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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 $0$.

MrsHaar
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  • 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? – Nate Eldredge Sep 12 '19 at 13:13
  • I meant dense $G_{\delta}$ linear subspaces. – MrsHaar Sep 12 '19 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. – Nate Eldredge Sep 12 '19 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. – MrsHaar Sep 12 '19 at 13:52
  • When you write "distinct", do you mean "disjoint"? – Goldstern Sep 12 '19 at 15:02
  • Hm. Suppose you start with a dense subspace of uncountable dimension whose complement is not Haar null. It should be possible to write it as an uncountable disjoint union of dense subspaces (after removing 0), Won't that give a counterexample? – Nate Eldredge Sep 12 '19 at 15:09
  • @NateEldredge True, in principle my intuition says the same. Though I'm having trouble doing so. – MrsHaar Sep 12 '19 at 15:15
  • @Goldstern after churning out some work, I managed to show that distinct can indeed be replaced with disjoint! – MrsHaar Sep 12 '19 at 16:54
  • I meant dense $G_{\delta}$ linear subspaces --- Do you want the subspaces to actually be dense and $G_{\delta},$ or to just contain a dense and $G_{\delta}$ set? Possibly relevant: Are proper linear subspaces of Banach spaces always meager? AND Does there exist a linearly independent and dense subset? – Dave L Renfro Sep 12 '19 at 19:55
  • @DaveLRenfro Interesting, I have indeed read the second post but the first one was new to me and very nice. As for the question, I posted all the details I know about the situation, as well as counter-examples in the case (not covered by my question) where $I$ is a singelton; I also made my assumptions as clear as possible. Thanks Dave :) – MrsHaar Sep 13 '19 at 08:19
  • @MrsHaar Please, edit your question in order to avoid contradicting conditions: dense $G_\delta$-sets in a Polish space always have non-empty intersection, so the first and third condition cannot hold simultaneously. Also for any dense $G_\delta$-sets $A,B$ in a Polish group, the sum $A+B$ is always equal $X$ (because for any $x\in X$ the dense $G_\delta$-sets $A$ and $x-B$ have non-empty intersection). So, the last condition follows automatically from the first one. – Taras Banakh Sep 13 '19 at 11:24
  • @TarasBanakh Indeed, it should have read the intersection of all the sets should be empty (not any pair; thanks for noticing that...otherwise the claim would contradict the Baire category theorem). Also, it is true that the last condition follows from the first one, I had originally put it there to make things clearer but I'll move it to the "facts" section in his case. – MrsHaar Sep 13 '19 at 12:17
  • @MrsHaar The connectedness of $X_i$ automatically follows from the connectedness of any infinite-dimensional linear subspace with removed point. So, the connectendness of $X_i$ can be removed from the assumptions. Also, the set $I$, can it be assumed to be equal to $\mathbb R\setminus{0}$? Because in the problem you ask about the complement $X\setminus\bigcup_{i\in \mathbb R\setminus {0}}X_i$, not $X\setminus\bigcup_{i\in I}X_i$. And what is the role of $\mathbb R\setminus{0}$ as the index set? Can in be any uncountable set $I$? – Taras Banakh Sep 13 '19 at 14:33
  • @MrsHaar the condition $\bigcap_{i\in I}X_i$ implies that the cardinality of $I$ should be at least $\mathrm{cov}(\mathcal M)$, the smallest cardinality of a cover of $\omega^\omega$ by meager sets. – Taras Banakh Sep 13 '19 at 14:40
  • @TarasBanakh That's interesting, why is this? Also, how can we use this fact to help establish (or disprove) the conclusion? – MrsHaar Sep 13 '19 at 14:46

1 Answers1

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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 $\{x_i\}_{i<n}$. 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.

Taras Banakh
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  • This is a great answer. Thanks Taras, very much honestly. – MrsHaar Sep 13 '19 at 15:57
  • @MrsHaar You are welcome. I simplified a bit the construction to obtain a family of open sets with the required properties. – Taras Banakh Sep 13 '19 at 16:02
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    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! – MrsHaar Sep 13 '19 at 16:09
  • @TarasBanakh Can you think of "reasonable" sufficient conditions on $X$ such that MrsHaar's claim holds? – ABIM Nov 14 '19 at 15:41
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    @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. – Taras Banakh Nov 14 '19 at 16:19
  • Oh this result I didn't know. I'll look into it. Thanks – ABIM Nov 14 '19 at 16:33