18.100B/C: Fall 2008
Solutions to Homework 7
1.(a)
For a given
ε
and
x
,
δ
f
(
x, ε
) is the supremum of a set of positive values. Thus,
δ
f
(
x, ε
)
>
0
if and only if this set is nonempty. So
δ
f
(
x, ε
)
>
0 if and only if there exists some
δ >
0 such that
for all
t
∈
X
,
d
X
(
x, t
)
< δ
implies
d
Y
(
f
(
x
)
, f
(
t
))
< ε
. That this statement is true for every
ε >
0
is exactly the definition of continuity of
f
at
x
, so we’re done.
(b)
First suppose that
f
is unformly continuous and choose
ε >
0.
Then there exists some
δ >
0 such that for any
x, y
∈
X
we have
d
Y
(
f
(
x
)
, f
(
y
))
< ε
whenever
d
X
(
x, y
)
< δ
. It follows that
δ
f
(
x, ε
)
≥
δ
for every
x
∈
X
, so inf
x
∈
X
δ
f
(
x, ε
)
≥
δ >
0.
Conversely, suppose inf
x
∈
X
δ
f
(
x, ε
)
>
0, and define
δ
=
1
2
·
inf
x
∈
X
δ
f
(
x, ε
).
It follows that
0
< δ < δ
f
(
x, ε
) for every
x
∈
X
. By the properties of the supremum, this means that for each
x
∈
X
there exists some
c > δ
such that
c
belongs to the set of which
δ
f
(
x, ε
) is the supremum,
i.e.
d
X
(
x, t
)
< c
implies
d
Y
(
f
(
x
)
, f
(
t
))
< ε
for every
t
∈
X
. Since
δ < c
we have that
{
t
∈
X

d
X
(
x, t
)
< δ
} ⊂ {
t
∈
X

d
X
(
x, t
)
< c
}
and thus
d
X
(
x, t
)
< δ
implies
d
Y
(
f
(
x
)
, f
(
t
))
< ε
for all
x, t
∈
X
. But this is exactly the definition of uniform continuity for
f
, so we’re done.
(We introduced a factor of
1
2
in this last paragraph not suggested by the hint. It may be possible
to set
δ
= inf
x
∈
X
δ
f
(
x, ε
) directly, without the
1
2
factor, and still produce a valid proof. The one
problem to watch out for is the following situation: for fixed
ε >
0, what if
δ
f
(
x, ε
) =
d
is constant
but
d
negationslash∈ {
δ >
0
 ∀
t
∈
Xd
X
(
x, t
)
< δ
=
⇒
d
Y
(
f
(
x
)
, f
(
t
))
< ε
}
for some
x
(i.e., the supremum is not
achieved)?)
(c)
For any
x, t
∈
X
we have

f
(
x
)
−
f
(
t
)

=

x
2
−
t
2

=

x
−
t
 · 
x
+
t

< δ
·
(2
x
+
δ
)
,
so 2
xδ
+
δ
2
is an upper bound for
{
f
(
x
)
−
f
(
t
)
 
t
∈
X, d
X
(
x, t
)
< δ
}
. It addition, the limit as
t
approaches
x
+
δ
from below of

f
(
x
)
−
f
(
t
)

is equal to
δ
·
(2
x
+
δ
), so it is in fact the least upper
bound, as desired.
(d)
It follows from part (c) that
δ
f
(
x, ε
) is the supremum of the set of
δ
such that 2
xδ
+
δ
2
< ε
.
This supremum will simply be the larger of the two roots of the equation 2
xδ
+
δ
2
=
ε
, and one can
quickly solve the quadratic to find that this root is
δ
f
(
x, ε
) =
√
x
2
+
ε
−
x
. Since we have for every
x
∈
X
that
δ
f
(
x, ε
)
2
+ 2
xδ
f
(
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 Fall '10
 Prof.KatrinWehrheim
 Supremum, Metric space, dx, Uniform continuity, infimum

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