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I Proving non homeomorphism between a closed interval & ##\mathbb{R}##

  1. Jul 10, 2017 #1
    I was trying to show that a closed interval ##[a,b]## and ##\mathbb{R}## cannot be homeomorphic. I would like to know whether this can actually be considered as a proof. It is the following:

    - The closed interval ##[a,b]## can be written as ##[a,p] \cup [p,b]##, where ##a \leq p \leq b##.
    - ##\mathbb{R}## can be written as ##(- \infty, q) \cup (s, \infty)##, where ##s < q##.

    Let ##[a, b] = A## and ##[p,b] = B##.
    If there is a homeomorphism ##f## from ##[a, b]## to ##\mathbb{R}##, then

    - ##\mathbb{R} = f(A) \cup f(B)##

    Each point on ##f(A) \cap f(B)## is the image of one, and only one, point which is in both ##A## and ##B##. Considering the extreme case, there will be only one point on ##A \cap B##, namely ##\text{{p}}##. On the other hand, ##f(A) \cap f(B)## will have more than one point (possibly infinite points) as it is the intersection of two open intervals ##f(A)## and ##f(B)## whose union is ##\mathbb{R}##.
    So ##f## cannot be an injection, which contradicts ##f## being a homeomorphism.
     
    Last edited: Jul 10, 2017
  2. jcsd
  3. Jul 10, 2017 #2
    I assume you mean to write [itex]A=[a,p][/itex]. How do you know that [itex]f(A)[/itex] and [itex]f(B)[/itex] are of the form [itex](-\infty,q)[/itex] and [itex](s,\infty)[/itex], respectively?
     
  4. Jul 10, 2017 #3
    You are correct. I should have only said that ##\mathbb{R} = f(A) \cup f(B)##. We don't know the form of ##f(A)## nor ##f(B)##.
     
  5. Jul 10, 2017 #4
    Okay, but then you can't conclude [itex]f(A)\cap f(B)=(s,q)[/itex].
     
  6. Jul 10, 2017 #5
    Yes. I'm going to edit my post.
     
  7. Jul 10, 2017 #6
    Wait, by a suitable choice of the function ##f##, ##f(A)## and ##f(B)## would have those forms, wouldn't?
     
  8. Jul 10, 2017 #7
    You don't get to choose [itex]f[/itex]. You have to prove that no such [itex]f[/itex] is a homeomorphism.
     
  9. Jul 10, 2017 #8
    Plase, take a look at the opening post again. I have edited it.
     
  10. Jul 10, 2017 #9
    I think I still have the same objection. Why are [itex]f(A)[/itex] and [itex]f(B)[/itex] open intervals?

     
  11. Jul 10, 2017 #10
    Because ##\mathbb{R}## is open, and thus it has to be the union of two open intervals?
     
  12. Jul 10, 2017 #11
    Having [itex]\mathbb{R}=A\cup B[/itex] doesn't mean that [itex]A[/itex] and [itex]B[/itex] are open. What if, say, [itex]A=(-\infty,0][/itex] and [itex]B=[0,\infty)[/itex]?
     
  13. Jul 10, 2017 #12
    In this case, as ##A## is closed and ##(- \infty, 0]## is not, ##f## would not be a bijection. Similarly for ##B## and ##f(B)##.
     
  14. Jul 10, 2017 #13
    [itex](-\infty,0][/itex] is a closed subset of [itex]\mathbb{R}[/itex].
     
  15. Jul 10, 2017 #14
    I have to leave now, but here are a couple hints/solution sketches in case you get stuck:

    Show that a continuous injection [itex] f:[a,b]\to\mathbb{R} [/itex] has to be (strictly) monotonic. Examine [itex]f(a)[/itex] to violate surjectivity.

    Alternatively, recall the following form of the intermediate value theorem: If [itex]J\subset\mathbb{R}[/itex] is an interval and [itex]f:J\to\mathbb{R}[/itex] is continuous, then [itex]f(J)[/itex] is an interval. It can be used as follows: [itex]f([a,b))[/itex] must be an interval in [itex]\mathbb{R}[/itex], but it is also the punctured line [itex]\mathbb{R}\setminus\{f(b)\}[/itex] by bijectivity. Contradiction.
     
  16. Jul 10, 2017 #15
    Thanks for the hints
     
  17. Jul 11, 2017 #16
    Would another way be noticing that any bijection from a closed interval maps to a closed set? (While ##\mathbb{R}## is open.)
     
  18. Jul 11, 2017 #16
    No, [itex]\mathbb{R}[/itex] is closed too, as a subspace of itself (open does not imply not closed). Also, mere bijections don't preserve openness/closedness- you're using the fact that [itex]f^{-1}[/itex] is continuous when you say that [itex]f[/itex] takes closed sets to closed sets.

    If you're familiar with compactness, you could just say [itex] [a,b][/itex] is compact while [itex]\mathbb{R}[/itex] isn't and this would show the stronger statement that there is no continuous surjection [itex][a,b]\to\mathbb{R}[/itex], but it's better to do things with your bare hands when learning this stuff.
     
  19. Jul 11, 2017 #17
    See my last post.
     
  20. Jul 11, 2017 #18

    WWGD

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    Ah, sorry. You can then use the connectivity number: removal of any one point will disconnect the Real line, while the same is not the case for [a,b]. Can you see that?EDIT: along the lines of post 14, consider this and the Euler number.
     
    Last edited: Jul 11, 2017
  21. Jul 11, 2017 #19
    Yep, this is basically my second hint in post 14 (restating IVT in connectivity terms).
     
  22. Jul 11, 2017 #20
    Yes, I find it more easy to show they are not homeomorphic by arguments of compactness. But the thing is that I want to prove it without using compactness.
    How does one show this?

    Sorry, I don't see.
     
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