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... **


  • 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.
  • 3
    $\begingroup$ For those of us who do not have access to the paper, could you recall the definition of ($\sigma$-)porous? $\endgroup$
    – Wojowu
    Oct 9, 2019 at 12:53
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    $\begingroup$ 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. $\endgroup$
    – Wlod AA
    Oct 9, 2019 at 19:15
  • $\begingroup$ A modified conjecture: .... ...... Then $\ D\ $ contains an infinite-dimensional linear subspace. $\endgroup$
    – Wlod AA
    Oct 9, 2019 at 19:18
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    $\begingroup$ 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. $\endgroup$
    – Wlod AA
    Oct 9, 2019 at 19:24
  • $\begingroup$ Don't worry about it, the feedback was still very helpful! $\endgroup$
    – AIM
    Oct 23, 2019 at 22:45

1 Answer 1


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 $X-D$.

In fact, let me note that almost the exact same construction shows that your claim regarding Corollary 3.4 of the paper is incorrect, at least when understood in terms of the dimension of the subspace spanned. The only claims the paper makes are about the Hausdorff dimension of that set.

  • $\begingroup$ This is a very interesting/simple/illustrative counter-example. Thank you. $\endgroup$
    – AIM
    Oct 9, 2019 at 13:21
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    $\begingroup$ I always thought that $\ell_\infty$ wasn't separable... $\endgroup$
    – Asaf Karagila
    Oct 10, 2019 at 8:42

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