Page 194 - DMTH503_TOPOLOGY

Topology

Notes

f = f  (g  g  ... g (n 1) )

0
0
1
n

n 1
Write S =  g r
n

r 0
Now S can be regarded as partial sums bounded continuous function defined on X. Since the
n
space of bounded real valued function is complete and
2

1   n  1 2  n
g (x)  .   and     1,
n

3   3 3 
3

n 0
the sequence S  converges confirmly on X to g (say) when |g(x)|  1.
n
n
 2 
f (x)     S  converges uniformly on F to f say
n  3  n 0
Hence g = f on F.
Thus g is a continuous extension of f to X which satisfies the given conditions.
21.2 Summary
 Tietze extension theorem:
Suppose (X, ) is a topological space. The space X is normal iff every continuous real
function f defined on a closed subspace F of X into a closed interval [a, b] has a continuous
extension f* X [a, b]
21.3 Keywords

Closed Set: A subset A of a topological space X is said to be closed if the set X – A is open.
Continuous Map: A function f : R R is said to be continuous if for each a R and each positive

real number , there exists a positive real number  such that x a    f(x) f(a)   .
Normal Space: A topological space (X, T) is said to be a normal space iff it satisfies the following
axioms of Urysohn: If F ad F are disjoint closed subsets of X then there exists a two disjoint
1 2
subsets one containing F and the other containing F .
1 2
21.4 Review Questions

1. Show that the Tietze extension theorem implies the Urysohn lemma.
2. Let X be metrizable. Show that the following are equivalent:

(a) X is bounded under every metric that gives the topology of X.
(b) Every continuous function : X R is bounded.

(c) X is limit point compact.

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