Suppose $X\subseteq \mathbb R^2$ is nowhere compact ($X$ has no compact neighborhood) and non-empty.

Can $X$ be densely embedded into the plane?

In other words, is there a dense set $X'\subseteq \mathbb R ^2$ such that $X'\simeq X$?

I believe the answer is yes if $X$ is zero-dimensional. Is it still true for totally disconnected $X$?

Also I am primarily interested in Polish spaces.

  • $\begingroup$ Isn't $\mathbb Z\times\{0\}$ a counterexample? Perhaps I'm misunderstanding what you mean with "nowhere compact". $\endgroup$
    – Wojowu
    Nov 18, 2019 at 23:15
  • $\begingroup$ Not if $X$ is the $x$-axis together with all rational coordinate points below it...? $\endgroup$ Nov 18, 2019 at 23:15
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    $\begingroup$ @Wojowu nowhere compact would have to look more like the rationals; discrete spaces are locally compact $\endgroup$ Nov 18, 2019 at 23:16
  • $\begingroup$ @D.S.Lipham How does it fail your definition of nowhere compact though? $\mathbb Z\times\{0\}$ doesn't have a compact neighbourhood. Perhaps you meant no point of $X$ has a compact neighbourhood in $X$? $\endgroup$
    – Wojowu
    Nov 18, 2019 at 23:19
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    $\begingroup$ Is Erdos space homeomorphic to a dense subset of the plane? $\endgroup$ Nov 19, 2019 at 19:50

1 Answer 1


According to the definition here a topological space is nowhere compact if every compact subset has empty interior. Assuming that's that you mean by "nowhere compact", the following set $X$ seems to be a counterexample for your first question, without "totally disconnected":

$$X=\{(x,y)\in\mathbb R\times\mathbb R:x^2+y^2\le1\}\setminus\{(x,y)\in\mathbb Q\times\mathbb Q:x^2+y^2\lt1\}$$

Let $S=\{(x,y)\in\mathbb R\times\mathbb R:x^2+y^2=1\}\subset X$. If $X$ is embedded in the plane, then the image of $S$ is a simple closed curve. Since $X\setminus S$ is connected, its image is either entirely inside or entirely outside that simple closed curve. In either case, the image of $S$ is not dense in the plane.

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    $\begingroup$ BTW this is also a Polish space. $\endgroup$
    – YCor
    Nov 19, 2019 at 7:54
  • $\begingroup$ thank you. now what about for totally disconnected polish spaces? $\endgroup$ Nov 19, 2019 at 18:29
  • $\begingroup$ Totally disconnected Polish spaces can have arbitrary high (even infinite) dimension, see Theorem 3.9.3 in the book of van Mill elsevier.com/books/… $\endgroup$ Nov 19, 2019 at 19:37
  • $\begingroup$ @TarasBanakh yes but my question is: Can a totally disconnected Polish subspace of the plane be densely embedded into the plane. $\endgroup$ Nov 19, 2019 at 21:52

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