# 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

$\top = x \wedge y$
$\top \preceq x$ and $\top \preceq y$

By definition of greatest element:

$x \preceq \top$

Thus by definition of antisymmetry:

$\top = x$

$\blacksquare$