A detail in the proof of Stone representation Theorem

Question

Let $(\mathcal{B},\sqcap,\sqcup,\leq)$ be a Boolean algebra. Let $x,y\in\mathcal{B}$. I want to prove the following implication: $$x\sqcap y'\leq 0\Rightarrow x\leq y$$ where $y'$ is the complement of $y$.

I have checked that this works in the case $\mathcal{B}=\mathcal{P}(X)$, with $\sqcap$=intersection, $\sqcup$=union, $\leq=\subseteq$, $0=\varnothing$.....but I can't prove it for a generic Boolean algebra.

score 1 · Accepted Answer · answered Nov 08 '13 at 10:18

1

Hint: $x=x\sqcap1=x\sqcap(y\sqcup y')=(x\sqcap y)\sqcup(x\sqcap y')=\ldots$

answered Nov 08 '13 at 10:18

egreg

238,574

I can't understand how did you apply distributive law – Danae Kissinger Nov 08 '13 at 10:20
@DanaeKissinger Oh, sorry! I mixed up the symbols. – egreg Nov 08 '13 at 10:22

score -1 · Answer 2 · edited Feb 03 '14 at 18:35

-1

Since $$x⊓y′≤0$$ $$(x⊓y′)⊔y ≤0 ⊔ y = y$$ $$(x⊔y)⊓(y'⊔y)≤ y$$ $$(x⊔y)⊓1 ≤ y$$ $$(x⊔y)≤ y$$ $$x ≤ y$$

edited Feb 03 '14 at 18:35

Cameron Buie

102,994

answered Feb 03 '14 at 18:32

Dr. P.K.Sharma

1

A detail in the proof of Stone representation Theorem

2 Answers2