Boundedness and continuous functions

[orion]

I am working my way through elementary topology, and I have thought up a theorem that I am having trouble proving so any help would be greatly appreciated.
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Theorem: Let A ⊂ ℝⁿ and B ⊂ ℝ^m and let f: A → B be continuous and surjective. If A is bounded then B is bounded.
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You should not assume that A is compact.

Thanks in advance for any help.

 

[Niladri Dan]

If you think about the projective bijection between [0,1] and R then we can see that here B=R is not bounded...

 

[PeroK]

I am working my way through elementary topology, and I have thought up a theorem that I am having trouble proving so any help would be greatly appreciated.
----------------------
Theorem: Let A ⊂ ℝⁿ 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:

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

 

[orion]

Thanks. Hmm. This causes me to think because I really thought I had proven it this morning using the sequential definition of continuity and the fact that f is surjective.

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^-1to 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?

 

[micromass]

Thanks. Hmm. This causes me to think because I really thought I had proven it this morning using the sequential definition of continuity and the fact that f is surjective.

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^-1to 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

 

[orion]

Ah, ok. Now it's starting to make sense. But compactness is a topological concept. So if we come at compactness from Heine-Borel, the important addition to my theorem to make it true would be to modify it to this:

A closed and bounded ⇒ B closed and bounded.

Would that make it true (in ℝⁿ)?

 

[micromass]

Ah, ok. Now it's starting to make sense. But compactness is a topological concept. So if we come at compactness from Heine-Borel, the important addition to my theorem to make it true would be to modify it to this:

A closed and bounded ⇒ B closed and bounded.

Would that make it true (in ℝⁿ)?

Yes, that would be true.

 

[orion]

Thank you! That has been a great help!

 

[orion]

I have one other question to explore this a little further. Is there an example of a closed set that is not bounded (other than a sequence of distinct points)?

 

[micromass]

I have one other question to explore this a little further. Is there an example of a closed set that is not bounded (other than a sequence of distinct points)?

$[0,+\infty)$.

 

[orion]

Yep, that does it. Thanks, again!