Bourbaki

Exercise 16, § 1

Tag 03UC, permanent. Cite it as [Bourbaki, Tag 03UC].

Theory of Sets, III, 217

Bourbaki marks this one as harder than the rest.

A lattice E is said to be distributive if it satisfies the following two conditions :

(D') sup⁡(x,inf⁡(y,z))=inf⁡(sup⁡(x,y),sup⁡(x,z)),\qquad \sup (x, \inf (y, z)) = \inf (\sup (x, y), \sup (x, z)), (D'') inf⁡(x,sup⁡(y,z))=sup⁡(inf⁡(x,y)inf⁡(x,z))\qquad \inf (x, \sup (y, z)) = \sup (\inf (x, y) \inf (x, z))

for all x,y,zx, y, z in E. A totally ordered set is a distributive lattice.

(a) Show that each of the conditions (D), (D') separately implies the condition

(D) sup⁡(inf⁡(x,y), inf⁡(y,z), inf⁡(z,x))\quad \sup (\inf (x, y), \ \inf (y, z), \ \inf (z, x))

=inf⁡(sup⁡(x,y), sup⁡(y,z), sup⁡(z,x))= \inf (\sup (x, y), \ \sup (y, z), \ \sup (z, x))
for all x,y,zx, y, z in E.

(b) Show that the condition (D) implies the condition

(M) \qquad If x⩾zx \geqslant z, then sup⁡(z,inf⁡(x,y))=inf⁡(x,sup⁡(y,z)).\sup (z, \inf (x, y)) = \inf (x, \sup (y, z)).

Deduce that (D) implies each of (D′) and (D″), and hence that the three axioms (D), (D′), and (D″) are equivalent (to show, for example, that (D) implies (D′), take the least upper bound of xx and each side of (D), and use (M)).

(c) Show that each of the two conditions

(T′) inf⁡ (z,sup⁡ (x,y))⩽sup⁡ (x,inf⁡ (y,z)),\qquad\qquad \inf\,(z, \sup\,(x, y)) \leqslant \sup\,(x, \inf\,(y, z)), (T″) inf⁡ (sup⁡ (x,y),sup⁡ (z,inf⁡ (x,y)))=sup⁡ (inf⁡ (x,y),inf⁡ (y,z),inf⁡ (z,x))\inf\,(\sup\,(x, y), \sup\,(z, \inf\,(x, y))) = \sup\,(\inf\,(x, y), \inf\,(y, z), \inf\,(z, x))

for all xx, yy, zz in E is necessary and sufficient for E to be distributive. (To show that (T′) implies (D″), consider the element

inf⁡ (z,sup⁡ (x,inf⁡ (y,z))).)\inf\,(z, \sup\,(x, \inf\,(y, z))).)
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