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

davidge
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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.
 
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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?
 
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Infrared said:
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)##.
 
Okay, but then you can't conclude f(A)\cap f(B)=(s,q).
 
Infrared said:
Okay, but then you can't conclude f(A)\cap f(B)=(s,q).
Yes. I'm going to edit my post.
 
Wait, by a suitable choice of the function ##f##, ##f(A)## and ##f(B)## would have those forms, wouldn't?
 
You don't get to choose f. You have to prove that no such f is a homeomorphism.
 
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Infrared said:
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.
 
I think I still have the same objection. Why are f(A) and f(B) open intervals?

davidge said:
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}##.
 
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  • #10
Infrared said:
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?
 
  • #11
davidge said:
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)?
 
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  • #12
Infrared said:
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)##.
 
  • #13
davidge said:
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}.
 
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  • #14
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.
 
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  • #15
Infrared said:
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
 
  • #16
Would another way be noticing that any bijection from a closed interval maps to a closed set? (While ##\mathbb{R}## is open.)
 
  • #17
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.
 
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  • #18
WWGD said:
Hint: Heine -Borel theorem.

See my last post.
 
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  • #19
Infrared said:
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.
 
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  • #20
WWGD said:
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).
 
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  • #21
Infrared said:
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.
WWGD said:
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?

Infrared said:
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.
 
  • #22
davidge said:
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?
 
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  • #23
Oh, got it now. Thanks.
 
  • #24
Infrared said:
Yep, this is basically my second hint in post 14 (restating IVT in connectivity terms).
Edited to acknowledge.
 
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  • #25
davidge said:
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?
 
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  • #26
WWGD said:
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.
 
  • #27
davidge said:
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.
 
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  • #28
WWGD said:
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.
 
  • #29
davidge said:
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.
 
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  • #30
WWGD said:
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".
 
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  • #31
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?
 
  • #32
davidge said:
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?
 
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  • #33
Krylov said:
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?
 
  • #34
davidge said:
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.
 
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  • #35
Infrared said:
I think the theorem you're talking about is that the image of a compact space under a continuous map is compact.
Yes
Infrared said:
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.
 
  • #36
davidge said:
Yes

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.

Like Infrared pointed out: what if ##Y##={##pt##}, a singleton? Or consider ##[-1,1)## under ##f(x)=x^2 ##.
 
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  • #37
WWGD said:
Like Infrared pointed out: what if ##Y##={##pt##}, a singleton? Or consider ##[-1,1)## under ##f(x)=x^2 ##.
Oh, I see. So what can we conclude from all this? Connectedness is a stronger condition than compactness when determining whether or not two spaces are homeomorphic?
 
  • #38
davidge said:
Connectedness is a stronger condition than compactness when determining whether or not two spaces are homeomorphic?

No, neither compactness nor connectedness implies the other. (0,1) is connected but not compact. \{0,1\} is compact but not connected.

Edit: Also, sorry but I can't understand your attempted proof that the image of a non-compact space is non-compact.
 
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  • #39
davidge said:
Oh, I see. So what can we conclude from all this? Connectedness is a stronger condition than compactness when determining whether or not two spaces are homeomorphic?
I think, if I( understood you correctly) that you are trying to argue ## (A \rightarrow B) \rightarrow (\neg A \rightarrow \neg B )## e.g., if compact is sent to compact then non-compact is mapped into non-compact, through continuous functions.
 
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  • #40
Infrared said:
No, neither compactness nor connectedness implies the other. (0,1) is connected but not compact. \{0,1\} is compact but not connected.
I see
Infrared said:
sorry but I can't understand your attempted proof that the image of a non-compact space is non-compact.
Suppose there is a surjection ##f## from a non compact space ##X## to another space ##Y##. As ##f## is a surjection, each ##y \in Y## will be the image of at least one ##x \in X##. So we can write ##Y## as the image of ##X## under ##f##, can't we? If so, it follows that ##Y## is also not compact, by the arguments shown in my post #34.
WWGD said:
I think, if I( understood you correctly) that you are trying to argue ## (A \rightarrow B) \rightarrow (\neg A \rightarrow \neg B )## e.g., if compact is sent to compact then non-compact is mapped into non-compact, through continuous functions.
Excuse me. I was actually trying to ask whether we should test for connectedness or compactness when we want to know whether two spaces are homeomorphic.
 
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  • #41
davidge said:
I see
Suppose there is a surjection ##f## from a non compact space ##X## to another space ##Y##. As ##f## is a surjection, each ##y \in Y## will be the image of at least one ##x \in X##. So we can write ##Y## as the image of ##X## under ##f##, can't we? If so, it follows that ##Y## is also not compact, by the arguments shown in my post #34.

Excuse me. I was actually trying to ask whether we should test for connectedness or compactness when we want to know whether two spaces are homeomorphic.
Ah, sorry. I can't see a general argument to be made, I just think it is more of a case-by-case basis.
 
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  • #42
davidge said:
Suppose there is a surjection ##f## from a non compact space ##X## to another space ##Y##. As ##f## is a surjection, each ##y \in Y## will be the image of at least one ##x \in X##. So we can write ##Y## as the image of ##X## under ##f##, can't we? If so, it follows that ##Y## is also not compact, by the arguments shown in my post #34.

I really don't understand this argument at all, so I'll just guess at where you're going wrong. Writing a space as an infinite union of open sets does not make a space non-compact. Not being compact would only follow if this open cover had no finite sub-cover.
 
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  • #43
Infrared said:
Writing a space as an infinite union of open sets does not make a space non-compact. Not being compact would only follow if this open cover had no finite sub-cover.
Yes, but in most cases, a open cover doesn't have a finite sub-cover only when there are infinite elements in the index set, i.e. when a infinite number of unions is needed to cover the space.
 
  • #44
WWGD said:
Ah, sorry. I can't see a general argument to be made, I just think it is more of a case-by-case basis.
Oh, ok
 
  • #45
davidge said:
Yes, but in most cases, an open cover doesn't have a finite sub-cover only when there are infinite elements in the index set, i.e. when a infinite number of unions is needed to cover the space.

I don't know what you mean by 'most' here. You're interested in the case that the image is compact, which means you can find a finite sub-cover..

In any event, this thread has drifted from the topic question. I'd suggest making a new thread.
 
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  • #46
Infrared said:
I don't know what you mean by 'most' here. You're interested in the case that the image is compact, which means you can find a finite sub-cover..

In any event, this thread has drifted from the topic question. I'd suggest making a new thread.
Ok. Thanks.
 
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