- #1

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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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- Thread starter deanslist1411
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- #1

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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!

- #2

cristo

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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.

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- #4

morphism

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So for future reference, it would be good if, in addition to showing your work and thoughts, you posted the problems correctly!

- #5

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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

Office_Shredder

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To start 1, write down the definition of compact, then suppose f(A) is not compact and see what you can find

- #7

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You just reposted the question? Seems somewhat unnecessary...

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