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.
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 Summer '10
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 Graph Theory, Republic of Ireland, Westminster system

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