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Let $X$ be a compact Hausdorff space. Show that the cone in $X$ is homeomorphic to the compactification of $X \times [0,1)$. If $A$ is closed in $X$, show that $X/A$ is homeomorphic to the compactification of $X\setminus A$.

I don't know even how to begin to solve this question, it's seems so hard, anyone could help me, please.

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    To the one point compactification, you mean; there are many compactifcations!2012-09-30
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    Try to define a map $\mathrm{Cone}(X)\to \mathrm{Compatification}(X\times[0,1))$ There are not many natural ways to do this... in fact, there is pretty much only one—and that one will do the trick. Similarly for the other problem: there is exactly *one* reasonable map you can define, and it is the homeo you are looking for. When you have the correct maps figured out, you will need to prove they are homeos, of course: for that, you will need to use the fact that your space is compact and Hausdorff, because for that sort of...2012-09-30
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    ...spaces there is a simple way of checking if a map is an homeo.2012-09-30
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    yes, I think the one point compactification is better in this case. Speaking of the map, I know that, the problem is that, I don't know how to find this map, I new on this topic (compactifications).2012-09-30
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    It is not that «the one point compactification is better in this case»: it is the only one which will make the statements true!2012-10-01

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Here’s a start on the second problem; you should be able to use some of the ideas to help you with the first, as well.

Suppose that $X$ is a compact Hausdorff space, and $A\subseteq X$ is closed. Let $Y=X\setminus A$, and let $Y^*=Y\cup\{p\}$ be the one-point compactification of $Y$. Finally, let $Z=X/A$, let $q:X\to Z$ be the quotient map, and let $a\in Z$ be the point corresponding to $A$ in $X$. You want to prove that $Y^*$ is homeomorphic to $Z$.

The first step is figure out what the homeomorphism should be. $Z=q[Y]\cup\{a\}$, where $q[Y]$ is a homeomorphic copy of $Y$, and $Y^*=Y\cup\{p\}$, where $Y$ is a homeomorphic ‘copy’ of $Y$ sitting inside $Y^*$. Thus, each of the spaces $Z$ and $Y^*$ consists of a copy of $Y$ together with one extra point. This suggests that we should try the function

$$h:Y^*\to Z:y\mapsto\begin{cases} q(y),&\text{if }y\in Y\\ a,&\text{if }y=p\;, \end{cases}$$

which sends each point of $Y$ to its copy in $Z$ and sends the extra point $p$ in $Y^*$ to the extra point $a$ in $Z$. Now you just have to check that this $h$ really is a homeomorphism.

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    I'm struggling to prove the continuity of h. It's seems I have to create another map from X to Y*.2012-09-30
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    @user42912, to prove continuity, simply show that the preimage of an open set is open!2012-10-01
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    @user42912: As Mariano said, just show that if $U\subseteq Z$ is open, then $g^{-1}[U]$ is open in $Y^*$. There are just two cases: $a\in U$, and $a\notin U$. If $a\notin U$, then $h^{-1}[U]=q^{-1}[U]$, and you know that $q$ is continuous. If $a\in U$, use what you know about the quotient map $q$ and what you know about the nbhds of $p$ in the one-point compactification.2012-10-01
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    @Bryan To prove continuity, let U/A a neighborhood of a in Z, so h-¹(u)= V, where v = (U-A) ∪ {p}, So I have to proof that V is open in the topology of Y*,i.e., V-{p} is open in X-A and its complement is compact in X-A, i.e., (X-A)-V is compact in X-A. Am I right? if yes I couldn't proof the compacity part.2012-10-03
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    @user42912: Almost right. You know that $V\setminus\{p\}=U\setminus A$ is open in $X$, because $A$ is closed and $U$ is open. Now you have to show that the complement of $V\setminus\{p\}$ is compact, but that’s easy: $V\setminus\{p\}$ is open in $X$, so its complement is closed in $X$, and every closed set in a compact Hausdorff space is compact.2012-10-03
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    but we have to prove that the complement of V\{p} is compact in X\A, you proved in x.2012-10-04
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    @user42912: It’s the same thing: $Y$ is open in $X$, so any open cover in $Y$ of a subset of $Y$ is also an open cover in $X$ of that same subset of $Y$. It’s compact in $X$ if and only if it’s compact in $Y$.2012-10-04
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    @BrianM.Scott Thank you, I think I solved the item b, one question, can I use this item to prove directly the item a?2012-10-06
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    @user42912: Yes, you can; just pick the right set for $A$.2012-10-06
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    @BrianM.Scott can I do the cone like this: CX = $X\times [0,1]/X\times \{1\}$ where A is $X\times \{1\}$2012-10-06
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    @user42912: Yes, that’s the right $A$.2012-10-06