Prove every finite lattice has a greatest element - without induction I have to prove that every finite lattice (L, ≤) has a greatest element. I have seen a lot of proofs proving this by using induction, however, I have to prove it without induction since our institution says that in an induction proof we can't be sure that we're still dealing with a lattice in the hypothesis of induction part.
So I guess a constructive proof would be a solution. Can anyone help me with this or show me an example of how it could possibly be done?
Thanks.
 A: You can still prove it by induction, you just have to be careful about what it is you are proving by induction.  What you want to prove by induction on $n$ is that if $(L,\leq)$ is a lattice and $A\subseteq L$ is a subset with $n$ elements, then there is an element $x\in L$ such that $x\geq a$ for all $a\in A$.  Given a finite lattice $(L,\leq)$, you can then apply this statement with $n=|L|$ and $A=L$ to get that $L$ has a greatest element.  Since you do not assume that $A$ is a lattice but merely that it is a subset of a lattice (and your upper bound $x$ might be in $L\setminus A$), you avoid the difficulty you mentioned.
A: If you have already proved (e.g. by induction) that every (nonempty) finite partially ordered set has a maximal element, then you can argue as follows to show that a finite lattice $(L,\le)$ has a greatest element.
Let $a$ be a maximal element of $L;$ I claim that $a$ is the greatest element of $L.$ Consider any element $x\in L;$ I have to show that $x\le a.$ We know that $a\le x\vee a,$ but $a\not\lt x\vee a$ because $a$ is maximal, so $a=x\vee a$ and $x\le x\vee a=a.$
