Two spaces are not homeomorphic

1. Jun 3, 2010

beetle2

Prove that no continuous surjective function $f : ]0; 1] \rightarrow R$can be injective.

My questions is can I use a proof by contradiction and assume that there is a injection

Then I can use the fact that if there was an injection ie theres a bijective function then the two spaces would be homeomorphic to each other.

And then show the the two spaces are not homeomorphic so impossible?

2. Jun 3, 2010

Dickfore

Re: Homeomorphism

If there's a bijection, they would be isomorphic as far as I can tell. I think that it has something to do with the continuity and that the segment is closed.

3. Jun 3, 2010

quasar987

Re: Homeomorphism

Of course.

4. Jun 3, 2010

lavinia

Re: Homeomorphism

the inverse image of the positive integers has a limit point

5. Jun 3, 2010

beetle2

Re: Homeomorphism

Then I will try and show that both spaces are homeomorhic.

6. Jun 3, 2010

Dickfore

Re: Homeomorphism

Suppose that f is continuous and injective. Maybe it's easiest to prove that it cannot be a surjection, i.e. there is at least one real number which is not the image of any point on the segment [0, 1].

7. Jun 3, 2010

beetle2

Re: Homeomorphism

But the segment is (0.1]

8. Jun 3, 2010

Dickfore

Re: Homeomorphism

You only need one boundary point contained in the finite region, so i think you can still do it.

9. Jun 3, 2010

beetle2

Re: Homeomorphism

So If I assume that f is continuous and injective:

take
$a \in X a = 1$such that $f(a) \in Y$
And as $f$is continuous and injective I can find for any $\epsilon > 0$
a real element $y \in Y$ such that the inverse $f^{-1}(y+ \epsilon) - f^{-1}(y) + f^{-1}(y) - f^{-1}(y-\epsilon)$ is in $X$
How ever since $f^{-1}f(a)$ is a boundary point this is impossible

showing there is at leat one real number which is not the image of any point on the segment (0, 1].

10. Jun 4, 2010

quasar987

Re: Homeomorphism

I'm sorry, I read your post a little too fast. Your second sentence contains a false statement: the existence of a continuous bijective function does not imply the existence of a homeomorphism.

To solve this problem, try supposing that there is a continuous injective and surjective function f:]0,1]-->R. Then f([½,1])=[a,b] for some real numbers a and b. Suppose that f(1)=b (the case f(1)=a is similar). Show that no x>b is hit by f. In other words, $f^{-1}(]b,\infty[)=\emptyset$ and thus f is not surjective: a contradiction.

11. Jun 4, 2010

boneill3

Re: Homeomorphism

I appreciate all your help guys.

12. Jun 5, 2010

Jamma

Re: Homeomorphism

Your map can't exist in the first place; the image of a compact subset under a continuous map is compact, and the reals are not compact.

Other than that, I suppose you could also say that if your map was injective, then since bijective maps from compact spaces to Hausdorff spaces are homeomorphisms, then you would have a contradiction since [0,1] is not homeomorphic to the reals, since for example [0,1] is compact; although this is the point I made in the first paragraph.

13. Jun 6, 2010

Dickfore

Re: Homeomorphism

Prove it.

14. Jun 6, 2010

Office_Shredder

Staff Emeritus
Re: Homeomorphism

The domain isn't compact anyway (it doesn't contain 0). As for proving the reals are NOT compact, that's easy since all you need to do is come up with a counterexample to the definition.

quasar has the right idea.... basically, for there to be a continuous bijection it has to be monotone (increasing or decreasing). Then f(1) taking a finite value causes serious problems

15. Jun 6, 2010

Landau

Re: Homeomorphism

A subset of R^n is compact iff it is closed and bounded. ]0,1] is not compact because it is not closed. R is not compact because it is not bounded.

16. Jun 7, 2010

lavinia

Re: Homeomorphism

the negative integers -m map to an infinite decreasing or increasing sequence in (0,1] that either has a limit point or converges to 0. If there is a limit point the inverse can not be a homeomorphism. but then the positive integers must have a limit point.

17. Jun 14, 2010

Bacle

Re: Homeomorphism

Maybe you can use the fact that a continuous bijection between [0,1] and R
would be (is) a homeo. (cont. bijection, compact, Hausd. , yada yada)

Then, if h is that homeo., the restriction:

h<sup>^</sup>:(0,1]-->R-{f(0)} is also a homeo.

but R-{f(0)} is disconnected, and (0,1] (as a subspace) is not.

18. Jun 14, 2010

quasar987

Re: Homeomorphism

This looks like an argument against the existence of a homeomorphism between R and [0,1] (which is direct by compactness by the way). The OP was trying to find an argument against the existence of a homeo between R and (0,1].

19. Jun 14, 2010

Bacle

Re: Homeomorphism

I was not claiming this is a solution; just mentioning
a result that seemed closed to what the OP was asking
for; I thought it may help give him an idea.

20. Jun 19, 2010

Jamma

Re: Homeomorphism

Sorry, I thought that the ]0,1] was a typo to mean [0,1], in which case I'd obviously have been correct. I've never seen this (very weird) notation, and would use (0,1].

Looking at it though, it does make a bit of sense, its like the brackets are indicating visually where the boundaries lie on the real line, and it can be annoying having things like (0,1) around when there are also 2-tuples.

21. Jun 19, 2010

quasar987

Re: Homeomorphism

The notation is universal in french books. I've seen it on rare occasions in english books, perhaps by authors trying to popularize it.

22. Jun 19, 2010

some_dude

Re: Homeomorphism

Since f is continuous, the preimage $$A = f^{-1}((f(1), +\infty))$$ must be open in (0, 1]. Ditto for $$B = f^{-1}((-\infty, f(1))$$.

Then $$(0, 1) \subset A \cup B$$ and the surjectivity assures neither A nor B are empty. And this along with their openness implies $$A \cap B \ne \emptyset$$, which implies $$A \cup B$$ is connected, and so $$f(A \cup B)$$ is connected (and equal the whole real line), and therefore $$f(1) \in f(A \cup B)$$. So f is not injective.

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