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Question. Is there a topological space $X$ with $X \cong X+2$ and $X \not\cong X+1$?

Here, $X+n$ denotes the disjoint union (i.e. coproduct) of $X$ with $n$ isolated points.

This question is similar to MO/218113 and MO/225896. I am pretty sure that it is easier, though. Perhaps it already works with a nasty topology on $\mathbb{N}$?

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Martin Brandenburg
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1 Answers1

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A reference to such a space and a brief description can be found in this answer; there is a more thorough description in this answer. Briefly, the space is obtained by taking two copies of $\beta\Bbb N$, the Čech-Stone compactification of $\Bbb N$, and identifying the remainders in the obvious way.

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Brian M. Scott
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  • @Martin: You’re welcome. I’m not familiar with the $\cup_\infty$ notation, I’m afraid. – Brian M. Scott Dec 15 '15 at 20:08
  • Doesn't the same argument work the Alexandroff compactification $\mathbb{N}^+$ and the pushout (aka gluing) $X=\mathbb{N}^+ \sqcup_{\infty} \mathbb{N}^+$? But adding one isolated point doesn't change this space, in my opinion: We can use $1+\mathbb{N} \cong \mathbb{N}$. – Martin Brandenburg Dec 15 '15 at 20:10
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    @Martin: You mean two disjoint simple sequences with the same limit? No, that won’t work, because you don’t have preservation of pairs as in the lemma at the answer linked above. In other words, you’re right: in this case adding one point doesn’t change the space. – Brian M. Scott Dec 15 '15 at 20:14
  • But I can use the same homeomorphism for $X=\beta\mathbb{N} \cup_{\beta\mathbb{N} \setminus \mathbb{N}} \beta\mathbb{N}$. – Martin Brandenburg Dec 15 '15 at 20:15
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    @Martin: No, you can’t. Work through the argument at the second link. – Brian M. Scott Dec 15 '15 at 20:15
  • Ok. Let's apply Stone duality. $X$ is the spectrum of the boolean ring $R={(a,b) \in \mathbb{F}2^{\mathbb{N}} \times \mathbb{F}_2^{\mathbb{N}} : a_n = b_n \text{ for almost all } n}$, and $X+1$ is the spectrum of $R \times \mathbb{F}_2$. Now, there is at least no obvious isomorphism $R \times \mathbb{F}_2 \cong R$. However, $\mathbb{N}^+ \sqcup{\infty} \mathbb{N}^+$ is the spectrum of the subring $R' \subseteq R$ consisting of those $(a,b)$, where $a,b$ are eventually constant (the value at $\infty$). And this offers an isomorphism $R' \cong R' \times \mathbb{F}_2$. – Martin Brandenburg Dec 15 '15 at 20:26
  • @Martin: I’m sorry, but you’re simply not speaking my language. – Brian M. Scott Dec 15 '15 at 20:28
  • Well, I'm an algebraic geometer / algebraist, so I like to think in terms of commutative rings. I just wanted to reformulate what you have said, in order to get some intuition why $X+1 \cong X$ fails. – Martin Brandenburg Dec 15 '15 at 20:29
  • @Martin: Ah, okay. Clearly we’ve very different kinds of intuition here! (Not really surprising under the circumstances.) – Brian M. Scott Dec 15 '15 at 20:33