8/13/2019 Sp.metrice4
1/3
The University of Sydney
MATH 3901
Metric Spaces 2004
Tutorial 4
PROBLEM SET 4
1. If a sequence (xn) in a metric space Xis convergent and has
limitx, show thatevery subsequence (xnk) of (xn) is convergent and
has the same limit x.
Solution.
Let >0. Since xn x as n , there is N such that n > N
implies
d(xn, x)< .
Thus, ifnk > N, thend(xnk , x)< . Hence (xnk) x as k .
2. Let M be the metric subspace consisting of all sequences x=
(xk) withat most finitely many nonzero terms, where d is given
by
d(x, y) = supkN
|xk yk|.
Show thatMis not closed. [Hint: Try to produce a sequence in
Mwhich doesnot converge to a point in M.]
Solution.
For n= 1, 2, . . . , let
x(n) =
1, 1
2, . . . ,
1
n, 0, 0, . . .
.
Then (x(n)) is a sequence in M. Let x be defined by
x=
1, 1
2, . . . ,
1
n, . . .
.
Thenx butx / M. Moreover, for n > N, we see that
d(x(n), x) = sup 1
n+ 1,
1
n+ 2, . . .
=
1
n+ 10, we choose N >1/ so that for any n > N,
d(x(n), x)< 1
N+ 10, we
must show that there is a >0 such that ifx satisfies |x x0|
< , then
|f(x) f(x0)| < .
Now, for such , we consider /3. Since fn f, there is N such that
for alln N, d(fn, f)< /3; that is, for all x [a, b], we have for
all n N that
|fn(x) f(x)| < /3.
Since fN is continuous at x0, for such /3, there exists > 0
such that ifx
satisfies|x x0| < , then
|fN(x) fN(x0)| < /3 .
Hence ifx satisfies |x x0| < , then
|f(x) f(x0)| < |f(x) fN(x)| + |fN(x) fN(x0)| + |fN(x0)
f(x0)|
<
3+
3+
3= .
Hence fis continuous on [a, b].
