Is a Finite Lattice also a Complete Lattice?

[XodoX]

I'm not sure if I am using the right terms here, but:

When X is a finite set and R is a relation...

If (X,R) is a lattice, then (X,R) is also a complete lattice.



Does this make sense? The question then is, why is is also automatically complete. I don't understand that.

 

[micromass]

Yes, it makes sense and it is true.

A lattice is an ordered set where every finite subset has a supremum and an infimum. A complete lattice is an ordered set where every subset has a supremum and an infimum.

Clearly, if the underlying set $X$ is finite, then every subset of $X$ is finite. Thus every subset has a supremum and an infimum since every finite subset has a supremum and an infimum.

 

[XodoX]

Yes, it makes sense and it is true.

A lattice is an ordered set where every finite subset has a supremum and an infimum. A complete lattice is an ordered set where every subset has a supremum and an infimum.

Clearly, if the underlying set $X$ is finite, then every subset of $X$ is finite. Thus every subset has a supremum and an infimum since every finite subset has a supremum and an infimum.

Makes sense. Does this prove it, though ? I mean, is there any calculation to prove this, or you just basically say that's how it is.

 

[economicsnerd]

It depends on the definition you've used. Some books only require, for a poset to be a lattice, that every pair $x,y$ of elements have a supremum $x\vee y:= \bigvee\{x,y\}$ and infimum $x\wedge y:= \bigwedge\{x,y\}$. If this is so in your course, you still need to prove (by induction) that every finite set in a lattice has a supremum and infimum.

 

[XodoX]

Thank you! Didn't think of proof by induction.