# Haar-null union of dense subsets

Let $$\{X_i\}_{i \in \mathbb{R}-\{0\}}$$ be a set of subsets of a separable infinite-dimensional Fréchet space $$X$$ and $$I$$ be uncountable. Moreover, suppose that

• (Dense $$G_{\delta}$$) $$X_i$$ is a dense $$G_{\delta}$$ subset of $$X$$ not containing $$0$$,
• (Almost Contains a Linear Subspace) For each $$i$$, there exists a dense linear subset $$E_i\subset X$$ satisfying $$E_i-\{0\}\subseteq X_i$$
• (Disjoint) $$\bigcap_{i \in I} X_i=\emptyset$$,
• (Not a Cover) $$\cup_{i \in I} X_i \neq X-\{0\}$$,

Can we conclude that: $$X - \bigcup_{i \in \mathbb{R}-\{0\}} X_i,$$ is Haar-null, or at-least it is finite-dimensional?

I have never seen this type of result and am pretty new to this type of thing but I ask here since it seems beyond the level of math-stack exchange.

Relevant Definitions: Haar-null set: A subset $$A\subseteq X$$ is Haar-null if there exists a Borel probability measure $$\mu$$ on $$X$$ and a Borel subset $$A\subseteq B$$ satisfying $$\mu\left( B+x \right)=0 \qquad (\forall x \in X).$$

Facts:

• I do know that $$X=X_i -X_i$$ upon applying the Baire category theorem. (Also from the comments the Pettis Lemma). This means that every element in $$X$$ can be represented as a sum of elements from each $$X_i$$.
• In the case (not covered by my question) where $$I$$ is a singleton, this paper gives a counter-example.

Intuitions:

As intuition, it can be seen here, that if $$X$$ is locally compact, then a Borel set is Haar-null if and only if it is of Haar-measure $$0$$.

• I think the question, as it currently stands, is vacuous. By the Pettis lemma, the only residual linear subspace of a Polish vector space $X$ is $X$ itself. Does this answer your question, or did you mean to write something different? Sep 12, 2019 at 13:13
• I meant dense $G_{\delta}$ linear subspaces. Sep 12, 2019 at 13:35
• It's still the same - a dense $G_\delta$ is residual. Indeed, Pettis says that in any topological vector space, a proper linear subspace having the property of Baire must be meager. Sep 12, 2019 at 13:36
• Ah, I found the bug. It is not supposed to be $G_{\delta}$ nor linear. Each $X_i$ is equal to a dense subset of $X$, which is equal to a dense linear subspace with the $0$ removed. Excuse me for the earlier conclusion. Sep 12, 2019 at 13:52
• When you write "distinct", do you mean "disjoint"? Sep 12, 2019 at 15:02

In the Frechet space $$X:=\mathbb R^\omega$$ consider the dense linear subspace $$L_0:=\{(x_n)_{n\in\omega}\in\mathbb R^\omega:|\{n\in\omega:x_n\ne0\}|<\omega\}.$$

Fix a countable base $$\{V_n\}_{n\in\omega}$$ of the topology of the space $$L_0$$ and in each set $$V_n$$ choose a point $$x_n$$, which is not contained in the linear hull of the set $$\{x_i\}_{i. Then $$\{x_n\}_{n\in\omega}$$ is a dense linearly independent set $$\{x_n\}_{n\in\omega}$$ in $$X$$. For every $$n\in\mathbb N$$ consider the linear hull $$L_n$$ of the set $$\{x_m\}_{m\ge n}$$ and observe that $$\{x_m\}_{m\ge n}$$ and $$L_n$$ are dense in $$X$$, and $$\bigcap_{n\in\omega}L_n=\{0\}$$.

Consequently, for every non-zero element $$x\in X$$ we can find a number $$n_x\in \omega$$ such that $$x\notin L_{n_x}$$.

It is easy to see that the closed convex set $$F:=[1,\infty)^\omega$$ in $$X=\mathbb R^\omega$$ is not Haar-null but is disjoint with the dense linear subspace $$L_0$$ of $$X$$.

For any $$x\in X\setminus\{0\}$$ consider the open subset $$W_x:=X\setminus(F\cup \cup\{x,0\})$$ and observe that $$L_{n_x}\setminus\{0\}\subset W_x\subset X\setminus\{x,0\}$$, which implies $$\bigcap_{x\in X\setminus \{0\}}W_x=\emptyset$$.

Also $$X\setminus \bigcup_{x\in X\setminus\{0\}}W_x\supset F$$ is not Haar-null.

So, the family of dense open (and hence $$G_\delta$$) sets $$(W_x)_{x\in X\setminus\{x\}}$$ has the properties required in the question.

• This is a great answer. Thanks Taras, very much honestly. Sep 13, 2019 at 15:57
• @MrsHaar You are welcome. I simplified a bit the construction to obtain a family of open sets with the required properties. Sep 13, 2019 at 16:02
• Excellent, both the new and old constructions are very clear. Thanks. Unfortunately this makes what I was looking for harder, but fortunately it makes it more of a challenge =more fun :) Have a great day! Sep 13, 2019 at 16:09
• @TarasBanakh Can you think of "reasonable" sufficient conditions on $X$ such that MrsHaar's claim holds? Nov 14, 2019 at 15:41
• @AIM_BLB Maybe try to inspect what happens for the Hilbert space $\ell_2$ or other reflexive Banach spaces? In $\ell_2$ the positive cone is Haar null by some result of Matouskova. Nov 14, 2019 at 16:19