# Dense $G_{\delta}$ set with $\sigma$-porous complement is cofinite?

Let $$X$$ be a separable Banach space and $$D\subseteq X$$ be a

• proper, connected, and dense $$G_{\delta}$$ subset of $$X$$,

• $$X-D$$ is $$\sigma$$-porous.

Then is $$X-D$$ contained in a finite-dimensional subspace $$E$$ of $$X$$?

**This seems at least plausible since Corollary 3.4 of this paper shows that the codimension of a prous set in (finite-dimensions) is less than the entire space... **

Background/Definition(s)

• σ-porous: A set $$A$$ is $$\sigma$$-porous if it can be covered by a countable number of porous sets.
• porous: A set $$A$$ is porous if for each $$a \in A$$, every $$\lambda\in (0,1)$$, and every $$\epsilon>0$$, there exists some $$b \in X-A$$ satisfying $$d(a,b)< ε \mbox{ and } B(b,\lambda d(a,b)) \cap A=\emptyset.$$
• Facts: I can be shown that if $$X=\mathbb{R}^n$$ then any porous set is of Lebesgue measure $$0$$, Haar-null, and nowhere dense. However, it can be shown that (even in this finite-dimensional setting) there exist sets which are either Lebesgue measure $$0$$, Haar-null, or nowhere dense but fail to be $$\sigma$$-porous.
• For those of us who do not have access to the paper, could you recall the definition of ($\sigma$-)porous? Oct 9, 2019 at 12:53
• There are easy infinitely countable discrete subsets $\ A\subseteq\ell^2\$ which are not contained in any finite-dimensional linear subspace. Of course, each such $\ A\$ is $\sigma$-porous, while $\ D:=\ell^2\setminus A\$ is a dense connected $G_\delta$- set. This would be a counter-example. (I'll propose a modification of the OP's conjecture below). Possibly, I have missed something the big way. Oct 9, 2019 at 19:15
• A modified conjecture: .... ...... Then $\ D\$ contains an infinite-dimensional linear subspace. Oct 9, 2019 at 19:18
• My comments were a knee-jerk reaction meaning that I didn't take time to see the actual answer written by @Wojowu six hours earlier, sorry. Oct 9, 2019 at 19:24
– AIM
Oct 23, 2019 at 22:45

Let $$X=\ell_\infty$$ and let $$D$$ be the complement of the set $$\{(x_n)\in\ell_\infty:x_n\neq 0\text{ for at most one n}\}$$. This last set is clearly closed, so $$D$$ is open, in particular $$G_\delta$$. It is further clearly a proper, connected, dense subset. Let us now show that $$X-D$$ is porous, hence $$\sigma$$-porous.
Let us take $$\alpha=1/4$$ and arbitrary $$x\in X,r>0$$. The ball $$B(x,r/2)$$ necessarily contains a point $$y$$ all of whose coordinates have absolute value at least $$r/4$$. Now the ball $$B(y,r/4)$$ is contained in $$B(x,r)$$ and is clearly disjoint from $$X-D$$. Thus we conclude $$X-D$$ is porous.
Finally, $$X-D$$ is not contained in a subspace of finite dimension, since $$(1,0,0,\dots),(0,1,0,\dots),(0,0,1,\dots)$$ is an infinite linearly independent subset of $$X-D$$.
• I always thought that $\ell_\infty$ wasn't separable... Oct 10, 2019 at 8:42