# in a lattice, does the GLB and LUB of each element need to be contained in the lattice itself?

I'm confused about this definition on wikipedia. Do the GLB and LUB need to be contained in $L$ for it to be a lattice?

if they must be contained, what's the GLB or LUB of two elements $(x,y) \in L$ where $x$ is the LUB or GLB of $L$ itself and happens to be incomparable with $y$? wouldn't the result of that operation need to be outside $L$ if it were to exist? does that mean it's not a lattice?

If the result can't be contained in $L$ then it needs to be outside of $L$, then I guess I would need to define some superset of $L$ that has the same operations as $L$ for the GLB and LUB to live on? in that case is this still a lattice?

• Yes, the GLB and LUB must be in the lattice. If $x$ is the LUB of $L$, then $x$ cannot be incomparable with $y$ - by definition of LUB you must have $x \geq y$. Similarly if $x$ is the GLB of $L$. – kccu May 5 '17 at 16:35
• I'm not sure I understand your question. $x \wedge x=x$ always, so what conditions are you imposing on $y$? – kccu May 5 '17 at 16:59
• Oh, certainly not. Those are usually called $0$ and $1$ in a lattice ($0$ is the GLB and $1$ is the LUB), but they need not exist. For instance, the integers are a lattice with the usual ordering, but there is no least or greatest integer. A finite lattice will have a $0$ and $1$. – kccu May 5 '17 at 17:11
• Yes, a finite lattice always has a $0$ and a $1$. – kccu May 5 '17 at 17:13