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I was reading Even-Zohar’s paper "On sums of generating sets in $\mathbb{Z}_2^n$", which I found very interesting. Here is the link to the arXiv version of the paper.

I’m particularly interested in Section 2.3, which discusses compressions of subsets of $\mathbb{Z}_2^n$. To fully understand this section, it is necessary to read Section 2.2 as well. Fortunately, that part is not very difficult, and one can gain some relevant background knowledge from my previous post here.

I’m trying to understand the proof of Theorem 2.4 (Sumset Compression), which states:

If $A, B \subseteq \mathbb{Z}_2^n$ and $I \subseteq [n]$, then $$C_I(A) + C_I(B) \subseteq C_I(A + B).$$

The proof proceeds by induction, but one part of it doesn’t make sense to me. I’ve spent a week trying to understand it but haven’t succeeded. The author claims that:

The case $n > |I|$ is implied by the case $n = |I|$:

$$ C_I(A) + C_I(B) = \bigcup_{H_c \in \mathbb{Z}_2^n / H_I} \left( (C_I(A) + C_I(B)) \cap H_c \right) $$ $$ = \bigcup_{H_c \in \mathbb{Z}_2^n / H_I} \bigcup_{H_a + H_b = H_c} \left( (C_I(A) \cap H_a) + (C_I(B) \cap H_b) \right) $$ $$ = \bigcup_{H_c \in \mathbb{Z}_2^n / H_I} \bigcup_{H_a + H_b = H_c} \left( C_I(A \cap H_a) + C_I(B \cap H_b) \right) $$ $$ \subseteq \bigcup_{H_c \in \mathbb{Z}_2^n / H_I} \bigcup_{H_a + H_b = H_c} C_I\left( (A \cap H_a) + (B \cap H_b) \right). $$

I am wondering how the last inclusion follows, namely $C_I(A \cap H_a) + C_I(B \cap H_b)\subseteq C_I((A\cap H_a)+(B\cap H_b))$. It is not at all clear to me. Has anyone else read this paper and can explain the reasoning behind this step?

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  • $\begingroup$ This is by the assumption that the inclusion in the result holds for $n=\lvert I\rvert$ (applied to a relevant coset, since e.g. $H_a$ is a coset with dimension $\lvert I\rvert$). $\endgroup$ Commented Jun 1 at 4:46
  • $\begingroup$ @ThomasBloom, Thanks a lot for your attention! The author gives the same reasoning behind this inclusion. Anyway, in this induction argument, $n$ is fixed and we only induct on $(|I|, h(A) + h(B))$, so how would the case $|I| < n$ follow from $I = [n]$? Moreover, if $|I| < n$, then $I$ may not be of the form ${1, \dots, k}$, so perhaps some hidden property of compression is also being applied here? Hopefully what I'm saying makes sense. In any case, I would greatly appreciate a detailed explanation of this step, and I'd be happy to accept your answer as it has been bothering me for a while. $\endgroup$ Commented Jun 1 at 7:17
  • $\begingroup$ @ThomasBloom, So our goal is to prove that $C_I(A \cap H_a) + C_I(B \cap H_b) \subseteq C_I\big((A \cap H_a) + (B \cap H_b)\big)$. First case: If $h(A \cap H_a) + h(B \cap H_b) < h(A) + h(B)$, then the inclusion follows by the induction hypothesis. Second case: If $h(A \cap H_a) + h(B \cap H_b) \geq h(A) + h(B)$, then $h(A \cap H_a) = h(A)$ and $h(B \cap H_b) = h(B)$, which implies that $A \subseteq H_a$ and $B \subseteq H_b$. $\endgroup$ Commented Jun 1 at 22:17
  • $\begingroup$ @ThomasBloom, By the definition of compression, the desired inclusion then reduces to $IS(|A|, H_a) + IS(|B|, H_b) \subseteq IS(|A+B|, H_a + H_b)$, and it is not clear how to prove this. This is probably different from what you suggested, which, to be honest, I didn’t understand, so if you get a chance, please write out your argument. Many thanks! $\endgroup$ Commented Jun 1 at 22:17
  • $\begingroup$ This is not the part of the proof that uses induction - the inductive part is to reduce to the case when $A,B$ are both compressed. Then the proof explains how to deduce the result when $n=\lvert I\rvert$. The final part (which you have quoted from here) explains how to deduce the result when $n>\lvert I\rvert$. $\endgroup$ Commented Jun 2 at 6:44

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