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We say that a topological space $(X,\tau)$ is point-removal insensitive if for all $x\in X$ we have $X\cong X\setminus \{x\}$.

If $X,Y$ are point-removal insensitive, does this imply that $X\times Y$ with the product topology is point-removal insensitive?

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    $\begingroup$ Maybe call these spaces "puncturable" or "holeable" (or "punctured"?). Just thinking out loud.. $\endgroup$
    – Henno Brandsma
    Commented Jun 11, 2018 at 22:16
  • $\begingroup$ You might look into Max Pitz's papers: arxiv.org/search/?searchtype=author&query=Pitz%2C+M. He has done quite a bit of work with homeomorphism types of one-point deleted subsets. I think he uses the phrase "deck of cards" to describe (the homeomorphism classes of) all subspaces $X\setminus \{x\}$. $\endgroup$
    – Forever Mozart
    Commented Jun 12, 2018 at 0:00

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Let $X = {\mathbb R} \setminus {\mathbb Z}$, that is, $X$ is the free union of countably many open intervals. Then $X$ is point-removable insensitive because the removal of any point leaves a free union of countably many open intervals. However, $X \times X$ is not point-removal insensitive because every component of $X \times X$ just a product of intervals and is, therefore, simply connected, but if any point is removed from $X \times X$, that point's component is not simply connected.

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