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

[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.

 

[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?

 

[davidge]

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?

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 $f(B)$.

 

[Infrared]

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

 

[davidge]

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

Yes. I'm going to edit my post.

 

[davidge]

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

 

[Infrared]

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

 

[davidge]

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

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

 

[Infrared]

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

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 $\mathbb{R}$.

 

[davidge]

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

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

 

[Infrared]

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

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

 

[davidge]

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

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

 

[Infrared]

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

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

 

[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.

 

[davidge]

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.

Thanks for the hints

 

[davidge]

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

 

[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.

 

[Infrared]

Hint: Heine -Borel theorem.

See my last post.

 

[WWGD]

See my last post.

Ah, sorry. You can then use the connectivity number: removal of anyone 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.

 

[Infrared]

Ah, sorry. You can then use the connectivity number: removal of anyone point will disconnect the Real line, while the same is not the case for [a,b]. Can you see that?

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

 

[davidge]

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.

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.

Ah, sorry. You can then use the connectivity number: removal of anyone point will disconnect the Real line, while the same is not the case for [a,b]. Can you see that?

How does one show this?

$f([a,b))$ must be an interval in $\mathbb{R}$, but it is also the punctured line $\mathbb{R}\setminus\{f(b)\}$ by bijectivity. Contradiction.

Sorry, I don't see.

 

[Infrared]

Sorry, I don't see.

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?

 

[davidge]

Oh, got it now. Thanks.

 

[WWGD]

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

Edited to acknowledge.

 

[WWGD]

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.

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?

 

[davidge]

What happens when you remove ( one- or- more of) the endpoints of $[a,b]$, is the resulting space connected?

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.

 

[WWGD]

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.

Yes, but we are considering connectedness, not openness; the space remains connected.

 

[davidge]

Yes, but we are considering connectedness, not openness; the space remains connected.

What definition of connectedness are you thinking of? The one I know states that a space is disconnected if it can be expressed as the union of two open spaces, such that their intersection is empty.

 

[WWGD]

What definition of connectedness are you thinking of? The one I know states that a space is connected if it can be expressed as the union of two open spaces, such that their intersection is empty.

Well, yes, the definition I know of for connectedness is a "negative definition" , in that a space is connected if there exists no disconnection of the space. But tyou need to tighten your definitoon, otherwise, $(0,1) \cup (2,3)$ is connected.

 

[davidge]

Well, yes, the definition I know of for connectedness is a "negative definition" , in that a space is connected if there exists no disconnection of the space. But tyou need to tighten your definitoon, otherwise, $(0,1) \cup (2,3)$ is connected.

Oops, I edited my last post. I meant "disconnected" instead of "connected".

 

[davidge]

I think I have found a problem. Consider $(0, 2 \pi]$ and $[-1,1]$, and the function $\text{Sin(x)}$. There seems to be a surjection from $(0, 2 \pi]$ to $[-1,1]$.

- By compactness theorem, this can't be true, since the former is not compact while the latter is.
- By connectedness, this is true, since both are connected.

How do we solve this problem?

 

[S.G. Janssens]

By compactness theorem, this can't be true, since the former is not compact while the latter is.

What is the "compactness theorem"? What does it say?

 

[davidge]

What is the "compactness theorem"? What does it say?

Well, a half-open space requires infinite unions of closed spaces, approaching its open end element, but never reaching it. So isn't it not compact?

 

[Infrared]

I think I have found a problem. Consider $(0, 2 \pi]$ and $[-1,1]$, and the function $\text{Sin(x)}$. There seems to be a surjection from $(0, 2 \pi]$ to $[-1,1]$.

- By compactness theorem, this can't be true, since the former is not compact while the latter is.
How do we solve this problem?

I think the theorem you're talking about is that the image of a compact space under a continuous map is compact. This says nothing about when the domain is not compact. It is of course possible for a non-compact space to map surjectively onto a compact one. Take the constant function $(0,1)\to\{1\}$ for a simpler example.

 

[davidge]

I think the theorem you're talking about is that the image of a compact space under a continuous map is compact.

Yes

This says nothing about when the domain is not compact

Why not? Can you point out where I'm wrong in the following? Suppose $X$ is a non compact space and suppose we have a surjection $f$ from $X$ to another space $Y$. Then

$$X = \bigcup_{i \in I} U_i = \bigcup_{i \in F \subset I} U_i \Longleftrightarrow F = I$$

Also, $Y = f(X)$. That is, $Y$ is the image of $X$ under $f$. But $$ f(X) = f \bigg(\bigcup_{i \in I} U_i \bigg) = f \bigg(\bigcup_{i \in F \subset I} U_i \bigg)$$ with $F = I$, which means $Y$ is also not compact.