Unformatted text preview: a compressed subset S of the metric space R such that S has no limit point.
Look at the set
Z Þ n 1
n Z .
n
d. Is it true that if neither of two subsets A and B of a metric space X is compressed then the set A Þ B
cannot be compressed?
No! Again, look at the set
Z Þ n 1
n Z .
n
e. Prove that if A is a finite subset of a metric space X and B is a subset of X and B is not compressed then
the set A Þ B is not compressed.
Choose a number 1 0 such that whenever x and y belong to B and x y we have
d x, y
1 . We know that, since neither of the sets A and B has a limit point, nor does the set
A Þ B. Using the fact that A is finite and the fact that no member of A is a limit point of A Þ B,
choose a number 2 0 such that, whenever x A we have
AÞB
x.
B x, 2
We now define to be the smaller of the two numbers 1 and 2 and we observe that
whenever x and y belong to A Þ B and x y we have d x, y
.
17. We shall say that a metric space X is strongly complete if every compressed subset of X has a limit point.
a. Prove that every strongly complete metric space is complete.
Suppose that X is a strongly complete metric space and that x n is a Cauchy sequence in X.
From the Cauchy condition we see that, unless the sequence x n is constant from some point
on, the range of x n is compressed and has a limit point to which the sequence x n must
converge.
b. Prove that every compact metric space is strongly complete.
Every compressed set is infinite and every infinite subset of a compact space must have a limit
point. 236 c. Prove that the metric space Z of integers is strongly complete but not compact.
The metric space Z has no compressed subsets and therefore cannot have a compressed
subset that fails to have a limit point. Therefore Z is strongly complete. The fact that Z fails to
be compact is obvious: No subset of Z can have a limit point, even though Z is infinite.
d. Prove that if d is the discrete metric on an infinite set X then the metric space X, d is strongly complete
but not compact.
We can repeat, almost word for word, what we said about the space Z.
e. Improve the principal theorem on uniform continuity by proving that every continuous function from
a strongly complete metric space X to a metric space Y must be uniformly continuous on X.
a. Solution: Suppose that X is a strongly complete metric space and that f is a continuous function
from X to a metric space Y. To obtain a contradiction we shall assume that the function f fails to be
uniformly continuous on X. Choose a number 0 such that for every number 0 there exist
points x and t in the space X such that d t, x and d f t , f x
.
For every positive integer n we now choose points that we shall call t n and x n in X such that
d t n , x n and
.
d f tn , f xn
Now we define
S t n n 1, 2, Þ x n n 1, 2,
and, using the fact that S is compressed and the fact that the space X is strongly complete, we choose a
limit point u of the set S.
Using the fact that...
View
Full
Document
This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.
 Fall '08
 STAFF
 Math, Calculus

Click to edit the document details