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Let $(X,d)$ be a non-complete metric space. Let $\tilde X$ be its completion. Assume $\tilde X \setminus X$ is a finite set. (That is, $X$ has a finite number of "holes").

Is it possible for $X,\tilde X$ to be homeomorphic?

In general, a (non-complete) metric space can be homeomorphic to its completion. However, all the examples I have seen so far are with infinite number of holes. (Note that $\mathbb{R} \setminus \{0\} \times \mathbb{R} \times \mathbb{R} \times \mathbb{R} \times \dots$ actually has an infinite number of holes).

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Asaf Shachar
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1 Answers1

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Yes, this is possible. Let $\tilde X$ be any infinite-dimensional Banach space, and $X=\tilde X\setminus \{0\}$. A theorem of Klee asserts that these two spaces are homeomorphic. See

  1. V. Klee, Two topological properties of topological linear spaces, Israel J. Math. 4 (1964), 211-220.
  2. K. Goebel and J. Wośko, Making a hole in the space, Proc. Amer. Math. Soc. 114 (1992), 475-476.
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    BTW, we can even remove a $\sigma$-compact subset and still have it homeomorphic (using absorbers etc.). It's a fascinating area of infinite dimensional topology. – Henno Brandsma Jun 05 '16 at 13:57