I asked the following on MSE a few weeks ago but I did not get any answer :
<http://math.stackexchange.com/questions/1039593/carlsons-translatability-are-theses-characterisations-equivalent>


Given a translation-invariant ideal $\mathcal{I}$ on a commutative group $G$ and it's dual filter $\mathcal{I}^*$, why is
$$ (\forall I \in \mathcal{I})(\exists I' \in \mathcal{I})(\forall g_1, g_2 \in G)(\exists h \in G)\bigg((I+g_1)\cup( I+g_2) \subseteq ( I' + h)\bigg)$$
equivalent to
$$(\forall F \in \mathcal{I}^*)(\exists F' \in \mathcal{I}^*)(\forall g_1, g_2 \in G)(\exists h \in G)\bigg((F'+g_1)\cup( F'+g_2) \subseteq ( F + h)\bigg).$$



Reference : "strong measure zero and strongly meager sets", Timothy Carlson (p.582).


**Edit**

Suppose $\kappa$ is a cardinal number and $\mathcal{U}$ is a set of subsets of a group $G$. $\mathcal{U}$ is $\kappa$-translatable if for every $A$ in $\mathcal{U}$ there is a $B$ in $\mathcal{U}$ such that the union of any $\kappa$ many translates of $B$ is contained in a translate of $A$.


Here is what Carlson says : 

> Notice that if $\mathcal{U}$ is a filter then $\mathcal{U}$ is $\kappa$-translatable iff for every $A$ in the dual ideal there is a $B$ in the dual ideal such that the intersection of any $\kappa$ translates of $B$ contains a translate of $A$. Hence, $\mathcal{U}$ is $\kappa$-translatable iff for every $A$ in the dual ideal there is $B$ in the dual ideal such that the union of any $\kappa$ translates of $A$ is contained in a translate of $B$.


The second iff is a bit obscur to me.

This result is used a two or three articles but none of them give more precision on why this result should be true. I might be splitting hair here, but Carlson's result is not transparently clear to me.