Proof that any finite subset of a lattice has a supremum.

Question

Let $C\subseteq D$ where $C = \{c_{1}, c_{2},\ldots, c_{n}\}$ and $D$ is a lattice. I believe that this supremum is $((c_1 \vee c_2)\vee c_3)\vee \ldots \vee c_n)$ (sorry about the notation). However I cannot prove that this is the supremum, I tried assuming it wasn't but couldn't find a contradiction. All help is welcome, thank you very much for your attention.

Answer 1

Proceed by induction on $n$ to prove that $s = (\ldots(((c_1 \vee c_2) \vee c_3) \vee \ldots \vee c_n)$ is indeed the supremum. We know that in the case of $n = 1$, $s = c_1$ is a supremum of $n$.

Now assume true for all values $< n$. Then by definition of the meet (which is denoted by $\vee$) we have that $s \geq c_i$ for every $i$. Now, suppose $a \geq c_i$ for every $i$. Define $b = c_1 \vee c_2 \vee \ldots \vee c_{n-1}$; by the inductive hypothesis, $b$ is the supremum of $\{c_i\}_{i = 1}^{n-1}$, so we have $a \geq b$. Now, since $a \geq b$ and $a \geq c_n$, we have that $a \geq b \vee c_n$ by definition of the meet. But $b \vee c_n = s$, so we have $a \geq s$, i.e. $s$ is a supremum.

Answer 2

This proof is really a template of a proof, that you have certainly seen before, and will see again.

If $A$ is a collection of sets and $*$ is an associative binary operation on $A$, then $*$ can be extended to any finite sequence of elements. If $*$ is also commutative and idempotent (in the sense that $a*a=a$), then $*$ can be extended to finite subsets of $A$.

This is simply by induction, where the fact $*$ is well-defined for two means you can extend it further.

You have probably seen this in the proof of DeMorgan law extended to any finite number of sets; and you may have seen this in the proof that any finite intersection of open sets is open; and you probably saw this in the proof that a finite union of finite sets is a finite set; and you may have seen this in other places through mathematics.

In this case, you use the fact that taking supremum is well-defined for two elements; it is associative and commutative, and in fact idempotent ($a\lor a=a$).