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If possible, give an explicit space $Y \subset \mathbb{R}^n$ such that $Y$is homeomorphic to the formal one-point compactification $X^* = X \cup \{ \infty \}$ (with the topology $\{U\subset X | U \text{ open in } X \} \cup \{X^* - C | C \subset X \text{ compact } \}$) of the following spaces $X$ or argue why there can not be such a space $Y \subset \mathbb{R}^n$

$a) X= [0,1)$

$b) X = (0,1)$

$c) ... h)$

I was thinking that if I could get help on understand the first part $a)$ I might be able to continue with my seven other spaces.

Bernard
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Olba12
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    In cases like these I like to think of the one point compactification as adding a piece of string to every limit point of $X\subset \mathbb R^n$ not in $X$, tying the ends of the strings together and then pulling it all tight to a point. For example/hints the answer to part a) is $Y\cong [0,1]$ and for b) $Y\cong S^1$ is a circle. Once you know what the answer is these things are easier to prove. – Dan Robertson Jul 18 '17 at 18:42
  • Im trying to understand your visualisation. If we take $X = [0,1)$ then $1$ is a limit point. Adding a piece of string? Do you mean like $1+\epsilon$ so that we end up with $[1,1+\epsilon]$? Would appreciate if you could elaborate a bit more. @DanRobertson – Olba12 Jul 18 '17 at 19:02
  • When I say add a piece of string I'm talking about adding a line not in the space (e.g. Say you embed it in $\mathbb R^3$ but I don't really want to talk about embeddings). The $[0,1)$ case isn't worth applying this method to as there is only one limit point. For $(0,1)$ you have points at $0,1$ so imagine you have a line (let's say it's black) and you attach a piece of red string to each end (where 0 and 1 are). Now thread the other ends of your red string through a needle and pull on them. This pulls your line into a circle shape and pulls 0,1 together into the same point. You get $S^1$ – Dan Robertson Jul 18 '17 at 19:47
  • Another example is to take $X$ to be an open ball in the plane. Then it's boundary is a circle. Attach pieces of string all around this boundary (say take them into a third dimension if you like) and hold all the ends together. This looks like a big cone above your circle made out of string. Now pull the string tight to pull the circle into a point and you get a sphere, the one point compactification of the open disc (which is homeomorphic to the plane) – Dan Robertson Jul 18 '17 at 19:50
  • Another way to see one point compactification is as taking the boundary of a subspace and gluing it all together into one point. – Dan Robertson Jul 18 '17 at 19:51
  • How would I think about the one point compactification of $[0,1]$ then? @DanRobertson – Olba12 Jul 18 '17 at 19:59
  • $[0,1]$ is already compact so it doesn't make much sense to talk about its one point compactification. This one point compactification would be homeomorphic to $[0,1]\cup[2]$. This can be determined from the definition – Dan Robertson Jul 18 '17 at 20:07
  • Yes I understand it doesnt make sense. But I was trying to understand what would happend if $X$ was given to be compact. By reading at https://math.stackexchange.com/questions/971130/one-point-compactification-of-a-compact-space Henno Brandsma claims that it is incompatible with the usual definition of compactification, hence my question, since I dont understand why! @DanRobertson – Olba12 Jul 18 '17 at 20:13
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    If $X$ is compact then as any closed subset of a compact space is compact we find that ${X^\ast-C|C\subset X\text{ compact}}$ is precisely ${U\cup{\infty}| U\subset X \text{ open}}$ and so the one point compactification is simply the disjoint union with a point. – Dan Robertson Jul 18 '17 at 20:16

1 Answers1

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A space $X$ has a one-point compactification $Y$ if and only if it is a locally compact Hausdorff space which is not itself compact. Furthermore, the one-point compactification is unique up to a homeomorphism. Thus, given $X$ locally compact Hausdorff (but not compact), you need to find $Y$ such that

  1. $X$ is a homeomorphic to subspace $X'$ of $Y,$
  2. $Y-X'$ is a single point,
  3. $Y$ is compact Hausdorff.

(a) $X=[0,1)$: This space is locally compact Hausdorff, so it has a one-point compactification. Note that $X$ is a subspace of $Y=[0,1],$ the set $Y-X=\{1\}$ is a single point and $Y$ is compact. Thus, up to homeomorphism, $Y$ is the one-point compactification of $X.$

(b) $X=(0,1)$: Hint. $(0,1)$ is homeomorphic to the punctured circle $S^1-\{p\},$ where $p$ is any point of $S^1.$

positron0802
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  • I have a theorem "close" to yours saying "For every locally compact Hausdorff space $X$, there is a Hausdorff space $Y$ with a subspace $X_1$ homeomorphic to $X$ and $Y - X_1 = 1$. $Y$ is unique up to homeomorphism and is called the one-pont compactification of $X$." – Olba12 Jul 18 '17 at 18:41
  • @Olba12 Yes, I edited. Usually when one says that "$X$ is a subspace of $Y$" one includes the possibility of being homeomorphic to a subspace of $Y.$ You need to find a space $Y$ which contains an homeomorphic copy $X'$ of $X,$ such that $Y$ is compact and $Y-X'$ is a single point. – positron0802 Jul 18 '17 at 18:46
  • But It doesnt say that $Y$ has to be compact? – Olba12 Jul 18 '17 at 18:47
  • Nor does it say that $X$ can not as well be compact. – Olba12 Jul 18 '17 at 18:50
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    The formal one-point compactification $X^=X \cup {\infty}$ is compact, so any $Y$ homeomorphic to $X^$ must be compact. – positron0802 Jul 18 '17 at 18:53
  • If $X$ is compact already, then $Y$ consist of a copy $X'$ of $X$ plus an isolated point $Y-X'={a}.$ – positron0802 Jul 18 '17 at 18:59
  • Why is a problem, I dont se why $X$ compact causes a problem. – Olba12 Jul 18 '17 at 19:42
  • It is not a problem, it's just less interesting. If $X$ is already compact then its one-point compactification is just merely adding to it an isolated point. While if $X$ is not compact and $Y$ is is its one-point compactification, then that point is a limit point of $X,$ so that $\overline{X}=Y.$ But you're right, is is not necessary that $X$ not be compact. – positron0802 Jul 18 '17 at 19:52
  • @Olba12. By technical definition a complactication is a dense embedding into a compact space. If S is compact, then the embedding of S into S union {oo} is not dense. – William Elliot Jul 18 '17 at 19:53
  • So there can not be a one-point compactification of $[0,1]$ because then $[0,1] \cup { \infty }$ would not be dense? @WilliamElliot – Olba12 Jul 18 '17 at 19:55
  • @positrón0802 As there is a 1-pt compactification of Q even though it is not locally compact, your claim needs to be amended. Are you talking about Hausdorf compactifications only? – William Elliot Jul 18 '17 at 19:57
  • @WilliamElliot Actually didn't know that. I only know the definition from Munkres. How should it be stated? – positron0802 Jul 18 '17 at 20:00
  • @positrón0802 The technical definition is a compactification is an embedding f:X -> Y with f(X) a dense subset of Y. – William Elliot Jul 19 '17 at 11:33
  • @Olba12. By the technical definition it is not a compactification. However the construction of a compact set, called the one point compactification, is still valid though of little significance. – William Elliot Jul 19 '17 at 11:43
  • The definition of a compactification of $S$ is a homeomorphic embedding $c:S\to T$ where $T$ is a compact Hausdorff space AND the image $c(S)$ is dense in $T. $ – DanielWainfleet Jul 20 '17 at 00:48