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.
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?WWGD said:Like Infrared pointed out: what if ##Y##={##pt##}, a singleton? Or consider ##[-1,1)## under ##f(x)=x^2 ##.
davidge said: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 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 seeInfrared said:No, neither compactness nor connectedness implies the other. [itex](0,1)[/itex] is connected but not compact. [itex] \{0,1\}[/itex] is compact but not connected.
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.Infrared said:sorry but I can't understand your attempted proof that the image of a non-compact space is non-compact.
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 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.
Ah, sorry. I can't see a general argument to be made, I just think it is more of a case-by-case basis.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.
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.
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.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.
Oh, okWWGD 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.
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.
Ok. Thanks.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.