Question

If A is a set, write A={(x,x):x∈A}.

Prove: If S is a strict partial order on A, then S∗=S∪A is a partial order on A .

Answer 1

ANSWER:-

Suppose S is a relation on a set A, and  is asymmetric if:

∀x∈A∀y∈A ((x,y)∈S→(y,x)∉S)

The first point of the exercise was to demonstrate that if R is asymmetric then it is also antisymmetric.:

Show that if S is a strict partial order, then it is also asymmetric.

A partial order is a binary relation on a set, say, A that is reflexive, antisymmetric and transitive.

And the proof is the following:

Suppose that S is a strict partial order, and suppose that for some x,y∈Ax, (x,y)∈S and (y,x)∈D

Then by transitivity of S, (x,x)∈S which contradicts the fact that S is irreflexive.

Therefore, S is asymmetric.

What exactly I am not understanding is why if (x,y)∈S and (y,x)∈S

since Sis antisymmetric. At least, x=y

. Then I also cannot understand why it follows immediately that it's asymmetric.

Why did mathematicians invented all that Simple: strict partial orders and partial orders are exactly the same thing.

If A is a set, write Δ(A)={(x,x):x∈A}

If S is a strict partial order on A,

then S∗=S∪Δ(A)S∗=S∪Δ(A) is a partial order on AA (prove it).

If S is a partial order on A,

then S∗=S∖Δ(A) is a strict partial order on A

also that (S∗)∗=S and (S∗)∗=S,

so there is a very well behaved bijection between the strict and the lax partial order relations on A.