Existence of Non-Empty Finite Suprema in Join Semilattice

Theorem
Let $\struct {S, \preceq}$ be a join semilattice.

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

Then $A$ admits a supremum in $\struct {S, \preceq}$.

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

Base case

 * $\forall A \subseteq S: \card A = 1 \implies \exists \sup A$

where $\card A$ denotes the cardinality of $A$.

Let $A \subseteq S$ such that
 * $\card A = 1$

By Cardinality of Singleton:
 * $\exists a: A = \set a$

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

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

Induction Hypothesis

 * $n \ge 1 \land \forall A \subseteq S: \card A = n \implies \exists \sup A$

Induction Step

 * $\forall A \subseteq S: \card A = n + 1 \implies \exists \sup A$

Let $A \subseteq S$ such that:
 * $\card A = n + 1$

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:
 * $f \restriction_{\N_{< n} }: \N_{< n} \to f^\to \sqbrk {\N_{< n} }$ is an injection.

By definition:
 * $f \restriction_{\N_{< n} }: \N_{< n} \to f^\to \sqbrk {\N_{< n} }$ is a surjection.

By definition:
 * $f \restriction_{\N_{< n} }: \N_{< n} \to f^\to \sqbrk {\N_{< n} }$ is a bijection.

By definition of set equivalence:
 * $\N_{< n} \sim f^\to \sqbrk {\N_{< n} }$

By definition of cardinality of finite set:
 * $\card {f^\to \sqbrk {\N_{< n} } } = n$

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

By Subset Relation is Transitive:
 * $f^\to \sqbrk {\N_{< n} } \subseteq S$

By Induction Hypothesis:
 * $\exists \map \sup {f^\to \sqbrk {\N_{< n} } }$

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

By definition of mapping:
 * $\map f n \in A$

By definition of subset:
 * $\map f n \in S$

Thus:

Thus $A$ admits a supremum in $\struct {S, \preceq}$.