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If $(L, \land, \lor, 0, 1)$ is a lattice, and there exists a unary operation ' on $L$ such that

  1. (x \lor x')=1, and

  2. (x \land x')=0

both hold, is the unary operation ' an isomorphism between the semilattices $(L, \land)$ and $(L, \lor)$? If we add the condition that ' is an involution (i. e. $x''=x$), is ' an isomorphism?

1 Answers 1

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No.

Consider this lattice:

                                    1                                    / \                                   /   \                                  /     \                                 d       w                                / \     / \                               b   c   y   z                                \ /     \ /                                 a       x                                  \     /                                   \   /                                    \ /                                     0 

and make ' exchange $0\leftrightarrow 1$, $d\leftrightarrow w$, $b\leftrightarrow y$, $c\leftrightarrow z$, $a\leftrightarrow x$. It is easy to verify that r\land r' = 0 and r\lor r' = 1 for all $r$; but ' does not define an isomorphism from $(L,\land)$ to $(L,\lor)$, since for example b'\lor c' = y\lor z = w, but (b\land c)' = a' = x\neq w. Nor does it define an isomorphism going the other way, since the map is self-invertible.

Added. A smaller example:

                                     1                                     / \                                    /   \                                   b     y                                   |     |                                   a     x                                    \   /                                     \ /                                      0 

But a'\lor b' = x\lor y = y, (a\land b)' = a' = x.

  • 0
    I meant "Nontrivial meet"$\neq 0$, "nontrivial join$\neq 1$. The second example shows you don't need $x\lor y\notin\{x,y\}$ for$a$counterexample to work.2012-02-16