# Nowhere compact subsets of the plane

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.

• Isn't $\mathbb Z\times\{0\}$ a counterexample? Perhaps I'm misunderstanding what you mean with "nowhere compact". – Wojowu Nov 18 '19 at 23:15
• Not if $X$ is the $x$-axis together with all rational coordinate points below it...? – Bjørn Kjos-Hanssen Nov 18 '19 at 23:15
• @Wojowu nowhere compact would have to look more like the rationals; discrete spaces are locally compact – D.S. Lipham Nov 18 '19 at 23:16
• @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 $X$? – Wojowu Nov 18 '19 at 23:19
• Is Erdos space homeomorphic to a dense subset of the plane? – Ramiro de la Vega Nov 19 '19 at 19:50

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.