Page 182 - DMTH503_TOPOLOGY

Topology

Notes  f(x) = 1  x  X – B ( X – B = H )
1
 f(x) = 1  x  B

 f(B) = {1}
Thus we have shown that f(A) = {0}, f(B) = {1}.
Step (ii): Secondly, we shall prove that f is continuous. Let a  R be arbitrary then [0, a) and (a, 1]
are open sets in R with usual topology.
–1
–1
Write G = f ([0, a)), G = f ((a, 1]).
1 2
Then G , G can also be expressed as
1 2
G = {x  X : f(x)  [0, a)}
1
= {x  X : 0  f(x) < a}
= {x  X : f(x) < a}
According to the construction of f

0  f(x)  1  x  X
G = {x  X : f(x)  (a, 1)}
2
= {x  X : a < f(x)  1}

= {x  X : a < f(x)}
= {x  X : f(x) > a}
Finally, G = {x  X : f(x) < a}, G = {x  X : f(x) > a}
1 2
H
We claim G  H , G   
t
2
1
1


t a t a
Any x  G  f(x) < a  x  H for some t < a
1 t
This proves that G  H
1 t

t a
x  G  f(x) > a  x is out side of H for t > a
t
2
H
 x    '
t

t a
H
Hence we get G =   '
t
2
t a

Since an arbitrary union of open sets is an open set and hence
H ,   '  X
H
t
t
t a t a


are open i.e., G , G  X are open, i.e.,
1 2
f ([0, a)), f ((a, 1]) are open in X.
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–1
 f is continuous
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