How to Prove It, Sec. 4.4, #18a

[IntroAnalysis]

Homework Statement

Suppose R is a partial order on A. B1\subseteqA, B2\subseteqA, and
\forallx\inB1\existsy\inB2(xRy),
and \forallx\inB2\existsy\inB1(xRy).

Prove that \forallx\inA, x is an upperbound of B1 if and only if
x is an upperbound of B2.

Homework Equations

R is a partial order on A if R is reflexive, transitive and antisymmetric.

Definition of an upper bound: Suppose R is a partial order on A, B\subseteqA,
and a \inA. Then a is called an upper bound for B if \forallx\inB(xRa).

The Attempt at a Solution

Is this way off base?
What if A = {(x,y)\inZ X Zl Z are integers}
R = {(x,y)\inZ X Z l n\inZ, y=2n, x=2n+1 and x≥y}
B1 = {1, 3, 5, 6, 8} B2 = {0, 2, 4, 7, 9}
A= {(1,0), (3,2), (5,4), (7,6), (9,8)}

→Let x be an arbitrary upper bound of B1. Since B1\subseteqA, then x\inA. \forallx\inB1\existsy\inB2(xRy). Also,
\forallx\inB2\existsy\inB1(xRy). Since
R is the relation x ≥ y, then (xRy) means x is an upper bound of B2.

\leftarrowLet x be an arbitrary upper bound of B2. Similar reasoning (or lack thereof) to show reverse if, then statement.

 

[IntroAnalysis]

Proof. let x \inA. Suppose x is an upperbound of B1. That means
yRx for all y\inB1. Let z\inB2. By assumption this z\inB2,
\existsw\inB1 such that zRw. Since w\inB1 and yRx for
all y \inB1, we know wRx. By transitive property, zRw and wRx implies zRx.

\leftarrowSimilarly, suppose x is an upperbound of B2.