Any non-empty finite subset $\{x_1,x_2,...,x_k\}$ of $B$ has a g.l.b and l.u.b in $B$

Question

Prove that any non-empty finite subset $\{x_1,x_2,...,x_k\}$ of $B$ has a g.l.b and l.u.b in $B$ where $(B,\leq)$ forms a lattice, i.e.

• $(B, \leq)$ is a partially ordered set
• Any two elements $x, y\in B$ have a g.l.b(greatest lower bound) $x \land y$ and an l.u.b(least upper bound) $x \lor y$

I've been thinking of induction, but I'm not sure that it'd work since we only have a partial order on $B$. Nevertheless,

1. Base case: Only one element, which is the g.l.b and l.u.b both
2. Induction hypothesis: Let's say the statement holds for sets of size $n-1$ and less
3. Consider a set of size $n$, namely $\{x_1, x_2,...,x_n\}$. $\{x_1, x_2,...,x_{n-1}\} \subset \{x_1, x_2,...,x_n\}$ has a g.l.b (say $x_g$) and an l.u.b (say $x_l$) in $\{x_1, x_2,...,x_{n-1}\}$. All that remains to be shown is that g.l.b($x_1,...,x_n$) = g.l.b($x_g,x_n$). Similarly for l.u.b.

I'm not sure how to proceed from here!

Answer 1

We proceed by induction on $k$.

1. Base case: $k=1$. $|X| = 1$, and $x_1$ is its g.l.b and l.u.b both, since $x_1 \leq x_1$.
2. Induction Hypothesis: Let the statement hold for $|X| \leq k-1$.
3. Induction Step: Consider $X = \{x_1,x_2,...,x_k\}$ = $\{x_1,...,x_{k-1}\} \cup {x_k}$. $\{x_1,...,x_{k-1}\}$ has a g.l.b, say $x_g \in B$ (by the induction hypothesis). Moreover, $x_g$ and $x_k$ have a g.l.b, say $x'_g$. It remains to show that $x'_g$ is in fact the g.l.b of $X$. We have that $x'_g \leq x_g,x_k$, and if $z\leq x_g,x_k$ then $z\leq x'_g$. In addition, $x_g \leq x_1,...,x_{k-1}$ and if $z \leq x_1,...,x_{k-1}$ then $z\leq x_g$. Combining the above two statements, $x'_g\leq x_1,...,x_k$ and if $z\leq x_1,...,x_k$ then $z\leq x'_g$.

Similarly for l.u.b!