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.
On
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.
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 $x\le x\vee a=a.$