Compact, connected, closed sets

In summary, if set A is compact, then f(A) is not compact. If set A is connected, then f(A) is connected. If set B is closed, then B inverse is closed.
  • #1
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1. If set A is compact, show that f(A) is compact. Is the converse true?

2. If set A is connected, show that f(A) is connected. Is the converse true?

3. If set B is closed, show that B inverse is closed.

Any help with any or all of these three would be greatly appreciated.

Stumped!
 
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  • #2
Welcome to the forums.

Note that, for homework questions, you need to show some work before we can help you. For example, a good way to begin would be to state the definitions of compact, closed,.. and all the other terms you are using here.
 
  • #3
If you're trying to prove these for arbitrary functions, good luck. I imagine that the problem stated that f is continuous.
 
  • #4
Also, by "B inverse" you probably meant "f-1(B)", i.e. the preimage (or inverse image) of B under f.

So for future reference, it would be good if, in addition to showing your work and thoughts, you posted the problems correctly!
 
  • #5
Sorry, here goes.
1. If set A is compact, show that f(A) is compact. Is the converse true?
Ans. f:M to N is continuous and A subset M is compact. Then f(A) is compact.
The converse is not necessarilly true. For Ex: F(x)=0 for every x in R(real #'s) and
k={0}. Then f^-1(k)=R is not compact.

2. If set A is connected, show that f(A) is connected. Is the converse true?
Ans. f:M to N is continuous and A subset M is connected. Then f(A) is connected.
The converse is not necessarilly true. For Ex: F(x)=x^2 and k=1.
Then f^-1(k)={-1,1} which is not connected.

3. If set B is closed, show that B inverse is closed.
Ans. f is continuous on B if f is continuous on every x sub 0 element B.
a. f is continuous on B
b. for every x sub n to x sub 0 in A. f(x sub n) approaches f(x sub 0)
c. for any u open in N, f^-1(u) is open in M
d. for any F closed in N, f^-1(F) is closed in M.


I hope this is better. If anyone can add to this I would be greatfull.
 
  • #6
Your answers for (1) and (2) are essentially just rewording the question. I'm not sure what the answer to three is supposed to be.

To start 1, write down the definition of compact, then suppose f(A) is not compact and see what you can find
 
  • #7
You just reposted the question? Seems somewhat unnecessary...
 

1. What does it mean for a set to be compact?

A set is compact if every open cover has a finite subcover. In other words, for any collection of open sets that covers the set, there exists a finite subset of those open sets that also covers the set.

2. How is compactness related to connectedness?

A set is compact if and only if it is both closed and connected. This means that a compact set cannot be split into two disjoint open sets, and all of its limit points are contained within the set.

3. What is the significance of compact sets in mathematics?

Compact sets have many important applications in mathematics, including in the study of continuity, convergence, and topological properties. They are also useful in proving the existence of solutions to equations and in optimization problems.

4. Can a set be compact but not closed?

No, a set cannot be compact if it is not closed. A closed set contains all of its limit points, and thus any open cover of the set must also contain all of its limit points. This means that a closed set will always have a finite subcover, making it compact.

5. How can you determine if a set is compact?

One way to determine if a set is compact is to check if it is closed and bounded in a metric space. In Euclidean space, this means that the set is closed and contained within a finite region. Another way is to check if every sequence in the set has a convergent subsequence that converges to a point within the set.

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