- #1

- 93

- 2

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Theorem: Let A ⊂ ℝ

^{n}and B ⊂ ℝ

^{m}and let f: A → B be continuous and surjective. If A is bounded then B is bounded.

------------------------

You should not assume that A is compact.

Thanks in advance for any help.

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- I
- Thread starter orion
- Start date

- #1

- 93

- 2

----------------------

Theorem: Let A ⊂ ℝ

------------------------

You should not assume that A is compact.

Thanks in advance for any help.

- #2

- 21

- 2

- #3

- 20,139

- 11,475

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Theorem: Let A ⊂ ℝ^{n}and B ⊂ ℝ^{m}and let f: A → B be continuous and surjective. If A is bounded then B is bounded.

------------------------

You should not assume that A is compact.

Thanks in advance for any help.

Counterexample:

##\tan : [0, \pi/2) \ \rightarrow \ [0, \infty)##

- #4

- 93

- 2

I guess I should explain what I'm trying to do. In fact, I am trying to prove compactness. I am trying to prove that a homeomorphism preserves compactness. I know the ordinary way of proving this is to take an open cover of B and using it's finite subcover and φ

So then what is it that makes the difference? Is it because a homeomorphism (as opposed to my function above) is bicontinuous or is it because it is bijective?

- #5

- 22,129

- 3,298

I guess I should explain what I'm trying to do. In fact, I am trying to prove compactness. I am trying to prove that a homeomorphism preserves compactness. I know the ordinary way of proving this is to take an open cover of B and using it's finite subcover and φ^{-1}to show that A must be compact. But I can show easily that if A is closed then B is closed and vice versa. I just want to show boundedness and then use Heine-Borel.

So then what is it that makes the difference? Is it because a homeomorphism (as opposed to my function above) is bicontinuous or is it because it is bijective?

Homeomorphisms don't preserve boundedness either. Post 3 is a homeomorphism between ##[0,2\pi)## and ##[0,+\infty)##. Succintly put: boundedness is not a topological concept

- #6

- 93

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A closed and bounded ⇒ B closed and bounded.

Would that make it true (in ℝ

- #7

- 22,129

- 3,298

A closed and bounded ⇒ B closed and bounded.

Would that make it true (in ℝ^{n})?

Yes, that would be true.

- #8

- 93

- 2

Thank you! That has been a great help!

- #9

- 93

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- #10

- 22,129

- 3,298

##[0,+\infty)##.

- #11

- 93

- 2

Yep, that does it. Thanks, again!

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