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Section 3.1
53
23. Two elements of a poset can have at most one least upper bound and here's why. Let £1,£2 each be
least upper bounds for elements
a
and
b.
Then £1 is an upper bound for
a
and
b,
so £2
~
£1 because £2
is
least.
Interchanging £1, £2 in the preceding statement gives £1
~
£2. So £1
=
£2 by antisymmetry.
Exercises 3.1
1. (a) [BB] Not a function;
f
contains two different pairs of the form (3,
).
(b) Not a function
with
domain
{I, 2, 3, 4}. There's no pair of the form (3,
).
(c) [BB] This is a function.
(d) Certainly not a function. There is more than one pair of the form (1,
):
In fact, there are four!
(e) This is a function.
2. (a) [BB] This is not a function unless each student at the University of Calgary has just one professor,
for if student
a
is taking courses from professors
b1
and
b2,
the given set contains
(a,
bI) and
(a,b2).
(b) Assuming that each student currently registered at the University of Calgary is taking at least one
course, then this is a function.
(c) Assuming some student
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory

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