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For a finite graph vertex set $W$ in a locally finite graph, let ${\partial }^{+}W$ be its exterior graph neighbours. The number of trifurcation vertices in percolation lying in $W$ is at most $|{\partial }^{+}W|$.

For each cluster meeting $W$, contract its connected components of a graph outside $W$ to terminal graph vertices, keeping only those adjacent to $W$. The resulting incidence graph is finite and connected. Each trifurcation graph vertex in $W$ separates at least three groups of terminals, because every infinite branch must leave the finite set $W$. Take a minimal subtree connecting all terminals. Each such trifurcation is unavoidable in that subtree and has degree of a vertex at least three. All leaves are terminals. The tree identity ${\sum }_v(\mathrm{deg}(v)-2)=-2$ bounds the number of these branching graph vertices by the number of leaves minus two, hence by the number of terminals. Different terminals, even across different clusters, can be assigned different exterior graph neighbours. Summing proves the bound. For square boxes in ${\mathbb{Z}}^2$, the boundary has order $n$ and the volume has order $n^2$.

If $B$ is a countably infinite set and $A$ is a finite set with at least two elements, the set of surjective functions $B\to A$ is uncountable. Fix finitely many values to cover $A$, then encode an arbitrary infinite binary sequence using two values of $A$ on the remaining inputs.

A pointwise periodic self-map has a uniform positive period exactly when its finite cycle lengths are bounded. A uniform period is divisible by every length; conversely, bounded lengths divide the least common multiple of $1,\dots ,M$ for some bound $M$. Arbitrarily many bounded-length cycles are allowed. Every such map on a finite set has a uniform period. On any infinite set containing a countably infinite subset, cycles of unbounded finite lengths and fixed points elsewhere give a counterexample. Thus, with the usual axiom of choice assumption, the sets on which every pointwise periodic self-map has a uniform period are exactly finite sets.

Consider the following finite set of subsets, ordered by subset containment:Every nonempty chain in a partial order has a greatest element within that chain in a partial order, hence an upper bound in $P$. For the empty chain in a partial order, choose $\varnothing \in P$. Thus the hypothesis of Zorn's lemma holds. The two maximal elements are $\{a,b\}$ and $\{b,c\}$: neither is a proper subset of another member of $P$. There is no greatest element, because it would have to contain both, and no member of $P$ contains $\{a,b,c\}$.

For the chain in a partial order $C=\{\varnothing ,\{b\}\}$, the member $\{b\}$ is already an upper bound, although it is not a maximal element of $P$. Both orange members are also upper bounds for $C$. A chain in a partial order need not have a unique upper bound, and the upper bound used to check the hypothesis need not be the eventual maximal element.

The restriction to chains in a partial order matters. The subset $\{\{a\},\{c\}\}\subseteq P$ has no upper bound in $P$, but its two members are incomparable, so it is not a chain in a partial order. Requiring every subset of $P$ to have an upper bound would be much stronger: taking that subset to be $P$ itself would force a greatest element.