# Sub-lattices and lattices.

I have read in a textbook that $\mathcal{P}(X)$, the power-set of $X$ under the relation ‘contained in’ is a lattice. They also said that $S := \{ \varnothing,\{ 1,2 \},\{ 2,3 \},\{ 1,2,3 \} \}$ is a lattice but not a sub-lattice. Why is it so?

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What is the original lattice? Is it $\mathcal{P}(\{ 1,2,3 \})$? – Haskell Curry Feb 22 '13 at 5:29
Perhaps they are saying $S$ is not a sublattice of $T=P(\{{1,2,3\}})$, and perhaps the reason is the meet of $\{{1,2\}}$ and $\{{2,3\}}$ is not the same in $S$ as it is in $T$. – Gerry Myerson Feb 22 '13 at 5:32

So now, if $L$ is a lattice and $S\subseteq L$ then $S$ is automatically a poset, indeed a subposet of $L$. But, even if with that poset structure it is a lattice it does not mean that it is a sublattice of $L$. To be a sublattice it must be that for all $x,y\in S$, the join $x\vee y$ computed in $S$ is the same as that computed in $L$, and similarly for the meet $x\wedge y$. This much stronger condition does not have to hold. Indeed, as noted by Gerry in the comment, the meet $\{1,2\}\wedge \{2,3\}$ computed in $\mathcal P({1,2,3})$ is $\{2\}$, while computed in the given subset it is $\emptyset$. None the less, it can immediately be verified that the given subset is a lattice since under the inclusion poset, all finite meets and joins exist.