Math 55 Problem Set 3
Neil Herriot with additions by Andrei Jorza
1. (i) d1(f, f ) = R1
0|f(x)−f(x)|dx =R1
00dx = 0. And, if f6=g, then
F(x) = |f(x)−g(x)|is not identically zero. Hence ∃x0so F(x0) = 2 >
0. And by continuity, ∃δsuch that ∀x, x0−δ < x < x0+δ, F (x)> . So,
d1(f, g ) = R1
0F(x)dx =Rx0−δ
0F(x)dx +Rx0+δ
x0−δF(x)dx R1
x0+δF(x)dx ≥
Rx0+δ
x0−δ, dx = 2δ > 0
0 = d1(f, g ) = R1
0|f(x)−g(x)|dx, implies |f(x)−g(x)|= 0 for all x,
and thus f=g.
(ii) d1(f, g ) is symmetric in definition, as |f−g|=|g−f|, and thus
equals d1(g, f ).
(iii) d1(f, h) +d1(h, g) = R1
0|f(x)−h(x)|+|h(x)−g(x)|dx ≥R1
0|f(x)−
g(x)|dx =d1(f, g ).
2. {fn}:∀ > 0,∃N(>1/) so that ∀n > N, x ∈Rwe have 0 <n
x2+n2≤
n
n2< . Thus, {fn}converges uniformly (and thus also point-wise) to
0 on all of R.
{gn}: For any given x, we can pick a sufficiently large N(>qx2
) so
that, n>Nimplies gnis close to, 1. Thus, {gn}converges point-wise.
But, for any n, if we pick x=nto get gn=1
2. This means that {gn}
does not converge uniformly to 1.
3. Let U={x:d(x, A)< d(x, B )}and Vdefined anaglgously, with d
the distance function defined in problem set 2 problem 1. Clearly, U
and Vare disjoint and contain Aand Brespectivly. Now, it remains
to show that they are open. Let x∈U,=1
3(d(x, B)−d(x, A)), and
y∈B(x). Then, d(y, B )−d(y, A)> d(x, B)−d(x, A)−2 > 0, and
we have y∈Uand Uopen. So Uand Vare as desired.
4. I claim that for U, V open, U⊂¯
U⊂V, there exists Wopen so that
¯
U⊂W⊂¯
W⊂V. To see this, consider ¯
Uand X−V, disjoint closed
sets. Then there exists W⊃¯
U,W0⊃X−Vopen and disjoint. But
then ¯
U⊂W⊂X−W0⊂X−V. Since X−W0is closed and contains
W, it also contains ¯
W. And we have U⊂¯
U⊂W⊂¯
W⊂V, as
desired.
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