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 $0$.