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

1. Jul 10, 2017

davidge

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. Jul 10, 2017

Infrared

I assume you mean to write $A=[a,p]$. How do you know that $f(A)$ and $f(B)$ are of the form $(-\infty,q)$ and $(s,\infty)$, respectively?

3. Jul 10, 2017

davidge

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 $f(B)$.

4. Jul 10, 2017

Infrared

Okay, but then you can't conclude $f(A)\cap f(B)=(s,q)$.

5. Jul 10, 2017

davidge

Yes. I'm going to edit my post.

6. Jul 10, 2017

davidge

Wait, by a suitable choice of the function $f$, $f(A)$ and $f(B)$ would have those forms, wouldn't?

7. Jul 10, 2017

Infrared

You don't get to choose $f$. You have to prove that no such $f$ is a homeomorphism.

8. Jul 10, 2017

davidge

Plase, take a look at the opening post again. I have edited it.

9. Jul 10, 2017

Infrared

I think I still have the same objection. Why are $f(A)$ and $f(B)$ open intervals?

10. Jul 10, 2017

davidge

Because $\mathbb{R}$ is open, and thus it has to be the union of two open intervals?

11. Jul 10, 2017

Infrared

Having $\mathbb{R}=A\cup B$ doesn't mean that $A$ and $B$ are open. What if, say, $A=(-\infty,0]$ and $B=[0,\infty)$?

12. Jul 10, 2017

davidge

In this case, as $A$ is closed and $(- \infty, 0]$ is not, $f$ would not be a bijection. Similarly for $B$ and $f(B)$.

13. Jul 10, 2017

Infrared

$(-\infty,0]$ is a closed subset of $\mathbb{R}$.

14. Jul 10, 2017

Infrared

I have to leave now, but here are a couple hints/solution sketches in case you get stuck:

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

Alternatively, recall the following form of the intermediate value theorem: If $J\subset\mathbb{R}$ is an interval and $f:J\to\mathbb{R}$ is continuous, then $f(J)$ is an interval. It can be used as follows: $f([a,b))$ must be an interval in $\mathbb{R}$, but it is also the punctured line $\mathbb{R}\setminus\{f(b)\}$ by bijectivity. Contradiction.

15. Jul 10, 2017

davidge

Thanks for the hints

16. Jul 11, 2017

davidge

Would another way be noticing that any bijection from a closed interval maps to a closed set? (While $\mathbb{R}$ is open.)

17. Jul 11, 2017

Infrared

No, $\mathbb{R}$ 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 $f^{-1}$ is continuous when you say that $f$ takes closed sets to closed sets.

If you're familiar with compactness, you could just say $[a,b]$ is compact while $\mathbb{R}$ isn't and this would show the stronger statement that there is no continuous surjection $[a,b]\to\mathbb{R}$, but it's better to do things with your bare hands when learning this stuff.

18. Jul 11, 2017

Infrared

See my last post.

19. Jul 11, 2017

WWGD

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
20. Jul 11, 2017

Infrared

Yep, this is basically my second hint in post 14 (restating IVT in connectivity terms).

21. Jul 11, 2017

davidge

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.

22. Jul 11, 2017

Infrared

The intermediate value theorem tells you that $f([a,b))$ is an interval. Since $f$ is an injection, $f(b)\notin f([a,b))$ as otherwise we would have $f(c)=f(b)$ for some $c\in[a,b)$, contradicting injectivity. Also, for any real $y\neq f(b)$, there is a $x\in[a,b)$ with $f(x)=y$ by surjectivity. Hence, $f([a,b))=\mathbb{R}\setminus\{f(b)\}$, which is not an interval. Contradiction.

Is any step still unclear?

23. Jul 11, 2017

davidge

Oh, got it now. Thanks.

24. Jul 11, 2017

WWGD

Edited to acknowledge.

25. Jul 11, 2017

WWGD

What happens when you remove ( one- or- more of) the endpoints of $[a,b]$, is the resulting space connected? By contrast, what happens when you remove any point from the Real line; is the resulting space connected?

26. Jul 11, 2017

davidge

I don't know how to use the concept of connectness in this case, as the resulting space e.g. $[a,b)$ is half-open, and by the definition of connectness the space must be open.