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Use binary encoding $b(S)={\sum }_{i\in S}{2}^{i-1}$ for vertices of the hypercube graph, and write $I_m=\{S:b(S)<m\}$. Let $e(\mathcal{A})$ count edges with both endpoints in a family. The exact edge-isoperimetric theorem for binary initial segments, due to Harper, Lindsey, Bernstein and Hart, iswhere $s_2(j)$ is the number of ones in the binary expansion. Equivalently,The equality for $I_m$ follows by assigning each internal edge to its larger endpoint: each one-bit of that endpoint can be cleared, giving an earlier endpoint still in $I_m$. This is the exact extremal statement, stronger than the logarithmic entropy bound.

We prove it by induction on $n$. The case $n=1$ follows by checking sizes zero, one and two. Fix a coordinate $i$ and delete it from the two sections ${\mathcal{A}}_0,{\mathcal{A}}_1$. The edge count splits asReplace each section by a binary initial segment of the same size. By induction its internal edges cannot decrease. The two replacement segments are nested, so their intersection size is the smaller section size, at least the previous intersection size. Thus this section compression in binary order cannot decrease $e(\mathcal{A})$.

Repeatedly perform any nontrivial section compression. The sum ${\sum }_{S\in \mathcal{A}}b(S)$ strictly decreases each time: in a fixed section, the original numerical order is exactly the numerical order after deleting its fixed coordinate. This nonnegative integer potential ensures termination at a family compressed in every coordinate section.

Classify such a terminal family. If an absent vertex $S$ precedes a present vertex $T$, they cannot agree in any coordinate, because their common section is compressed and its earlier vertex would then have to be present. Hence $T=[n]\setminus S$. There can be no vertex strictly between them: if present, it would also have to be the complement of $S$; if absent, it would have to be the complement of $T$. Thus they are consecutive complementary binary integers, forcingNo other inversion is possible for the same reason. Therefore a terminal family is either a binary initial segment or the sole exceptional formThis exception has size $2^{n-1}$. For $n\ge 2$, compared with the half-cube $I_{2^{n-1}}$, it removes a vertex of internal degree $n-1$ and adds a vertex joined only to the retained empty set. Its edge count is thus $e(I_{2^{n-1}})-(n-1)+1\le e(I_{2^{n-1}})$. The $n=1$ case was already handled. Since compression never decreased edges, every original family has at most the initial segment's edge count. This proves the theorem, including the exceptional terminal configuration rather than assuming all compressed families are initial segments.

For the remaining requests it is useful to obtain the entropy proof of cube edge-isoperimetry, including its equality cases. Inductively a family of size $m$ satisfiesIndeed let the two sections have sizes $a\ge b\ge 0$, $m=a+b>0$. Their internal edges are bounded inductively and the crossing edges are at most $b$, soHere $0{\log }_{2}0=0$ and $H_2(t)=-t{\log }_{2}t-(1-t){\log }_2(1-t)$ is the binary entropy function. Its concavity and its endpoint values at zero and one-half give $H_2(t)\ge 2t$ for $0\le t\le 1/2$. This proves the bound, starting from the zero-dimensional cube.

Define the isoperimetric number of a graph bySince every cube vertex has degree $n$, the internal-edge bound giveswhen $|A|\le 2^{n-1}$. A coordinate half-cube has one crossing edge per vertex, attaining ratio one. Hence, for $n\ge 1$,

Now establish the equality cases of entropy cube edge-isoperimetry. Strict concavity of $H_2$ makes $H_2(t)>2t$ for $0<t<1/2$. Equality in the internal-edge bound therefore requires either one section to be empty, or two equal-size sections. In the equal-size case equality in the crossing bound forces the sections to coincide; each section must also attain its inductive internal-edge bound. Induction gives exactly coordinate subcubes: the empty-section case fixes the coordinate, and the coincident-section case frees it. Conversely every coordinate subcube attains that bound.

At size $m=2^{n-1}$, an edge boundary of size $m$ means $e(A)=\frac{1}{2}m(n-1)=\frac{1}{2}m{\log }_{2}m$, so these equality cases apply. The subcube has dimension $n-1$ and fixes precisely one coordinate. ThereforeThese $2n$ coordinate half-cubes are the only extremizers of the requested size.