Page 82 - DMTH503_TOPOLOGY

Topology

Notes  1  1
   f (G)  f (G)

1
1



But f (G)  f (G) is always for C  C  C 




Combining the last two, [f (G)]° = f (G)
–1
–1
 f (G) is open in X.
-1
Example 5: Let f : R  R be a constant map.
Prove that f is continuous.
Solution: Let f : R  R be a map given by
f(x) = c  x  R. …(1)
Then evidently f is a constant map.
To show that f is continuous.
Let G  R be an arbitrary open set.

-1
By definition, f (G) = x R : f(x) G   …(2)

 R if c G,
1

From (1) and (2), f (G)  

  if c G,
 and R both are open sets in R and hence f (G) is open in R.
-1
Given any open set G in R, we are able to show that f (G) is open in R. This proves that f is a
-1
continuous map.
Example 6: Let T and U be any two topologies on R. Let
f : (R, T)  (R, U)
be a map given by f(x) = 1  x  R.

Then show that f is continuous.
Hint: take C = 1. Instead of writing
“Let G  R be an open set”, write

-1
“G  U and f (G)  T”.
Do these changes in the preceding solution.

7.2 Summary

 Let f : (X, T)  (Y, U) be a map.
The map f is said to be continuous at x  X is given any U open set H containing f(x ), a
0 0
T-open set G containing x s.t. f(G)  H.
0
If map is continuous at each x  X, then the map is called a continuous map.
 A function f : (X, T)  (Y, U) is said to be continuous on a set A  X if it is continuous at each
point of A.

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