What is an example of nondistributive semilattice which is not a lattice?

Question

What is an example of nondistributive semilattice which is not a lattice?

See Section Distributive semilattices from the Wikipedia page Semilattice.

EDIT: I added the condition of not being a lattice in order to avoid the question to be trivial.

Answer 1

I use the definition of distributive semilattice as in the linked wikipedia article:

A join-semilattice is distributive if for all $a, b$, and $x$ with $x \leq a \vee b$ there exist $a' \leq a$ and $b' \leq b$ such that $x = a' \vee b'$.

And indeed, there exist nondistributive semilattices which are not lattices.
I think the following is a minimal (in terms of cardinality of the underlying set) example:

(Here, I'm using $a+b$ to denote $a\vee b$.)
Clearly, $x<a\vee b$, but there $x\neq a'\vee b'$ for any elements $a',b'$ in the semilattice.

By the way, in case a semilattice is also a lattice, then it is a distributive semilattice iff it is a distributive lattice.

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Edit. After Keith Kearnes comment, I realized that the above is not the smallest nondistributive semilattice, that is, if $S$ has three elements, say $S=\{a, b, c\}$, with $a \vee b = c$, then, taking $x=a$ we have $x\leq c=a\vee b$, but $x\neq a'\vee b'$ for any $a'\leq a$ and $b'\leq b$.

This small example shows that actually any finite semilattice that is not a lattice is nondistributive.
Indeed, if there are $a,b$ as in the example above, and, taking $x=a$ we can find $a'\leq a$ and $b'\leq b$ with $x=a'\vee b'$, then $$a=x=a'\vee b',$$ whence $b'\leq a$, and so $b'$ is a lower bound of $a$ and $b$.
Now if the set of lower bounds of $a$ and $b$ is nonempty, then we can take its join, and that's $a\wedge b$.
Hence a finite distributive semilattice is always a lattice.

Answer 2

Suppose the elements of the lattice are $0$, $a$, $b$, $c$, $1$ which satisfy $0 < a, b, c < 1$, $a \wedge b = a \wedge c = b \wedge c = 0$, and $a \vee b = a \vee c = b \vee c = 1$. Then $a \vee (b \wedge c) = a \vee 0 = a$ but $(a \vee b) \wedge (a \vee c) = 1 \wedge 1 = 1$. There is another five element non distributive lattice.

Answer 3

(from https://en.wikipedia.org/wiki/Distributive_lattice#Characteristic_properties:)

"The pentagon lattice $N_5$ ($0<z<x<1,0<y<1$) is non-distributive: $$x ∧ (y ∨ z) = x ∧ 1 = x ≠ z = 0 ∨ z = (x ∧ y) ∨ (x ∧ z)".$$