4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Isn't $\mathbb Z\times\{0\}$ a counterexample? Perhaps I'm misunderstanding what you mean with "nowhere compact". $\endgroup$ – Wojowu Nov 18 '19 at 23:15
  • $\begingroup$ Not if $X$ is the $x$-axis together with all rational coordinate points below it...? $\endgroup$ – Bjørn Kjos-Hanssen Nov 18 '19 at 23:15
  • 1
    $\begingroup$ @Wojowu nowhere compact would have to look more like the rationals; discrete spaces are locally compact $\endgroup$ – D.S. Lipham Nov 18 '19 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 '19 at 23:19
  • 1
    $\begingroup$ Is Erdos space homeomorphic to a dense subset of the plane? $\endgroup$ – Ramiro de la Vega Nov 19 '19 at 19:50
5
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ BTW this is also a Polish space. $\endgroup$ – YCor Nov 19 '19 at 7:54
  • $\begingroup$ thank you. now what about for totally disconnected polish spaces? $\endgroup$ – D.S. Lipham Nov 19 '19 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/… $\endgroup$ – Taras Banakh Nov 19 '19 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. $\endgroup$ – D.S. Lipham Nov 19 '19 at 21:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.