# 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.