I know that Alexandroff compatification is unique, and if the Alexandroff compatification of two spaces are not homeomorphic, then the spaces can't be. Does uniqueness stand in n point compatifications? And what does homeomorphism (or not) between the compatifications tells us about the original spaces? Finally, if A has a 2 points compatification, and B doesn't (maybe has 1 point compatification) can we say A and B are not homeomorphic?
N-points compatification
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general-topology
compactness
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0@BenjaLim: See the above comment for (much belated) clarification. – 2013-06-13
1 Answers
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Uniqueness does not hold: for example, $(0,1)\cup (2,3)$ can be 2-point compactified into a circle (by adding $0=3$ and $1=2$) or into two disjoint circles (by adding $0=1$ and $2=3$).
However this is of no consequence for topological invariance. As long as a property is defined in terms of topology on $X$, it is invariant under homeomorphism. If you wish to make this more precise, you can rephrase the definition:
$X$ having an $n$-point compactification means that $X$ is homeomorphic to some space of the form $Y\setminus \{y_1,\dots,y_n\}$ where $Y$ is compact Hausdorff and $y_i\in Y$ are distinct.