Let be A a finite set and (A,<) a lattice. Prove that A has maximun and minimun.
First, I'm from Uruguay and I'm literally translating the terms so this are the defitions I have, in order to avoid misconceptions:
Let be $R$ a partial order relation in $A$, and $B\subseteq A$:
Supremum: $x$ is supremum of $B$ if and only if $x$ is an upper bound of $B$ and $\forall y$ such that $y$ is upper bound of B, it holds $xRy$.
Maximun: If the supremum is included on $B$.
Lattice: R is a lattice if and only if $\forall B\subseteq A$, $B$ finite, it holds $B$ has supremum and infimum.
This is my reasoning:
Since A is a lattice, then it holds that $\forall B \subseteq A$ finite $B$ has supremum and infimum. Since $A$ is finite, we can choose $B = A$ and by lattice definition $A$ will have supremum and infimum. Then, that supremum must belong to $A$ (I intuitively think this but I don't know how to formalize it). The same with the infimum.
I always suffer with stuff related to proving anything so any help will be appreciated.
If $A$ is a finite lattice then by the definition of lattice given in your question $A$ has a supremum and has an infimum.
This means that elements $x,y\in A$ exist such that $y\leq a\leq x$ for every $a\in A$.
This means exactly that $x$ is a maximum and $y$ is a minimum of lattice $A$.
So your reasoning is correct.
Of course it does. A supremum of a subset $B\subseteq A$ is by definition an element of $A$ that has a certain property.
Footnote:
The definition of lattice given by you is often practicized as definition of bounded lattice which makes the question a bit more easy because every bounded lattice has a maximum and a minimum. For this take infimum and supremum of the empty set.
In a weaker definition it is only demanded that non-empty finite sets (or equivalently pairs of elements) have a supremum and an infinum. Also if that definition is practicized then it is possible to prove the existence of a minimum and a maximum if the lattice is finite. E.g. see the answer of William.