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IntroAnalysis
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Homework Statement
Suppose R is a partial order on A. B1[itex]\subseteq[/itex]A, B2[itex]\subseteq[/itex]A, and
[itex]\forall[/itex]x[itex]\in[/itex]B1[itex]\exists[/itex]y[itex]\in[/itex]B2(xRy),
and [itex]\forall[/itex]x[itex]\in[/itex]B2[itex]\exists[/itex]y[itex]\in[/itex]B1(xRy).
Prove that [itex]\forall[/itex]x[itex]\in[/itex]A, 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[itex]\subseteq[/itex]A,
and a [itex]\in[/itex]A. Then a is called an upper bound for B if [itex]\forall[/itex]x[itex]\in[/itex]B(xRa).
The Attempt at a Solution
Is this way off base?
What if A = {(x,y)[itex]\in[/itex]Z X Zl Z are integers}
R = {(x,y)[itex]\in[/itex]Z X Z l n[itex]\in[/itex]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[itex]\subseteq[/itex]A, then x[itex]\in[/itex]A. [itex]\forall[/itex]x[itex]\in[/itex]B1[itex]\exists[/itex]y[itex]\in[/itex]B2(xRy). Also,
[itex]\forall[/itex]x[itex]\in[/itex]B2[itex]\exists[/itex]y[itex]\in[/itex]B1(xRy). Since
R is the relation x ≥ y, then (xRy) means x is an upper bound of B2.
[itex]\leftarrow[/itex]Let x be an arbitrary upper bound of B2. Similar reasoning (or lack thereof) to show reverse if, then statement.