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I'm confused about this definition on wikipedia. enter image description here 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?

The definition on wikipedia says nothing about this.

Let me know if any clarification is needed.

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    $\begingroup$ 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 $L$. $\endgroup$ – kccu May 5 '17 at 16:35
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    $\begingroup$ I'm not sure I understand your question. $x \wedge x=x$ always, so what conditions are you imposing on $y$? $\endgroup$ – kccu May 5 '17 at 16:59
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    $\begingroup$ 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 $1$. $\endgroup$ – kccu May 5 '17 at 17:11
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    $\begingroup$ Yes, a finite lattice always has a $0$ and a $1$. $\endgroup$ – kccu May 5 '17 at 17:13
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    $\begingroup$ Yes, that's true. $\endgroup$ – kccu May 5 '17 at 17:43

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