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If no $k$-set were shattered, the Sauer-Shelah lemma would bound the number of traces by ${\sum }_{i<k}\left(\genfrac{}{}{0ex}{}{2k}{i}\right)=(2^{2k}-(\genfrac{}{}{0ex}{}{2k}{k}))/2<2^{2k-1}$. This strict middle-layer gap contradicts the displayed lower bound. Therefore a family with such traces at every even size has unbounded finite VC dimension.

The Sauer-Shelah lemma says that a hypothesis class of finite VC dimension $d$ satisfiesIn particular, for $m\ge d\ge 1$,Here the binomial coefficients count small subsets. One way to obtain the last inequality is to set $t=d/m\le 1$: since $t^j\ge t^d$ for $j\le d$,Thus finite VC dimension replaces the unrestricted exponential number of labellings by a polynomial bound in the sample size.

Fix an $m$-point set $S$. The restriction of an intersection $c_1\cap c_2$ to $S$ is determined by the ordered pair of restrictions $c_1{|}_S,c_2{|}_S$. Some pairs can yield the same intersection, so the growth function of a hypothesis class satisfiesAssume first that $d\ge 1$. If $12d$ points were shattered by $H_2$, the Sauer-Shelah lemma would implyBut $e<3$, so $(12e{)}^2<36^2=1296<4096=2^{12}$. Raising this strict inequality to the positive integer power $d$ gives a contradiction. No $12d$-point set can therefore be shattered; neither can any larger set, since shattering passes to subsets. ConsequentlyThe printed strict inequality needs $d\ge 1$. For example, on a nonempty domain let $H$ contain only the identically zero classifier. Then $\mathrm{VC}(H)=\mathrm{VC}(H_2)=0$, whereas the stated conclusion would require $0<0$. More generally, a nonempty hypothesis class of VC dimension zero has only one distinct labelling function, since two different functions would shatter a point on which they differ. Its self-intersection class also has VC dimension zero.

View a set family as a set of vertices of the Boolean lattice. The operation $D_i$ acts independently on each pair $\{S,S\cup \{i\}\}$, where $i\notin S$: it keeps a full or empty pair and moves a sole upper member down. Thus downward coordinate compression preserves cardinality.

Fix a tracing set $Y$. If $i\notin Y$, compression leaves the trace of a set family on $Y$ unchanged. If $i\in Y$, write $B_0,B_1\subseteq \mathcal{P}(Y\setminus \{i\})$ for the two sections of the old trace, according to absence or presence of $i$. The absent-$i$ section of the new trace is $B_0\cup B_1$. Its present-$i$ section is contained in $B_0\cap B_1$: an upper member survives only when its lower partner was also present in the original family. ConsequentlyIn particular, downward coordinate compression cannot create new set shattering.

Repeatedly apply any changing downward coordinate compression. Each change strictly decreases ${\sum }_{A\in \mathcal{F}}|A|$, so the process terminates at a down-set $\mathcal{D}$ with the same cardinality and with no larger trace on any $Y$. The Sauer-Shelah lemma says that a family shattering no $k$-element set satisfiesIndeed, the compressed down-set also shatters no $k$-element set. If it contained any member of size at least $k$, all subsets of a $k$-subset of that member would belong to the down-set, contradicting this restriction. Every member therefore has size at most $k-1$, giving the bound. Equivalently, VC dimension at most $d$ gives size at most ${\sum }_{j=0}^d\left(\genfrac{}{}{0ex}{}{n}{j}\right)$. The family of all sets of size at most $d$ attains this bound.

There is a different extremal example when $n\ge 3$. Toggle a fixed coordinate in the family of sets of size at most two:The symmetric difference is a bijection, so this family has $1+n+\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ members. On each tracing set $Y$, it simply toggles coordinate $1$ if $1\in Y$; thus it preserves trace cardinality and set shattering. A triple has seven trace patterns, not eight. The new family differs from the original one because it contains $\{1,2,3\}$. For $n\le 2$ the specified cardinality is $2^n$, so the entire power set is the unique family of that size; the different example requires $n\ge 3$.

For the final bound, let $e_i(\mathcal{F})$ count pairs $S,S\cup \{i\}$ both in the family. Deleting coordinate $i$ identifies precisely these pairs, soSuppose every deletion lost at least two members, so $e_i(\mathcal{F})\ge 2$ for every $i$. The trace inequality and preservation of cardinality show that $e_i(\mathcal{D})\ge e_i(\mathcal{F})$ after downward coordinate compression. The down-set $\mathcal{D}$ contains the empty set and every singleton. For each $i$, its second edge in direction $i$ has an upper endpoint of size at least two. Downward closure then supplies a pair $\{i,j\}\in \mathcal{D}$. The pairs in $\mathcal{D}$ therefore cover all $n$ coordinates, so at least $\lceil n/2\rceil$ pairs are needed. This proves two cube edges in every direction force many vertices:Taking the contrapositive gives the required answer:The threshold is sharp for every $n\ge 2$. Take the empty set, all singletons and $\lceil n/2\rceil$ pairs covering every coordinate. For even $n$, use the disjoint pairs $12,34,\dots ,(n-1)n$; for odd $n$, pair the first $n-1$ coordinates and add $(n-1)n$. This down-set has exactly $\lceil 3n/2\rceil +1$ members and at least two edges in every direction, so every $(n-1)$-coordinate trace has size at most $|\mathcal{F}|-2$. Adding further members preserves those pairs and therefore preserves the failure whenever a larger cardinality is feasible. For $n=1$, the putative counterexample cardinality is three, exceeding the size of the Boolean lattice; the sharpness construction naturally starts at $n=2$.

A set family shatters a set $Z$ when its traces on $Z$ include all of $\mathcal{P}(Z)$. The Sauer-Shelah lemma says that a set family $\mathcal{F}\subseteq \mathcal{P}([n])$ which shatters no $k$-set satisfiesEquivalently, if $|\mathcal{F}|$ exceeds this sum, it shatters a $k$-set. Binomial coefficients with index greater than $n$ are zero; if $k>n$ the inequality reduces to the whole-power-set bound. The VC dimension formulation puts $k=d+1$.

We give the full inductive proof. Split $\mathcal{F}$ according to the last coordinate, deleting that coordinate from the present section, and write these two set families on $[n-1]$ as ${\mathcal{F}}_0,{\mathcal{F}}_1$. PutThe inclusion-exclusion principle gives $|\mathcal{F}|=|\mathcal{C}|+|\mathcal{D}|$. If $\mathcal{C}$ shattered a $k$-set, then $\mathcal{F}$ would shatter it too. If $\mathcal{D}$ shattered a $(k-1)$-set $Z$, both versions of each required trace, with and without the last coordinate, would occur in $\mathcal{F}$, which would shatter $Z\cup \{n\}$. Thus the two mathematical induction bounds apply:using Pascal's identity. For $n=0$ the set family has at most one member. For $k=1$, the intersection section must be empty, since any nonempty set family shatters the empty set. These supply the mathematical induction bases; the forbidden-empty-set case has an empty set family and an empty sum. Sharpness is attained by all subsets of size at most $k-1$, which cannot shatter a $k$-set.

Now fix $k\ge 1$ in the infinite-ground-set problem. Choose, using the hypothesis with $2k$, a set $Y$ of size $2k$ with at least $2^{2k-1}$ distinct traces. If its trace set family shattered no $k$-set, the Sauer-Shelah lemma would givea contradiction. Therefore some $Z\subseteq Y$ of size $k$ is shattered by the trace set family, and hence by $\mathcal{A}$ itself. Thus every finite shattering size occurs. The use of $2k$ makes the contradiction strict, even though the hypothesis only gives a half-power-set lower bound.

An infinite realization does not follow. For a concrete unbounded finite shattering without an infinite universal trace construction, partition $X=\mathbb{N}$ into disjoint finite blocksand takeEvery member is finite. Given $k$, choose $Y=B_k$; its trace set family is the full power set, so it has $2^k\ge 2^{k-1}$ members and is shattered. However, an infinite $Z$ cannot be contained in one finite block. Choose $x,y\in Z$ from different blocks. No member of $\mathcal{A}$ contains both, so $\{x,y\}$ is not a trace on $Z$. Consequently there need not be an infinite $Z$ whose traces contain every finite subset of $Z$.

The Sauer-Shelah lemma states that if $\mathrm{VC}(\mathcal{F})=v<\infty$, thenIn particular, the Sauer-Shelah growth bound gives $s(\mathcal{F},n)\le (n+1{)}^v$.

If $x_{1:n}$ is shattered by ${\bigcup }_{k=1}^r{\mathcal{F}}_k$, then all $2^n$ label vectors belong to the union of the label-vector sets generated by the individual classes. The shattering coefficient of a union and the Sauer-Shelah lemma therefore giveTaking base-two logarithms yieldsFor $r>1$, apply the stated elementary bound with $\alpha ={\log }_{2}r$ and $\beta =v\ge 3$:If $r=1$, the assertion follows directly from $n\le v$. Taking the largest shattered sample size proves

For a binary function class $\mathcal{F}$, the shattering coefficient isIts VC dimension is the largest $n$ for which $s(\mathcal{F},n)=2^n$, with value infinity if there is no largest such $n$. The Sauer-Shelah lemma states that, when $\mathrm{VC}(\mathcal{F})=d<\infty$,

The functions $x\mapsto x^T\beta$ form a real vector space of dimension $p$. Hence the VC dimension of a vector space givesApplying the Sauer-Shelah lemma and then the Sauer-Shelah growth bound,This is the growth bound for homogeneous linear classifiers.

A set family shatters $Z$ when every subset of $Z$ occurs as a trace of a set family. The largest finite shattered-set size is its VC dimension. The Sauer-Shelah lemma bounds the number of traces when no set of a given size is shattered.