Minimizing the Number of Unions

Let $\mathcal{A}=\{A_{1},...,A_{n}\}$ be a family of $k$-sets where the $A_{i}$ are distinct. We are interested in minimizing the size of the union-closed family generated by $\mathcal{A}$, namely

$$\text{\text{\textless}}\mathcal{A}\text{\textgreater}=\{A_{i_{1}}\cup A_{i_{2}}...\cup A_{i_{s}}|1\leq s\leq n,1\leq i_{1},i_{2},..,i_{s}\leq n\}.$$

It is natural to imagine that it is best to take the sets to be ‘as close together as possible’, so for example all $k$-sets from a $t$-set if the size is $\binom{t}{k}$. Note that the union-closed family generated by this family has size $|[t]^{(\geq k)}|$, where as usual $[t]$ denotes the set $\{1,2,...,t\}$ and for any set $X$ we write $X^{(\geq k)}$ for the family of all subsets of $X$ of size at least $k$. Our main aim is to prove this when $t$ is large.

This was conjectured by Roberts [7]. In fact, he conjectured a result that should hold for all sizes of $\mathcal{A}$ : we discuss this in the final section of the paper. In principle the answer could have depended on the size of the ground set, say $N$, but in fact it does not. The first possible case would be $k=2$, and for this Leck, Roberts and Simpson showed an exact result. Initial segments of colex are best (colex on $\mathbb{N}^{(k)}$ is the order in which $A<B$ if $\max(A\backslash B)<\max(B\backslash A)$ where $\mathbb{N}^{(k)}$ is the collection of all subsets of $\mathbb{N}$ of size $k$).

We give a new proof of this result. When $k\geq 3$ colex is not best. We discuss this at the end of the paper. We will use standard notation for set systems and graphs and their parameters. See Bollobás [1] for general background.

We will start by giving an overview of the proof of Theorem $1$. Let $N$ be the size of the ground set. The idea of the proof is that we will first reduce the ground set to size close to $t$. To reduce the size of the ground set we first show that it cannot be ”too big” and then we remove points belonging to few sets. If $N$ is not close to $t$ then after removing points belonging to few sets we get a set $X$ of size $N^{\prime}$ none of whose elements belong to too few of the remaining sets. Then we show that most subsets of $X$ are indeed unions. We can show that $N^{\prime}$ is close to $t$. Then unless $N$ is close to $t$ starting with $X$ we can add elements from the ground set one by one making many more unions. This will give more unions than the lower bound we are trying to prove. For the close to $t$ case, similar to above removing elements belonging to few sets we get a set $Y$ such that nearly all subsets of $Y$ are unions, but this time we will have $|Y|\geq t$. We will get too many unions unless we have both $|Y|=t$ and $Y$ being the whole ground set which is then the $[t]^{(k)}$ case.

We will first prove a few lemmas. For the following lemmas and the proof of Theorem 1 we will always use the notation that $n=\binom{t}{k}$ where $n,t,k$ are integers such that $t\geq k>1$ and that $\mathcal{A}=\{A_{1},..,A_{n}\}$ is our family of $k$-sets. We will also define $G=\cup_{i=1}^{n}A_{i}$. We may view $G$ as the ”ground set”.

The following lemma will be used to show that the ground set size cannot be too big.

Proof. We will prove a more general statement. Suppose that $l\in\mathbb{N}$ and $B_{1},..,B_{l}$ are subsets of $\mathbb{N}$ such that $|B_{i}|\leq k$ for all $i$. Let $T=\cup_{i=1}^{l}B_{i}$ and suppose $|T|\geq sk$. Finally let

$$\mathcal{F}=\{B_{i_{1}}\cup B_{i_{2}}...\cup B_{i_{r}}|1\leq r\leq l,1\leq i_{1},i_{2},..,i_{r}\leq l\}.$$

We will show that $|\mathcal{F}|\geq 2^{s}-1$. We will prove that there is some $S\subset T$ such that $|S|=s$ and for any non-empty $X\subset S$ there is some $B\in\mathcal{F}$ such that $B\cap S=X$. This will prove the above statement and hence the lemma. To show this we will induct on $s$.

For $s=1$ it is trivial as we may choose any $x\in B_{1}$ and set $S=\{x\}$.

Suppose it is true for some $s-1\geq 1$. Now consider the sets $B_{1},..,B_{l}$ and $T$. For every $x\in T$ we define $T_{x}=\{i|x\in B_{i}\}$. Pick an $x\in T$ with the smallest $|T_{x}|$. Now without loss of generality $x\in B_{1}$. Let $T^{\prime}=T\backslash B_{1}$. We have $|T^{\prime}|\geq(s-1)k$. Now consider the collection of sets

$$\mathcal{H}=\{B_{i}\cap T^{\prime}|i\not\in T_{x}\}.$$

Suppose that $y\in T^{\prime}$. We cannot have $T_{x}=T_{y}$ since $y\not\in B_{1}$ but $|T_{x}|\leq|T_{y}|$ so there must be some $i\in T_{y}\backslash T_{x}$. But then $B_{i}\cap T^{\prime}\in\mathcal{H}$ so $y\in\cup_{B\in\mathcal{H}}B$. Therefore

$$\cup_{B\in\mathcal{H}}B=T^{\prime}.$$

Now by induction there is some $S^{\prime}\subset T^{\prime}$ of size $s-1$ such that for every non-empty $X\subset S^{\prime}$ there is some $B$ which is a union of some sets in $\mathcal{H}$ such that $B\cap S^{\prime}=X$. In other words the projection of a union-closed family containing $\mathcal{H}$ onto $S^{\prime}$ contains all non-empty subsets of $S^{\prime}$. From the definition of $\mathcal{H}$ this means that there is some $B\in\mathcal{F}$ such that $B\cap S^{\prime}=X$ and $x\not\in B$. Now consider all possible unions of sets in the family

$$\{B_{i}|i\not\in T_{x}\}\cup\{B_{1}\}$$

and let $S=S^{\prime}\cup\{x\}$. We get that the projection of $\mathcal{F}$ onto $S$ contains all non-empty subsets of $S$ and $|S|=s$. This completes the induction step and the proof. ∎

We can see that $|[t]^{(\geq k)}|<2^{t}$ and so if $|\text{\text{\textless}}\mathcal{A}\text{\textgreater}|<|[t]^{(\geq k)}|$ then by Lemma 3 we must have $|G|\leq tk$. This is good as we have restricted the ground set which will make it easier to work with. The next step is showing that we can essentially remove all elements in the ground set that are in too few sets. For any $x\in G$ define $d_{x}=|\{i|x\in A_{i}\}|$. Also let

$$\mathcal{A}_{x}=\{A\subset G|x\in A,|A|\geq k,\forall i\ (x\in A_{i}\Rightarrow A_{i}\not\subset A)\}.$$

In other words $\mathcal{A}_{x}$ is the set of all subsets of $G$ of size at least $k$ for whom $x$ is one of the reasons that they are not in $\text{\textless}\mathcal{A}\text{\textgreater}$.

The following lemma will be used to show that after removing elements in too few sets, most subsets of the remaining elements are in fact in $\text{\textless}\mathcal{A}\text{\textgreater}$.

Proof. Consider the family $\mathcal{T}=\{B_{i}|x\in A_{i},B_{i}=A_{i}\backslash\{x\}\}$. We have that $|\mathcal{T}|=d_{x}$. We may assume that $G\backslash\{x\}=[p]$ for some $p$. Now we have $\mathcal{T}\subset[p]^{(k-1)}$. Notice that $\mathcal{A}_{x}$ consists of all subsets of $G$ containing $x$ that do not have a subset in $\mathcal{T}$. For $0\leq r\leq p-1$ we define the upper shadow of a family $\mathcal{F}\subset[p]^{(r)}$ to be

$$\partial_{+}\mathcal{F}=\{A\cup\{i\}|A\in\mathcal{F},i\in[p]\backslash A\}.$$

Also define $\partial_{+}^{k}\mathcal{F}=\partial_{+}\partial_{+}...\partial_{+}\mathcal{F}$ where $\partial_{+}$ is applied $k$ times ($\partial_{+}^{0}\mathcal{F}=\mathcal{F}$). For $N,r\in\mathbb{N}$ define the lex order $[N]^{(r)}$ to be the order in which $A<B$ if $\min(A\backslash B)<\min(B\backslash A)$. Now let $\mathcal{I}$ be the initial segment of lex on $[p]^{(k-1)}$ of size $|\mathcal{T}|$. For $1\leq r\leq p$ we define the lower shadow of a family $\mathcal{F}\subset[p]^{(r)}$ to be

$$\partial\mathcal{F}=\{A\backslash\{i\}|A\in\mathcal{F},i\in A\}.$$

Consider the map $g:[p]\rightarrow[p]$ given by $g(i)=p+1-i$. If $F:\mathcal{P}([p])\rightarrow\mathcal{P}([p])$ is given by $F(A)=g(A^{c})$ then $F$ is a bijection that takes any initial segment of lex into a corresponding initial segment of colex. For any $\mathcal{B}\subset[p]^{(r)}$ we have $F(\partial_{+}(\mathcal{B}))=\partial F(\mathcal{B})$. Applying the Kruskal-Katona theorem (Kruskal [5], Katona [4] and see Bollobás [1]) we see that $|\partial F(\mathcal{T})|\geq|\partial F(\mathcal{I})|$ and hence $|\partial_{+}\mathcal{T}|\geq|\partial_{+}\mathcal{I}|$.

Note that the upper shadow of an initial segment of lex is an initial segment of lex. We can now show by induction that for any $1\leq r\leq p-k+1$ we have that $|\partial_{+}^{r}\mathcal{T}|\geq|\partial_{+}^{r}\mathcal{I}|$. We have by above that this is true for $r=1$. If it is true for $r<p-k+1$ then if $\mathcal{I^{\prime}}$ is the initial segment of lex of length $|\partial_{+}^{r}\mathcal{T}|$ on $[p]^{(k-1+r)}$ then we know that $|\mathcal{I^{\prime}}|\geq|\partial_{+}^{r}\mathcal{I}|$. Also since the upper shadow of an initial segment of lex is an initial segment of lex we have that $\partial_{+}^{r}\mathcal{I}$ is an initial segment of lex on $[p]^{(k-1+r)}$ and hence $\partial_{+}^{r}\mathcal{I}\subset\mathcal{I}^{\prime}$. Now applying the Kruskal-Katona theorem

$$|\partial_{+}^{(r+1)}\mathcal{T|}\geq|\partial_{+}\mathcal{I^{\prime}}|\geq|\partial_{+}^{(r+1)}\mathcal{I}|.$$

So by induction $|\partial_{+}^{r}\mathcal{T}|\geq|\partial_{+}^{r}\mathcal{I}|$ for all $1\leq r\leq p-k+1$. Now if we define the total upper shadow of a family $\mathcal{F}\subset[p]^{(r)}$

$$U_{\mathcal{F}}=\{A\subset[p]|X\subset A\ \text{for some}\ X\in\mathcal{F}\}$$

then by above $|U_{\mathcal{T}}|\geq|U_{\mathcal{I}}|$. Notice also that $|\mathcal{A}_{x}|=|[p]^{(\geq k-1)}|-|U_{\mathcal{T}}|$. Since we want an upper bound on $|\mathcal{A}_{x}|$ we just need to show that $U_{\mathcal{I}}$ contains almost all sets in $[p]^{(\geq k-1)}$.

We know that in $[p]$ the number of $(k-1)$-sets containing a fixed element is $\binom{p-1}{k-2}$. We also know that $p=|G|-1$ So the number of $(k-1)$-sets containing at least one element from $[s]$ is at most $s\binom{|G|}{k-2}\leq d_{x}$ which also means that $s\leq p$. Note that in lex these sets are all before sets not containing any elements in $[s]$. so we must have that $\mathcal{I}$ contains all sets with at least one element from $[s]$ which means that the complement of $U_{\mathcal{I}}$ in $[p]^{(\geq k-1)}$ is a subset of $\mathcal{P}([p]\backslash[s])$. So we have

$$|\mathcal{A}_{x}|\leq|[p]^{(\geq k-1)}|-|U_{\mathcal{I}}|\leq|\mathcal{P}([p]\backslash[s])|\leq 2^{|G|-s}.$$

∎

The following will be a useful fact about getting many new unions.

Proof. Suppose that $x\in A\backslash\cup_{S\in\mathcal{H}}S$. If two sets $S_{1},S_{2}$ in $\mathcal{H}$ have $A\cup S_{1}=A\cup S_{2}$ then $S_{1}$ and $S_{2}$ can only differ on $A\cap(\cup_{S\in\mathcal{H}}S)$ which is a fixed set of size at most $|A|-1$. So at most $2^{|A|-1}$ different sets in $\mathcal{H}$ can give the same union with $A$. This means that

$$|\{A\cup S|S\in\mathcal{H}\}|\geq\frac{1}{2^{|A|-1}}|\mathcal{H}|.$$

However, no set containing $A$ is in $\mathcal{H}$ so we get the desired result. ∎

We now come to the heart of the proof.

Proof. Given $D_{k}$ suppose that $t>B$ and $|G|<t+D_{k}\log t$ where $B$ will be chosen later. Starting with $G$ and all of the $A_{i}$ we remove elements from $G$ one by one (and hence removing any of the $A_{i}$ containing those elements) if they are in less than $3\log(2tk)\binom{2t}{k-2}$ of the remaining sets $A_{i}$ until we get a $G^{\prime}$ which satisfies that no $x\in G^{\prime}$ is in less than $3\log(2tk)\binom{2t}{k-2}$ of the remaining sets. Notice that if $|G^{\prime}|\leq t-1$ then at some point in our process we were left with $G^{\prime\prime}\subset G$ such that $|G^{\prime\prime}|=t-1$. We can see that at that point we have removed at most $D_{K}\log t+1$ elements. This means that we have not removed that many sets either. In fact we have removed at most $3(D_{k}\log t+1)\log(2tk)\binom{2t}{k-2}$ sets. We will make sure that $B>3k$ to have that $\log(2tk)=\log t+\log(2k)\leq 2\log t$, $D_{k}\log t+1\leq 2D_{k}\log t$ and $2t\leq 3(t-k)$. Now we just need to make sure that

$$\frac{t-1}{\log^{2}t}>12t^{1/2}\cdot 3^{(k-2)}(k-1)D_{k}$$

which is possible as $(t-1)/t^{1/2}\geq\sqrt{t}-1=e^{\frac{1}{2}\log t}-1>\log^{3}t/48$ so we ensure that $B>e^{600\cdot 3^{(k-2)}(k-1)D_{k}}$. Now the number of sets removed when we are left with $G^{\prime\prime}$ is at most

$$3(D_{k}\log t+1)\log(2tk)\binom{2t}{k-2}\leq\frac{12\cdot 3^{(k-2)}D_{k}\log^{2}t(t-k)^{k-2}}{(k-2)!}<t^{-1/2}\binom{t-1}{k-1}.$$

This means that for large enough $B$ we could have only removed less than $\binom{t-1}{k-1}$ sets and hence when we get $G^{\prime\prime}$ we still have more than $\binom{t}{k}-\binom{t-1}{k-1}=\binom{t-1}{k}$ sets remaining. This is a contradiction as $|G^{\prime\prime}|=t-1$ and hence we have that $|G^{\prime}|\geq t$. Now we will show that if $B$ is large enough we can make sure that the vast majority of subsets of $G^{\prime}$ are indeed in $\text{\textless}\mathcal{A}\text{\textgreater}$. Define for any $x\in G^{\prime}$

$$\mathcal{A}_{x}^{\prime}=\mathcal{A}_{x}\cap\mathcal{P}(G^{\prime}).$$

From the proof of Lemma 4, we can see that the lemma is true for any $n$, not just $n=\binom{t}{k}$. We will apply Lemma 4 on $G^{\prime}$ and the remaining sets when we are left with $G^{\prime}$. For large enough $B$ we have $|G^{\prime}|\leq|G|<2t$ so if we define $d_{G^{\prime}}(x)=|\{A_{i}|x\in A_{i},A_{i}\subset G^{\prime}\}|$ we have $d_{G^{\prime}}(x)\geq 3\log(2tk)\binom{|G^{\prime}|}{k-2}$ for all $x\in G^{\prime}$. Now by Lemma 4

$$|\mathcal{A}_{x}^{\prime}|\leq 2^{|G^{\prime}|-[3\log(2tk)]}\leq 2^{|G^{\prime}|-2\log(2tk)}\leq\frac{2^{|G^{\prime}|}}{2tk}.$$

Now since every set in $G^{\prime(\geq k)}$ that is not in $\text{\textless}\mathcal{A}\text{\textgreater}$ must be in some $\mathcal{A}_{x}^{\prime}$ we have by the union bound

$\displaystyle|\text{\textless}\mathcal{A}\text{\textgreater}|\geq|G^{\prime(\geq k)}|-|\cup_{x\in G^{\prime}}\mathcal{A}_{x}^{\prime}|\geq 2^{|G^{\prime}|}-\sum_{i=0}^{k-1}\binom{|G^{\prime}|}{i}-|G^{\prime}|\frac{2^{|G^{\prime}|}}{2tk}.$

(1)

Since $|G^{\prime}|\geq t>3k$ we have $\binom{|G^{\prime}|}{i}<\frac{1}{2}\binom{|G^{\prime}|}{i+1}$ for $0\leq i\leq k-1$ because $|G^{\prime}|-i>2(i+1)$. Since exponential beats polynomial, for large enough $t$ we have $|G^{\prime}|^{k}<\frac{2^{|G^{\prime}|}}{2^{k+1}}$. We will take a large enough $B$ to have that $\sum_{i=0}^{k-1}\binom{|G^{\prime}|}{i}\leq\binom{|G^{\prime}|}{k}<|G^{\prime}|^{k}<\frac{2^{|G^{\prime}|}}{2^{k+1}}$. Now if $|G^{\prime}|>t+1$ by $(1)$ we have

$$|\text{\textless}\mathcal{A}\text{\textgreater}|\geq 2^{|G^{\prime}|}-\frac{2^{|G^{\prime}|}}{2^{k+1}}-\frac{2^{|G^{\prime}|}}{2}\geq\frac{2^{|G^{\prime}|}}{4}\geq 2^{t}$$

so we are done and if $|G^{\prime}|=t+1$ then $3|G^{\prime}|<4t\leq 2tk$ so again by $(1)$

$$|\text{\textless}\mathcal{A}\text{\textgreater}|\geq 2^{|G^{\prime}|}-\frac{2^{|G^{\prime}|}}{2^{k+1}}-\frac{2^{|G^{\prime}|}}{3}\geq\bigg{(}1-\frac{1}{8}-\frac{1}{3}\bigg{)}2^{|G^{\prime}|}>\frac{1}{2}2^{|G^{\prime}|}=2^{t}$$

and we are also done. So we may assume that $|G^{\prime}|=t$ and that $G^{\prime}=[t]$. Similar to when we considered $G^{\prime\prime}$ we have now removed at most $\frac{1}{t^{1/2}}\binom{t-1}{k-1}$ sets and since $n=\binom{t}{k}$ we know that at most $\frac{1}{t^{1/2}}\binom{t-1}{k-1}$ sets in $[t]^{(k)}$ are not in $\mathcal{A}$ so for any $x\in[t]$ we have

$$d_{G^{\prime}}(x)\geq\frac{\sqrt{t}-1}{\sqrt{t}}\binom{t-1}{k-1}\geq\frac{t-1}{2k}\binom{t-2}{k-2}\geq\lceil t^{1/2}\rceil\binom{t-2}{k-2}$$

for large enough $B$. Now by Lemma 4 on $G^{\prime}$ we have that

$$|\mathcal{A}_{x}^{\prime}|\leq 2^{t-t^{1/2}}<\frac{2^{t}}{t2^{k+1}}$$

for large enough $B$ since $t^{1/2}>\frac{\log t}{\log 2}+(k+1)$ for large enough $t$. Now we have that

$$|\cup_{x\in G^{\prime}}\mathcal{A}_{x}^{\prime}|\leq\sum_{x\in G^{\prime}}|\mathcal{A}_{x}^{\prime}|\leq\frac{2^{t}}{2^{k+1}}.$$

From before we have $\sum_{i=0}^{k-1}\binom{t}{i}\leq\frac{2^{t}}{2^{k+1}}$ and thus

$$|\text{\textless}\mathcal{A}\text{\textgreater}\cap[t]^{(\geq k)}|\geq 2^{t}-\frac{2^{t}}{2^{k+1}}-\frac{2^{t}}{2^{k+1}}=2^{t}\bigg{(}1-\frac{1}{2^{k}}\bigg{)}.$$

Now if $G\neq G^{\prime}$ there is some $A_{i}\in\mathcal{A},A_{i}\not\subset G^{\prime}$ so by Lemma 5

$$|\text{\textless}\mathcal{A}\text{\textgreater}|\geq\bigg{(}1+\frac{1}{2^{k-1}}\bigg{)}|\text{\textless}\mathcal{A}\text{\textgreater}\cap[t]^{(\geq k)}|\geq\bigg{(}1+\frac{1}{2^{k-1}}\bigg{)}2^{t}\bigg{(}1-\frac{1}{2^{k}}\bigg{)}>2^{t}$$

and we are done. If $G=G^{\prime}$ we must indeed have that $\mathcal{A}=[t]^{(k)}$ and so $|\text{\textless}\mathcal{A}\text{\textgreater}|=2^{t}-\sum_{i=0}^{k-1}\binom{t}{i}$. By the above we only have equality when $\mathcal{A}=X^{(k)}$ for some set $X$ where $|X|=t$. This proves the lemma. ∎

We now prove our main result. All we need to do is reduce the ground set size enough to apply Lemma $6$. To do this we will apply Lemmas $3,4$ and $5$.

Proof of Theorem 1. We will show that there is a $C_{k}$ such that if $t>C_{k}$ we have $|\text{\textless}\mathcal{A}\text{\textgreater}|\geq|[t]^{(\geq k)}|=2^{t}-\sum_{i=0}^{k-1}\binom{t}{i}$. If $|G|\geq tk$ then by Lemma 3 we have $|\text{\textless}\mathcal{A}\text{\textgreater}|\geq 2^{t}-1$ so we may assume that $|G|<tk$. Now let $s=2\lceil\log(2tk)\rceil$. Keep removing elements from $G$ one by one (and all of the sets $A_{i}$ containing them) until no element that is left is in less than $s\binom{tk}{k-2}$ of the remaining sets. Suppose that $G^{\prime}$ is the set of the remaining elements. We have removed at most $tks\binom{tk}{k-2}$ sets. Since $2k([t/2]+i)\geq tk$ for all $i\geq 1$ we have that

$$tk\binom{tk}{k-2}\leq\frac{(tk)^{k-1}}{(k-2)!}\leq(k-1)(2k)^{k-1}\binom{[t/2]+k}{k-1}.$$

We will make sure $C_{k}>2k$. Now let $p=2\lceil\log(2tk)\rceil(k-1)(2k)^{k-1}<3\log t(2k)^{k}$. Then we have removed at most $p\binom{[t/2]+k}{k-1}$ sets. But we have also removed at least $\binom{t}{k}-\binom{|G^{\prime}|}{k}$ sets. If $t>m\geq[t/2]+k$ then

$$\binom{t}{k}-\binom{m}{k}=\sum_{l=m}^{t-1}\binom{l}{k-1}\geq(t-m)\binom{[t/2]+k}{k-1}.$$

Since exponential beats polynomial take $C_{k}$ large enough so that

$$t-\bigg{[}\frac{t}{2}\bigg{]}-k\geq\frac{t}{3}=\frac{e^{\log t}}{3}>3\log t(2k)^{k}>p.$$

Then if $|G^{\prime}|\leq[t/2]+k$ then we have removed at least

$$\binom{t}{k}-\binom{[t/2]+k}{k}\geq(t-[t/2]-k)\binom{[t/2]+k}{k-1}>p\binom{[t/2]+k}{k-1}$$

sets which is impossible. So we must have $|G^{\prime}|\geq[t/2]+k$ and in fact if $|G^{\prime}|=t-r$ then we must have $r\leq p<3\log t(2k)^{k}$. We will first show that most subsets of $G^{\prime}$ are actually in $\text{\textless}\mathcal{A}\text{\textgreater}$.

Notice that if $x\in G^{\prime}$ then $d_{G^{\prime}}(x)\geq s\binom{tk}{k-2}\geq s\binom{|G^{\prime}|}{k-2}$. This means we can apply Lemma 4 to $G^{\prime}$. Define $\mathcal{A}_{x}^{\prime}$ as in Lemma 6. We obtain that for all $x\in G^{\prime}$ we have

$$|\mathcal{A}_{x}^{\prime}|\leq 2^{|G^{\prime}|-s}\leq\frac{2^{|G^{\prime}|}}{2^{2\log(2tk)}}\leq\frac{2^{|G^{\prime}|}}{2tk}.$$

This means that

$$|\cup_{x\in G^{\prime}}\mathcal{A}_{x}^{\prime}|\leq\sum_{x\in G^{\prime}}|\mathcal{A}_{x}^{\prime}|\leq|G^{\prime}|\frac{2^{|G^{\prime}|}}{2tk}\leq 2^{|G^{\prime}|-1}.$$

If $A\subset G^{\prime}$ and $|A|\geq k$ then $A\not\in\text{\textless}\mathcal{A}\text{\textgreater}$ if and only if there is some $x\in G^{\prime}$ such that $A\in A_{x}^{\prime}$. This means that

$\displaystyle|\text{\textless}\mathcal{A}\text{\textgreater}\cap\mathcal{P}(G^{\prime})|=2^{|G^{\prime}|}-|\cup_{x\in G^{\prime}}\mathcal{A}_{x}^{\prime}|-\sum_{i=0}^{k-1}\binom{|G^{\prime}|}{i}\geq 2^{|G^{\prime}|}-2^{|G^{\prime}|-1}-\sum_{i=0}^{k-1}\binom{|G^{\prime}|}{i}.$

(2)

We can easily bound the sum of the binomials since if $t>6k+2$ then we have $|G^{\prime}|>t/2>3k+1$ so just like in the proof of Lemma 6 we will take $C_{k}$ large enough so that $\sum_{i=0}^{k-1}\binom{|G^{\prime}|}{i}\leq\binom{|G^{\prime}|}{k}<|G^{\prime}|^{k}<2^{|G^{\prime}|-k-1}\leq 2^{|G^{\prime}|-2}$ since exponential beats polynomial. Now from (2) we have that $|\text{\textless}\mathcal{A}\text{\textgreater}\cap\mathcal{P}(G^{\prime})|\geq 2^{|G^{\prime}|-2}$. We will now show that there is a constant $D_{k}$ dependent on $k$ such that if $|G|\geq t+D_{k}\log t$ then we must have $|\text{\textless}\mathcal{A}\text{\textgreater}|\geq 2^{t}$. Let $\mathcal{H}_{0}=\text{\textless}\mathcal{A}\text{\textgreater}\cap\mathcal{P}(G^{\prime})$ and $G_{0}=G^{\prime}$. First of if $|G^{\prime}|\geq t+2$ then we are done so suppose that $|G^{\prime}|<t+2$. We will now show that we can get a lot more unions by adding sets with new elements. If we can pick some $x\in G\backslash G^{\prime}$ and a set $A_{i}$ containing $x$ then notice that by taking $\mathcal{H}_{1}=\mathcal{H}_{0}\cup\{A\cup A_{i}|A\in\mathcal{H}_{0}\}$ and $G_{1}=G_{0}\cup A_{i}$ we see that by Lemma 5

$$|\mathcal{H}_{1}|\geq\bigg{(}1+\frac{1}{2^{k-1}}\bigg{)}|\mathcal{H}_{0}|.$$

This is useful, because we have multiplied the total number of unions by a constant bigger than $1$ but dependent on $k$ and we have added at most $k$ new elements. We may keep on going as long as there are new elements and construct $\mathcal{H}_{2},\mathcal{H}_{3},...,\mathcal{H}_{q}$ and $G_{2},G_{3},...,G_{q}$ with $|H_{i}|\geq(1+\frac{1}{2^{k-1}})|H_{i-1}|$ and $|G_{i}\backslash G_{i-1}|\leq k$ for $1\leq i\leq q$ and $G_{q}=G$. This means that if there are too many elements in $G$ we will get that the total number of unions is too big. First of notice that $|G^{\prime}|=t-r>t-3\log t(2k)^{k}$. Now suppose that $|G|\geq t+2+(2k+3k\log t(2k)^{k})\log_{1+\frac{1}{2^{k-1}}}2$. This means that we can repeat the above process of adding a set and less than $k$ new elements at least $q\geq(2+3\log t(2k)^{k})\log_{1+\frac{1}{2^{k-1}}}2$ times so we have that

$$|\text{\textless}\mathcal{A}\text{\textgreater}|\geq|\mathcal{H}_{0}|\bigg{(}1+\frac{1}{2^{k-1}}\bigg{)}^{(2+3\log t(2k)^{k})\log_{1+\frac{1}{2^{k-1}}}2}\geq 2^{|G^{\prime}|-2}\cdot 2^{2+3\log t(2k)^{k}}\geq 2^{t}.$$

This means that there are constants $E_{k}$ and $D_{k}\geq 1$ dependent only on $k$ such that if $t>E_{k}$ and $|G|\geq t+D_{k}\log t$ we have that $|\text{\textless}\mathcal{A}\text{\textgreater}|\geq 2^{t}$. If $|G|<t+D_{k}\log t$ then since $D_{k}$ depends only on $k$ using Lemma 6 we get that there is a $B$ dependent only on $k$ such that if $t>B$ and $|G|<t+D_{k}\log t$ we have the desired result. Taking $C_{k}=\max(E_{k},B)$ proves the theorem. ∎

In the rest of this section we will prove Theorem 2 which solves the case $k=2$ completely. We will show that if $n\in\mathbb{N}$ and $t\geq 2$ is the smallest positive integer such that $n\leq\binom{t}{2}$ then $f(n,2)=2^{t}-2^{\binom{t}{2}-n}-t$. We note that $t=\lfloor\sqrt{2n}+\frac{3}{2}\rfloor$.