Existence of Non-Empty Finite Suprema in Join Semilattice

Theorem
Let $\left({S, \preceq}\right)$ be a join semilattice.

Let $A$ be a non-empty finite subset of $S$.

Then $A$ agmits a supremum in $\left({S, \preceq}\right)$.

Proof
We will prove by induction of the cardinality of finite subset of $H$.

Base case

 * $\forall A \subseteq S: \left\vert{A}\right\vert = 1 \implies \exists \sup A$

where $\left\vert{A}\right\vert$ denotes the cardinality of $A$.

Let $A \subseteq S$ such that
 * $\left\vert{A}\right\vert = 1$

By Cardinality of Singleton:
 * $\exists a: A = \left\{ {a}\right\}$

By definition of subset:
 * $a \in S$

Thus by Supremum of Singleton:
 * the supremum of $A$ exists in $\left({S, \preceq}\right)$.

Induction Hypothesis

 * $n \ge 1 \land \forall A \subseteq S: \left\vert{A}\right\vert = n \implies \exists \sup A$

Induction Step

 * $\forall A \subseteq S: \left\vert{A}\right\vert = n+1 \implies \exists \sup A$

Let $A \subseteq S$ such that
 * $\left\vert{A}\right\vert = n+1$

By definition of cardinality of finite set:
 * $A \sim \N_{< n+1}$

where $\sim$ denotes set equivalence.

By definition of set equivalence:
 * there exists a bijection $f: \N_{< n+1} \to A$

By Restriction of Injection is Injection:
 * $f \restriction_{\N_{< n}}: \N_{< n} \to f^\to\left({\N_{< n}}\right)$ is an injection.

By definition
 * $f \restriction_{\N_{< n}}: \N_{< n} \to f^\to\left({\N_{< n}}\right)$ is a surjection.

By definition
 * $f \restriction_{\N_{< n}}: \N_{< n} \to f^\to\left({\N_{< n}}\right)$ is a bijection.

By definition of set equivalence:
 * $\N_{< n} \sim f^\to\left({\N_{< n}}\right)$

By definition of cardinality of finite set:
 * $\left\vert{f^\to\left({\N_{< n}}\right)}\right\vert = n$

By definitions of image of set and subset:
 * $f^\to\left({\N_{< n}}\right) \subseteq A$

By Subset Relation is Transitive:
 * $f^\to\left({\N_{< n}}\right) \subseteq S$

By Induction Hypothesis:
 * $\exists \sup \left({ f^\to\left({\N_{< n}}\right)}\right)$

By definition $\N_{< n+1}$
 * $n \in \N_{< n+1}$

By definition of mapping:
 * $f \left({n}\right) \in A$

By definition of subset:
 * $f \left({n}\right) \in S$

Thus:

Thus $A$ admits a supremum in $\left({S, preceq}\right)$.