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

1. Oct 17, 2011

### IntroAnalysis

1. The problem statement, all variables and given/known data
Suppose R is a partial order on A. B1$\subseteq$A, B2$\subseteq$A, and
$\forall$x$\in$B1$\exists$y$\in$B2(xRy),
and $\forall$x$\in$B2$\exists$y$\in$B1(xRy).

Prove that $\forall$x$\in$A, x is an upperbound of B1 if and only if
x is an upperbound of B2.

2. Relevant 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$\subseteq$A,
and a $\in$A. Then a is called an upper bound for B if $\forall$x$\in$B(xRa).

3. The attempt at a solution
Is this way off base?
What if A = {(x,y)$\in$Z X Zl Z are integers}
R = {(x,y)$\in$Z X Z l n$\in$Z, 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$\subseteq$A, then x$\in$A. $\forall$x$\in$B1$\exists$y$\in$B2(xRy). Also,
$\forall$x$\in$B2$\exists$y$\in$B1(xRy). Since
R is the relation x ≥ y, then (xRy) means x is an upper bound of B2.

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

2. Oct 18, 2011

### IntroAnalysis

Proof. let x $\in$A. Suppose x is an upperbound of B1. That means
yRx for all y$\in$B1. Let z$\in$B2. By assumption this z$\in$B2,
$\exists$w$\in$B1 such that zRw. Since w$\in$B1 and yRx for
all y $\in$B1, we know wRx. By transitive property, zRw and wRx implies zRx.

$\leftarrow$Similarly, suppose x is an upperbound of B2.