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I Boundedness and continuous functions

  1. Jul 15, 2016 #1
    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 ⊂ ℝm 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.
     
  2. jcsd
  3. Jul 15, 2016 #2
    If you think about the projective bijection between [0,1] and R then we can see that here B=R is not bounded...
     

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    Last edited by a moderator: Jul 21, 2016
  4. Jul 16, 2016 #3

    PeroK

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    Counterexample:

    ##\tan : [0, \pi/2) \ \rightarrow \ [0, \infty)##
     
  5. Jul 16, 2016 #4
    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?
     
  6. Jul 16, 2016 #5

    micromass

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    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
     
  7. Jul 16, 2016 #6
    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)?
     
  8. Jul 16, 2016 #7

    micromass

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    Yes, that would be true.
     
  9. Jul 16, 2016 #8
    Thank you! That has been a great help!
     
  10. Jul 16, 2016 #9
    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)?
     
  11. Jul 16, 2016 #10

    micromass

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    ##[0,+\infty)##.
     
  12. Jul 16, 2016 #11
    Yep, that does it. Thanks, again!
     
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