Compactness Question

1. Dec 4, 2007

shapiro478

Say X is a topological space and F is in X. If C is contained in F and compact in F, then C is compact in X. This is obvious when you draw a picture, but proving it is a little more difficult.

By hypothesis, C is compact in F so C has a finite subcover M, with M being the union of m_1, m_2, m_3, and so on up through m_n. Now each m is in F since C is compact in F, and F is contained in X, so each m is in X.

But then M would also be contained in X since you union all the m's that are in X. So then C would have a finite subcover in X, so C is compact in X.

Are there flaws with this line or reasoning? Thank you for the insight.

2. Dec 4, 2007

JasonRox

The hypothese doesn't say that C is compact in F! The flaw is in the first line.

You have other errors too, but I'll ignore those.

Ok, what is the definition of compact? I want to see it explicitly written correctly in one of your posts.

Once you have that, it'll be much easier to show you what you need to do and easier for you do see how to prove such a thing.

3. Dec 4, 2007

rs1n

I don't see how it is obvious. If C is compact in F, then yes, any open cover of C in F has a finite open subcover in F. However, the finite open cover in F may not necessarily be open in X (as it depends on F). Fortunately, if A1, A2, ..., An is a open cover of C in F, then we can choose open sets A1', A2', ..., An' in X such that $$A_i = A_i' \cap F$$.

4. Dec 4, 2007

JasonRox

The statement isn't even true.

Look at [0,5] = F, so F is compact in R and let (1,2) = C. Is C compact in F or R? Why not?

5. Dec 4, 2007

shapiro478

JasonRox -- The hypothesis does say C is compact in F... "If C is contained in F and compact in F..." Compact means every open cover has a finite subcover

6. Dec 4, 2007

JasonRox

I thought I read if C is contained in F and F is compact, then C is compact.

Note: The definition of compact is not C is compact if it has a finite open cover. C is compact if every open cover of C, it has a finite open subcover of C.

7. Dec 4, 2007

shapiro478

so it looks good JasonRox? thanks for your help

8. Dec 5, 2007

rs1n

I'm not trying to be picky, but as JasonRox has mentioned earlier, you need a clear definition of compact: every open cover has a finite open subcover -- but where are those open sets? Perhaps we should say C is compact in F means every open cover M' in F (i.e. the sets in M' are open in F) has a finite open subcover M in F. To show that C is also compact in X you need to start with any open cover of C in X (i.e. start with a collection of sets which are open in X that cover C) and show that you can obtain a finite open subcover (i.e. a finite collection of open sets in X which cover C).

By "m is in F" I presume you mean "m is a subset of F" -- this isn't enough. You need the sets to also be open sets. However, a set that is open in F may not be open in X. For example, if X is the closed interval [0,1] and F is the closed interval [0,1/2], then the set (1/4,1/2] is open in F, however it is NOT open in X.

So again, suppose you have an open cover of C in X. How can we obtain a finite open subcover? And how can we use the fact that C is compact in F?

Here is a sketch to help you. Let M be an open cover (in X) of C. We wish to show that there exists a finite open subcover of C in X.

1) Consider the intersection of the open cover M (in X) of C with the set F. What do you get? Think about what open sets in F look like (if you consider F as a subspace of X).

2) Use part 1) and the fact that C is compact in F. What can you conclude with this information?

3) Use part 2) to find a finite open cover of C in X. I've also given you a hint in my previous post.

Last edited: Dec 5, 2007
9. Dec 5, 2007

shapiro478

rs1n -- This new approach looks a lot more promising.

I just have a question or two about this approach. M is a bunch (or even infinite number) of open subsets of X. When you intersect M with F, how do you know that the resultant intersection is still comprised of sets open in F? (is this an obvious fact or some theorem or what?)

Now this intersection (call it M') clearly still covers C, and it is comprised of open elements of F. So M' has a finite subcover since C is compact in F. With this in mind, I use what you suggested in the previous post about intersection the A's to generate the open cover on C in X. Can you just "choose" this A's by saying they exist?

10. Dec 5, 2007

Office_Shredder

Staff Emeritus
When you intersect M with F, it's open in F as:

If x is in both, then there is a ball around x contained in M inside X as M is open. Hence, given the same epsilon that works in X, any point in F within epsilon of x is contained in M, so the ball of radius epsilon in F is contained in M intersect F.

11. Dec 6, 2007

shapiro478

Office_Shredder -- I don't believe the epsilon-ball notion of open is valid here since we're working in some arbitrary topology "X", which is not necessarily the normal topology.

12. Dec 6, 2007

rs1n

This is how subspace topologies are defined. If you need a reference, check page 88 of Munkres' "Topology" (2nd ed.)

No, not quite. Look at the finite open subcover of C; it's open sets are all of the form $$M_i' = M_i\cap F$$ where $$M_i\in M$$ is an open set in X and $$1\le i \le n$$. The collection $$\{ M_i' \}$$ covers C in F, and the corresponding collection $$\{ M_i \}$$ covers C in X. You're not choosing just any $$M_i$$, you're choosing a specifically the $$M_i$$ such that $$M_i' = M_i\cap F$$. We know these $$M_i$$ exist because they are part of the open covering of C in X (and we started with this open covering).

13. Dec 7, 2007

Compactness is a "topological property". If X has property P, and Y is homeomorphic to X, then Y has property P. In this case, you are looking at $$C \subset Y \subset X$$. You can look at C in two different ways:C_Y as a subspace of Y, or C_X as a subspace of X. C_X = C_Y (as a set) are homeomorphic. The open sets are identical, so the identity is a homeomorphism.