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 $$ \mu\left( B+x \right)=0 \qquad (\forall x \in X). $$


  • 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.


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 $0$.

  • $\begingroup$ 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? $\endgroup$ Sep 12, 2019 at 13:13
  • $\begingroup$ I meant dense $G_{\delta}$ linear subspaces. $\endgroup$
    – MrsHaar
    Sep 12, 2019 at 13:35
  • $\begingroup$ 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. $\endgroup$ Sep 12, 2019 at 13:36
  • $\begingroup$ 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. $\endgroup$
    – MrsHaar
    Sep 12, 2019 at 13:52
  • $\begingroup$ When you write "distinct", do you mean "disjoint"? $\endgroup$
    – Goldstern
    Sep 12, 2019 at 15:02

1 Answer 1


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.

  • $\begingroup$ This is a great answer. Thanks Taras, very much honestly. $\endgroup$
    – MrsHaar
    Sep 13, 2019 at 15:57
  • $\begingroup$ @MrsHaar You are welcome. I simplified a bit the construction to obtain a family of open sets with the required properties. $\endgroup$ Sep 13, 2019 at 16:02
  • 1
    $\begingroup$ 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! $\endgroup$
    – MrsHaar
    Sep 13, 2019 at 16:09
  • $\begingroup$ @TarasBanakh Can you think of "reasonable" sufficient conditions on $X$ such that MrsHaar's claim holds? $\endgroup$
    – ABIM
    Nov 14, 2019 at 15:41
  • 1
    $\begingroup$ @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. $\endgroup$ Nov 14, 2019 at 16:19

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