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I came across a kind of separable metrizable space that is "far" from being completely metrizable. Before specifying what I mean with "far", I recall that a space is said to be Polish if it's separable and completely metrizable and that $2^\mathbb{N}$, known as the Cantor space, is the space of infinite binary sequences endowed with the product of the discrete topology on $2$.
Here is the definion of being strongly nowhere Polish spaces (I came up with the name)

A separable metrizable space $X$ is said to be strongly nowhere Polish if for every $d$ compatible metric on $X$, for every $U$ nonempty open subset of the completion of $(X,d)$ we have $2^\mathbb{N} \hookrightarrow U\setminus X$, i.e. the Cantor space embeds into $U\setminus X$

Here are my questions:

  1. Is this notion known to some extent? Has is been studied?
  2. Is $\mathbb{Q}$ with its usual topology strongly nowhere Polish?
  3. Is there an uncountable separable metrizable space which is nowhere Polish?
  4. Is there a separable metrizable space which is not Polish yet not strongly nowhere Polish?
  5. Are there interesting properties of these spaces that you can think of?

Thanks!

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    $\begingroup$ 4 is trivial to cook up using a disjoint union (e.g. $\mathbf{Q}$ + isolated point) $\endgroup$
    – YCor
    Mar 30, 2022 at 10:14
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    $\begingroup$ For (3), one example is $\mathbb Q + C$, where $C \subseteq \mathbb R$ is the Cantor set. This is nowhere completely metrizable (no open subset is a relative $G_\delta$, by Baire's theorem), but is $\sigma$-compact. So it is an $F_\sigma$ subspace of any completely metrizable space containing it, which means its complement is completely metrizable. (And the same argument gives a positive answer for question (2).) $\endgroup$
    – Will Brian
    Mar 30, 2022 at 12:22
  • $\begingroup$ @WillBrian by $\mathbb{Q}+C$ you mean disjoint union? $\endgroup$
    – Lorenzo
    Mar 30, 2022 at 12:42
  • $\begingroup$ @Lorenzo: No, I mean the union of all rational shifts of $C$. This won't be a disjoint union; it will give you a dense, $\sigma$-compact, meager subset of $\mathbb R$. $\endgroup$
    – Will Brian
    Mar 30, 2022 at 12:50

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