complete semilattice and its not subsemilattice/a counter-example

Question

Is there an example of a complete semi-lattice and its subset which is itself a complete semi-lattice without being sub-semi-lattice of the bigger one ?

Answer 1

Let $S$ be the semi-lattice in the picture below:

Now, let $A=\{2,3,18\}$. It is still a semi-lattice, but not a sub-semi-lattice of $S$, since $2 \vee 3 =12$ in $S$.

Code (this editor seems not adequate for this, but here it goes):
\usepackage{pstricks,pstricks-add,pst-node}
\begin{pspicture}(3,2)
\psset{unit=.75cm}
\psset{radius=0.1,labelsep=.1}
\Cnode(0,0){2} \nput{180}{2}{$2$}
\Cnode(2,0){3} \nput{0}{3}{$3$}
\Cnode(1,1){12} \nput{30}{12}{$12$}
\Cnode(1,2){18} \nput{90}{18}{$18$}
\ncline{12}{2}
\ncline{12}{3}
\ncline{12}{18}
\end{pspicture}

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Here's another example in which $A$ is still a sub-semi-lattice of $S$, but not a complete one.
In fact, in this example $S$ is a lattice and $A$ is a sub-lattice, but it works just the same.

Let $S = \wp_{fin}(\mathbb N) \cup \{\mathbb N, \mathbb Q\}$, where $\wp_{fin}(X)$ denotes the family of finite subsets of $X$.
Now, let $A = S \setminus \{\mathbb N\} = \wp_{fin}(\mathbb N) \cup \{\mathbb Q\}$.
Then $A$ is still a sub-lattice (whence a sub-semi-lattice) of $S$, but not a complete one, since $\bigvee \wp_{fin}(\mathbb N) = \mathbb N$, in $S$ but the result is $\mathbb Q$ in $A$.