Here is a nice example due to Bjorn Poonen, which I have taken from this survey paper of Bruhn and Schaudt. It is motivated by the following observations. Let $\mathcal{A}$ be a union-closed family. Call an element $x$ abundant if $x$ is contained in at least half of the members of $\mathcal{A}$.

Observation. If $\mathcal{A}$ contains a singleton $\{x\}$, then $x$ is abundant.

Proof. Partition $\mathcal{A}$ as $\mathcal{A}_x$ and $\overline{A}_x$ where $\mathcal{A}_x$ consists of the members of $\mathcal{A}$ containing $x$ and $\overline{A}_x$ are the members of $\mathcal{A}$ not containing $x$. Since $\{x\} \in \mathcal{A}$ and $\mathcal{A}$ is union-closed, the map $S \mapsto S \cup \{x\}$ is an injection from $\overline{A}_x$ to $\mathcal{A}_x$. Thus, $|\overline{A}_x| \leq |\mathcal{A}_x|$, as required. $\square$

Similarly, we also have the following.

Observation. If $\mathcal{A}$ contains a $2$-set $\{x,y\}$, then $x$ or $y$ is abundant.

This might lead one to conjecture the following false strengthening of Frankl's Union-closed Conjecture.

False Strengthening. Let $S$ be a non-empty member of $\mathcal{A}$ with the smallest number of elements. Then $S$ contains an abundant element.

Disproof. For two families $\mathcal{A}$ and $\mathcal{B}$ we let $$ \mathcal{A} \uplus \mathcal{B}=\{S \cup T: S \in \mathcal{A}, T \in \mathcal{B}\}. $$

Fix $k \geq 3$ and for each $i \in [k]$ define $$ \mathcal{B}^i=\{[k+1,3k] \setminus \{2i+2\}, [k+1,3k] \setminus \{2i+3\}\}. $$

Finally, define $$ \mathcal{A}^k=\{[k]\} \cup \bigcup_{i=1}^k(\{\emptyset, \{i\}, [k]\} \uplus \mathcal{B}^i) \cup (2^{[k]} \uplus [k+1,3k]). $$

It is easy to check that $\mathcal{A}^k$ is union-closed and that $[k]$ is the unique smallest set in $\mathcal{A}^k$. However, $|\mathcal{A}^k|=1+6k+2^k$ and each $i \in [k]$ is contained in only $1+(2k+2)+2^{k-1}$ members of $\mathcal{A}^k$. Thus, no element of $[k]$ is abundant. $\square$

This example highlights one of the obstacles in proving the Union-closed Conjecture; it is difficult to predict where abundant elements will be.

Here is a nice example due to Bjorn Poonen, which I have taken from this survey paper of Bruhn and Schaudt. It is motivated by the following observations. Let $\mathcal{A}$ be a union-closed family. Call an element $x$ abundant if $x$ is contained in at least half of the members of $\mathcal{A}$.

Observation. If $\mathcal{A}$ contains a singleton $\{x\}$, then $x$ is abundant.

Proof. Partition $\mathcal{A}$ as $\mathcal{A}_x$ and $\overline{A}_x$ where $\mathcal{A}_x$ consists of the members of $\mathcal{A}$ containing $x$ and $\overline{A}_x$ are the members of $\mathcal{A}$ not containing $x$. Since $\{x\} \in \mathcal{A}$ and $\mathcal{A}$ is union-closed, the map $S \mapsto S \cup \{x\}$ is an injection from $\overline{A}_x$ to $\mathcal{A}_x$. Thus, $|\overline{A}_x| \leq |\mathcal{A}_x|$, as required. $\square$

Similarly, we also have the following.

Observation. If $\mathcal{A}$ contains a $2$-set $\{x,y\}$, then $x$ or $y$ is abundant.

This might lead one to conjecture the following false strengthening of Frankl's Union-closed Conjecture.

False Strengthening. Let $S$ be a non-empty member of $\mathcal{A}$ with the smallest number of elements. Then $S$ contains an abundant element.

Disproof. For two families $\mathcal{A}$ and $\mathcal{B}$ we let $$ \mathcal{A} \uplus \mathcal{B}=\{S \cup T: S \in \mathcal{A}, T \in \mathcal{B}\}. $$

Fix $k \geq 3$ and for each $i \in [k]$ define $$ \mathcal{B}^i=\{[k+1,3k] \setminus \{2i+k-1\}, [k+1,3k] \setminus \{2i+k\}\}. $$

Finally, define $$ \mathcal{A}^k=\{[k]\} \cup \bigcup_{i=1}^k(\{\emptyset, \{i\}, [k]\} \uplus \mathcal{B}^i) \cup (2^{[k]} \uplus [k+1,3k]). $$

It is easy to check that $\mathcal{A}^k$ is union-closed and that $[k]$ is the unique smallest set in $\mathcal{A}^k$. However, $|\mathcal{A}^k|=1+6k+2^k$ and each $i \in [k]$ is contained in only $1+(2k+2)+2^{k-1}$ members of $\mathcal{A}^k$. Thus, no element of $[k]$ is abundant. $\square$

This example highlights one of the obstacles in proving the Union-closed Conjecture; it is difficult to predict where abundant elements will be.

Here is a nice example due to Bjorn Poonen, which I have taken from this survey paper of Bruhn and Schaudt. It is motivated by the following observations. Let $\mathcal{A}$ be a union-closed family. Call an element $x$ abundant if $x$ is contained in at least half of the members of $\mathcal{A}$.

Observation. If $\mathcal{A}$ contains a singleton $\{x\}$, then $x$ is abundant.

Proof. Partition $\mathcal{A}$ as $\mathcal{A}_x$ and $\overline{A}_x$ where $\mathcal{A}_x$ consists of the members of $\mathcal{A}$ containing $x$ and $\overline{A}_x$ are the members of $\mathcal{A}$ not containing $x$. Since $\{x\} \in \mathcal{A}$ and $\mathcal{A}$ is union-closed, the map $S \mapsto S \cup \{x\}$ is an injection from $\overline{A}_x$ to $\mathcal{A}_x$. Thus, $|\overline{A}_x| \leq |\mathcal{A}_x|$, as required. $\square$

Similarly, we also have the following.

Observation. If $\mathcal{A}$ contains a $2$-set $\{x,y\}$, then $x$ or $y$ is abundant.

This might lead one to conjecture the following false strengthening of Frankl's Union-closed Conjecture.

False Strengthening. Let $S$ be a non-empty member of $\mathcal{A}$ with the smallest number of elements. Then $S$ contains an abundant element.

Proof. For two families $\mathcal{A}$ and $\mathcal{B}$ we let $$ \mathcal{A} \uplus \mathcal{B}=\{S \cup T: S \in \mathcal{A}, T \in \mathcal{B}\}. $$

Fix $k \geq 3$ and for each $i \in [k]$ define $$ \mathcal{B}^i=\{[k+1,3k] \setminus \{2i+2\}, [k+1,3k] \setminus \{2i+3\}\}. $$

Finally, define $$ \mathcal{A}^k=\{[k]\} \cup \bigcup_{i=1}^k(\{\emptyset, \{i\}, [k]\} \uplus \mathcal{B}^i) \cup (2^{[k]} \uplus [k+1,3k]). $$

It is easy to check that $\mathcal{A}^k$ is union-closed and that $[k]$ is the unique smallest set in $\mathcal{A}^k$. However, $|\mathcal{A}^k|=1+6k+2^k$ and each $i \in [k]$ is contained in only $1+(2k+2)+2^{k-1}$ members of $\mathcal{A}^k$. Thus, no element of $[k]$ is abundant. $\square$

This example highlights one of the obstacles in proving the Union-closed Conjecture; it is difficult to predict where abundant elements will be.

Here is a nice example due to Bjorn Poonen, which I have taken from this survey paper of Bruhn and Schaudt. It is motivated by the following observations. Let $\mathcal{A}$ be a union-closed family. Call an element $x$ abundant if $x$ is contained in at least half of the members of $\mathcal{A}$.

Observation. If $\mathcal{A}$ contains a singleton $\{x\}$, then $x$ is abundant.

Proof. Partition $\mathcal{A}$ as $\mathcal{A}_x$ and $\overline{A}_x$ where $\mathcal{A}_x$ consists of the members of $\mathcal{A}$ containing $x$ and $\overline{A}_x$ are the members of $\mathcal{A}$ not containing $x$. Since $\{x\} \in \mathcal{A}$ and $\mathcal{A}$ is union-closed, the map $S \mapsto S \cup \{x\}$ is an injection from $\overline{A}_x$ to $\mathcal{A}_x$. Thus, $|\overline{A}_x| \leq |\mathcal{A}_x|$, as required. $\square$

Similarly, we also have the following.

Observation. If $\mathcal{A}$ contains a $2$-set $\{x,y\}$, then $x$ or $y$ is abundant.

This might lead one to conjecture the following false strengthening of Frankl's Union-closed Conjecture.

False Strengthening. Let $S$ be a non-empty member of $\mathcal{A}$ with the smallest number of elements. Then $S$ contains an abundant element.

Disproof. For two families $\mathcal{A}$ and $\mathcal{B}$ we let $$ \mathcal{A} \uplus \mathcal{B}=\{S \cup T: S \in \mathcal{A}, T \in \mathcal{B}\}. $$

Fix $k \geq 3$ and for each $i \in [k]$ define $$ \mathcal{B}^i=\{[k+1,3k] \setminus \{2i+2\}, [k+1,3k] \setminus \{2i+3\}\}. $$

Finally, define $$ \mathcal{A}^k=\{[k]\} \cup \bigcup_{i=1}^k(\{\emptyset, \{i\}, [k]\} \uplus \mathcal{B}^i) \cup (2^{[k]} \uplus [k+1,3k]). $$

It is easy to check that $\mathcal{A}^k$ is union-closed and that $[k]$ is the unique smallest set in $\mathcal{A}^k$. However, $|\mathcal{A}^k|=1+6k+2^k$ and each $i \in [k]$ is contained in only $1+(2k+2)+2^{k-1}$ members of $\mathcal{A}^k$. Thus, no element of $[k]$ is abundant. $\square$

This example highlights one of the obstacles in proving the Union-closed Conjecture; it is difficult to predict where abundant elements will be.