKG

It says that K is compact iff given ANY collection of closed subsets of K, then if any finite subcollection of those subsets has nonempty intersection, then the whole collection has nonempty intersection. A collection of sets is said to have the finite intersection property iff any finite subecollection has nonempty intersection. So you're asked to prove that K is compact iff any collection of closed subsets of K with the finite intersection property has nonempty intersection.

I'm sorry if the above doesn't clarify the question any better. Anyways, the main idea is to use the finite subcovering property of compact sets, together with DeMorgan's law.

3. Apr 8, 2006

But any collection of closed subsets of K with the finite intersection property has nonempty interesction by definition.

4. Apr 9, 2006

### AKG

Well, you can take that to be a definition for compact sets. If you do, then the thing you're asked to prove follows immediately by definition. Often, compactness is defined differently, and so you have to prove that this condition is equivalent, and is thus permissible as a definition. But, if you do indeed have this given as a definition for compactness, then unless I'm missing some subtle issue, there's really nothing to prove.

5. Apr 9, 2006

No, we defined compactness with finite subcover property. I'm not entirely sure what you mean by "K is compact iff any collection of closed subsets of K with the finite intersection property has nonempty intersection" since "a collection of sets is said to have the finite intersection property iff any finite subecollection has nonempty intersection".

But logically, what do I have to prove? (A and B => C) and (B=>C)=>A? I'm not entirely clear about the statement.

All my attempts have failed. Can you write down a first step?

Last edited: Apr 9, 2006
6. Apr 9, 2006

### AKG

I'm not entirely sure what you mean by "K is compact iff any collection of closed subsets of K with the finite intersection property has nonempty intersection" since "a collection of sets is said to have the finite intersection property iff any finite subecollection has nonempty intersection".

We say a collection of sets C has the finite intersection property iff for all finite subsets F of C, $\bigcap_{S\in F}S$ is nonempty.

We say that a collection of sets C has nonempty intersection iff $\bigcap_{S\in C}S$ is nonempty.

The stuff you need to prove is:

Suppose K is a compact space, and suppose C is any collection of closed subsets of K. If C has the finite intersection property, then C has nonempty intersection. Conversely, suppose K is some space. Suppose that if C is a collection of closed subsets of K with the finite intersection property also has nonempty intersection. Then K is compact.

This is equivalent to:

Suppose K is a compact space, and suppose C is any collection of closed subsets of K. If C has the finite intersection property, then C has nonempty intersection. Conversely, suppose K is not compact. Then there exists a collection C of closed subsets of K with the finite intersection property but with empty intersection.

So what you have to prove is of the form $A\Leftrightarrow(B\Rightarrow D)$ where A is the sentence "K is compact", B is the sentence "C is a collection of closed subsets of K with the finite intersection property", and D is the sentence, "C has nonempty intersection."

Like I said, the proof just requires the finite subcover formulation of compactness, together with DeMorgan's law. As a hint, consider some arbitrary collection

$$\mathcal{C} = \{C_i\}_{i\in I}\mbox{ where }(\forall i\in I)(C_i \subset K\mbox{ is closed})$$

Now consider the collection of open subsets of K:

$$\mathcal{O} = \{K - C_i : i \in I\}$$

This is just a collection of open sets, it need not necessarily cover K. What condition on $\mathcal{C}$ is necessary and sufficient for $\mathcal{O}$ to be an open cover of K?

Last edited: Apr 9, 2006
7. Apr 9, 2006

How did you conclude that K-C_i is open? It might be neither closed nor open.

8. Apr 9, 2006

### AKG

If X is a space, and A is a subset of X, then A is closed in X iff X-A is open in X. So K-Ci may not be an open subset of R, but it is an open subset of K. The problem probably is that you're used to thinking of compactness being a property of subsets of Rn, as opposed to a property of topological spaces. You're also probably thinking that an open cover is a collection of open subsets of Rn that cover K, rather than a collection of "open" subsets of K that cover K, where being open simply means that it is an element of K's topology, and not that it is a union of intervals (a,b), necessarily.

Anyways, stick to what you know for now. So instead, let's say that:

$$\mathcal{C} = \{C_i\}_{i\in I}\mbox{ where }(\forall i\in I)(C_i \subset \mathbb{R}\mbox{ is closed})$$

$$\mathcal{O} = \{\mathbb{R} - C_i : i \in I\}$$

I ask the same question: what condition on $\mathcal{C}$ is necessary and sufficient for $\mathcal{O}$ to be an open cover of K?

9. Apr 9, 2006

$$(\cup C_i)\cap K=\emptyset$$.

10. Apr 9, 2006

### AKG

Close, but not quite.

11. Apr 9, 2006

I don't think I have any unnecessary conditions, since if $$\cup C_i$$ does not share any points with K, then surely its complement contains K. However, arbitrary union of closed sets might be open, or not closed. So I'm guessing there's an extra condition of $$\cup C_i$$ being closed?

Last edited: Apr 9, 2006
12. Apr 9, 2006

### AKG

$$\bigcup_{i\in I}\left(\mathbb{R}-C_i\right) \supset K$$
iff
$$\left(\bigcup_{i\in I}\left(\mathbb{R}-C_i\right)\right)^C \subset K^C$$
iff
$$\bigcap_{i\in I}\left(\mathbb{R}-C_i\right)^C \subset K^C$$
iff
$$\bigcap_{i\in I}C_i \subset K^C$$
iff
$$\left(\bigcap_{i\in I}C_i\right) \cap K = \emptyset$$
iff
$$\bigcap_{i\in I}\left (C_i\cap K\right) = \emptyset$$

Last edited: Apr 9, 2006
13. Apr 9, 2006

Let K=[0,1] and C1, C2 to be 0 and 1. $$\bigcap_{i\in I}\left (C_i\cap K\right) = \emptyset$$, but R-C1 U R-C2 is not an open cover for K.

14. Apr 9, 2006

### AKG

Yes it is.

15. Apr 9, 2006