4130 HOMEWORK 5
Due Tuesday March 9
(1) A subset
I
of
R
is called an
interval
if for all
x, y
∈
I
and all
z
∈
R
, if
x < z < y
then
z
∈
I
.
Show that if
I
is a bounded interval, then (inf
I,
sup
I
)
⊂
I
. Using this, show that
I
must be one of the following four intervals:
(inf
I,
sup
I
)
[inf
I,
sup
I
)
(inf
I,
sup
I
]
[inf
I,
sup
I
]
.
Let
I
be a bounded interval. Then sup
I
and inf
I
exist. Suppose inf
I < z <
sup
I
.
Then there is some
x
∈
I
with inf
I
≤
x < z
. Indeed, if there was no such
x
, then
z
would be a lower bound for
I
, but
z
is greater than the greatest lower bound inf
I
.
Similarly, there is some
y
∈
I
with
z < y
≤
sup
I
. Therefore,
x < z < y
and so
z
∈
I
by definition of an interval.
Therefore, (inf
I,
sup
I
)
⊂
I
. Now suppose
w <
inf
I
. Then
w /
∈
I
because inf
I
is a lower bound for
I
.
Similarly, if
w >
sup
I
then
w /
∈
I
.
Therefore, we have
I
⊂
[inf
I,
sup
I
]. Altogether, we have
(inf
I,
sup
I
)
⊂
I
⊂
[inf
I,
sup
I
]
which leaves only the four given possibilities.
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 '08
 PROTSAK
 Topology, #, Empty set, Metric space, Closed set, Cauchy

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