Distributive Lattice

Question

Is below diagram is distributive lattice?

Answer 1

Yes. It is Distributive lattice.

Method 1 :

Theorem : A lattice $L$ is distributive iff every element has at most one relative complement in any interval. Or in Simple words, In every Sublattice of lattice $L$, Every element has at most one complement.

Complement $Y$ of any element $X$ is defined as following :

$X \vee Y = I$ And $X \wedge Y = O$ ...Where $I$ is the greatest element and $O$ is the least element.

So, here, $I = a$ and $O = e$ and $\vee$ represents $LUB$ and $\wedge$ represents $GLB$ .

So, Complement(s) of $a$ = {e}

Complement(s) of $b$ = None.

Complement(s) of $c$ = {d}

Complement(s) of $d$ = {c}

Complement(s) of $e$ = {a}

Complement(s) of $f$ = None.

And if you take any sublattice of $L$ then you can check that In every Sublattice, every element has at most one complement.

So,  It is indeed a Distributive lattice.

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Method 2 : 

Theorem : A lattice L is distributive iff L does not contain a pentagon or a diamond.

Here, the given lattice doesn't have $M_3$ or $N_5$ as Sublattice. Take 5-5 elements and check. 

One subset of elements which might confuse students is $\left \{ a,c,f,d,e \right \}$ ...But if we check then we find that It isn't a sublattice of the given lattice because Join of $d,f$ doesn't belong to this sublattice. $d \wedge f = b$ and $b$ isn't in the Subset $\left \{ a,c,f,d,e \right \}$.

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Method 3 :

Let $e =1, f = 2, d = 3, c = 4, b = 6, a = 12$... Hence, the Given lattice can be thought of as lattice $\left ( D_{12}, / \right )$

And 

Theorem : Every $(Dn, /)$ is Always a Distributive lattice. Where $D_n$ is the set of all positive divisors of $n$. and $n$ is some positive integer.

The set $D_n$ of all positive integer divisors of a fixed positive integer $n$, ordered by divisibility, is a distributive lattice.

If You seek a Formal Proof, Here it goes :

We already know that "The set $(D_n,/)$ is a  lattice." (also Can be proved easily)

So, In Order to Prove it Distributive, We need to Prove either one of the Two Distributive rules (Because If One holds then Other also Holds)

So, We need to Prove that $a \vee (b \wedge  c) = (a \vee b) \wedge (a \vee c)$

Comments

@Deepakk

Another question is 

if a diamond is not a lattice, then how the picture given in question is distributine?

In question it is not told 

a=LUB and e is GLB

So, can we not draw the picture like this?

Is not here complement of a and c is {d,e} for both?

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So sorry. Yes, you are right.
For 0 being reflexive, k can be $any$ number.

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if a diamond is not a lattice

Of course DIamond is a lattice. But It is Not a sublattice of the given lattice. 

how the picture given in question is distributine?

Already given three ways to prove that it is distributive. 

In question it is not told 

a=LUB and e is GLB

In the given lattice, $a$ is Greatest element and $e$ is least element.

 So, can we not draw the picture like this?

No. It is a Hasse diagram representation of the given Poset....Not some random Graph from graph theory. And Hasse diagram has some built-in Meanings.

Is not here complement of a and c is {d,e} for both? 

Second diagram is Not even a lattice. Let alone complements. 

Answer 2

A lattice L is Distributive if their elements have atmost one compliment. 

So in the above lattice 

Upper Bound = a

Lower Bound = e

e^c = a

 f^c = NO COMPLEMENT

c^c = d

b= NO COMPLEMENT

So given Lattice is Distributive.

Comments

A lattice L is Distributive if their elements have atmost one compliment. 

This logic is not correct.  If every element has at most one complement then lattice may or may not be distributive.