Top is Meet Irreducible

Theorem

Let $\left({S, \wedge, \preceq}\right)$ be a bounded above meet semilattice.

Then $\top$ is meet irreducible

where $\top$ denotes the greatest element in $S$.

Proof

Let $x, y \in S$ such that

$\top = x \wedge y$

By Meet Precedes Operands

$\top \preceq x$ and $\top \preceq y$

By definition of greatest element:

$x \preceq \top$

Thus by definition of antisymmetry:

$\top = x$

$\blacksquare$

Sources

• 1980: G. Gierz, K.H. Hofmann, K. Keimel, J.D. Lawson, M.W. Mislove and D.S. Scott: A Compendium of Continuous Lattices

• Mizar article WAYBEL_6:10