2
$\begingroup$

$L$ is a distributive lattice with top and bottom element($1$ and $0$ respectively). Show that if an element has a complement, the complement must be unique.

This is what I have so far, but am getting stuck on where to go, or if this is even the right route to take.

Let $a\in L$ and $b_1,b_2$ be two elements in $L$ s.t. $a\vee b_1=1, a\wedge b_1=0, a\vee b_2=1,a\wedge b_2=0$ Then by distribution $b_1\vee (a\wedge b_2)=b_1\vee 0=b_1$ and $(b_1\vee a)\wedge (b_1\vee b_2)=1\vee (b_1\vee b_2)=b_1\vee b_2$ So I have $b_1=b_1 \vee b_2$ but this does not help.

Any ideas would be great, it seems like I am close I just cant see the missing part.

$\endgroup$

2 Answers 2

2
$\begingroup$

You have $b_1=b_1\vee b_2$, equivalently, $b_2\leq b_1,$ and similarly $b_2=b_1\vee b_2$ so that $b_1\leq b_2$.

$\endgroup$
2
  • $\begingroup$ ok great, this is my first time with this stuff so its good to see the little tricks:) $\endgroup$
    – tmpys
    Oct 13, 2014 at 22:04
  • $\begingroup$ Glad that was helpful. $\endgroup$ Oct 14, 2014 at 3:05
1
$\begingroup$

You're on the right track.

By symmetry (exchanging $b_1$ and $b_2$), we also have $b_2=b_1\lor b_2$.

$\endgroup$
3
  • $\begingroup$ Great ty. I wish I could give two answers:0 $\endgroup$
    – tmpys
    Oct 13, 2014 at 22:04
  • $\begingroup$ Am I correct that you posted this first? $\endgroup$
    – tmpys
    Oct 13, 2014 at 22:05
  • $\begingroup$ ok I gave it to the other guy because you have 36k and he showed me that $b_1=b_1 \vee b_2 \implies b_2\le b_1$ I know it's obvious but I did not think about it. $\endgroup$
    – tmpys
    Oct 13, 2014 at 22:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.