# Answer to Question #15483 in Analytic Geometry for Sujata Roy

Question #15483

If lattice (C,<=) is a complemented chain, then

a) |C|<=1

b) |C|<=2

c) |C|>1

d) doesn't exist.

a) |C|<=1

b) |C|<=2

c) |C|>1

d) doesn't exist.

Expert's answer

Every chain is bounded: 0 and 1 belongs to C, or they coicide, so |C|=1.

If there is third element a, then its complement is such b, that inf(a,b)=0 and

sup(a,b)=1. But as it is chain, then we have a<=b or b<=a. If a<=b then

0=inf(a,b)=a and 1=sup(a,b)=b. Case b<=a is similar.

So there cannot be other third element in C.

|C|=2 or |C|=1 =>

If there is third element a, then its complement is such b, that inf(a,b)=0 and

sup(a,b)=1. But as it is chain, then we have a<=b or b<=a. If a<=b then

0=inf(a,b)=a and 1=sup(a,b)=b. Case b<=a is similar.

So there cannot be other third element in C.

|C|=2 or |C|=1 =>

**|C|<=2**
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