Once I heard that there exists two compact spaces $K_1$ and $K_2$ which are non-homeomorphic, but with $K_1\oplus K_1$ homeomorphic to $K_2\oplus K_2$ (where $\oplus$ denotes the topological sum).
Is it true?
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what is the topological sum? Do you mean disjoint union? – oxeimon May 29 '15 at 03:28
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That's interesting, I hope somebody can come up with a good answer. – Gregory Grant May 29 '15 at 03:28
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@oxeimon: The discrete union. – Brian M. Scott May 29 '15 at 03:28
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@BrianM.Scott If it means disjoint union then it seems shocking that the result is true. – Gregory Grant May 29 '15 at 03:30
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1Yes, I meant the disjoint union. – Luis Jun 06 '15 at 16:32
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K. Sundaresan, Banach spaces with Banach-Stone property, in Studies in Topology (N.M. Stavrakas & K.R. Allen, eds.), Academic Press, New York, $1975$, pp. $573$-$580$, contains an example of a compact Hausdorff space $X$ such that if $Y$ and $Z$ are the results of adding one and two isolated points, respectively, to $X$, then $X\cong Z\not\cong Y$, where $\cong$ denotes homeomorphism. Thus, $X\oplus X\cong X\oplus Z\cong Y\oplus Y$, even though $X\not\cong Y$. In On an example of Sundaresan, Topology Proceedings, Vol. $5$ ($1980$), pp. $185$-$186$, I gave a simpler proof of some of the properties of $X$; this paper is freely available here [PDF].
Briefly, $X$ is obtained by pasting together the remainders of two copies of $\beta\omega$ in the natural way. Let $X=\omega^*\cup(\omega\times 2)$, where $\omega^*=\beta\omega\setminus\omega$, and let $\pi:X\to\beta\omega$ be the obvious projection; the topology on $X$ is the coarsest making $\pi$ continuous and each point of $N=\omega\times 2$ isolated. For $i\in 2$ let $N_i=\omega\times\{i\}$. Intuitively $Y$, obtained by adding an isolated point to $X$, is not homeomorphic to $X$ because the extra point must be added to one of the ‘tails’ $N_0$ and $N_1$, and this ‘skews’ the pasting-together of the two copies of $\omega^*$ to form $X$; in $Z$, on the other hand, we can think of one of the new points as extending $N_0$ and the other, $N_1$, so that the two copies of $\omega^*$, being similarly ‘shifted’, still line up correctly. The actual argument can be found in the linked paper.
Brian M. Scott
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@Luis: An ordinal is the set of smaller ordinals, so $\omega=[0,\omega)$. You can replace $\omega$ everywhere by $\Bbb N$ if you like (bearing in mind that for me $\Bbb N$ includes $0$). – Brian M. Scott Jun 07 '15 at 02:03
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$\pi$ is such that $\pi|_{\beta\omega\setminus\omega}=id$ and $\pi((n,i))=n$, for all $n\in\omega$ and $i=0,1$. The accumulation point $\omega$ is in $\beta\omega$. We build a basis in $X$ containing: (i) the open sets of $\beta\omega\setminus\omega$ not containing $\omega$; (ii) the sets ${(k,0):k\geq n_0}\cup{(k,1):k\geq n_1}\cup W$, for each $n_0,n_1\in\omega$ and $W\subset\beta\omega\setminus\omega$ open neighborhood of $\omega$. (iii) the sets ${(n,i)}$, for each $n\in\omega$ and $i\in2$. This topology is the coarsest satisfying the properties we wish. Am I thinking correctly? – Luis Jun 07 '15 at 02:55
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@Luis: No, because there is no point of $\beta\omega$ corresponding to the ordinal $\omega$. $\beta\omega$ (or $\beta\Bbb N$) is the Čech-Stone compactification of the discrete space of natural numbers. If you're not familiar with it, you'll probably have a hard time making much sense of the example. – Brian M. Scott Jun 07 '15 at 03:12
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I do have a bit of familiarity with Cech-Stone compactification. I will be working on it. But, at least, is it $\pi$, as I described in the previous comment, the correct projection? – Luis Jun 07 '15 at 03:32
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