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

[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.

 

[WWGD]

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 $f(x)=x^2$.

 

[davidge]

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?

 

[Infrared]

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.

 

[WWGD]

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.

 

[davidge]

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

I see

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.

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.

 

[WWGD]

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.

 

[Infrared]

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.

 

[davidge]

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.

 

[davidge]

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

 

[Infrared]

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.

 

[davidge]

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.