this is apparently an easy question, for which i'm kind of struggling. I have to show that $(\mathbb Q \cap [0,\sqrt{2}],\leq)$ is a bounded lattice but not a complete lattice.
Could someone help me please ?
Same for an other question which is : show by reccurence on the number of elements that any finite part of a lattice is bounded.
I'm blocked.
Could someone help me for this question too please ?
Thanks a lot in advance !
Perhaps you know that $(\mathbb Q,\leq)$ is a chain, and that every chain is a lattice.
If you don't know that, then prove it.
Also, every subset of a chain is, again, a chain, and so it's also a lattice (once again, prove it if it is not clear enough to you).
Now there is a problem with this question, which is that you want to prove that set to be a bounded lattice.
In Lattice Theory, a bounded lattice is a lattice which has a minimum element (usually denoted by $0$ or $\bot$) and a maximum element (usually denoted by $1$ or $\top$). Other notations may occur.
But your lattice $L$ doesn't have a maximum element.
Indeed, if $a \in L$, i.e., $a$ is a rational number such that $a \leq \sqrt{2}$ (and thus, $a<\sqrt{2}$), then, there's still some rational number $b$ such that $a<b<\sqrt{2}$, and so $a$ is not the maximum.
Perhaps you'r interested in other notion of boundedness, which is based in the distances between elements of a set, and then the lattice would be bounded in that sense (since $\sup \{d(x,y) : x,y \in L\}$ is finite).
To prove that the lattice is not complete, it is enough the above observation that it doesn't have a maximum element (all complete lattices are bounded in the lattice theoretic sense).