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13. Let A subset X; let f : A arrow Y be continuous; let Y be Hausdorff. Show that if f may be
extended to a continuous function g : A arrow Y, then g is uniquely determined by f.
Solution
Let f(x) and g(x) be two continuous functions on clA such that they agree on A . Let
C={x?clA:f(x)=g(x)} .
Then, in particular, A?C . According to Theorem 18.4 (f,g) is continuous, and C=(f,g)?1(?) ,
where ?={(y,y)|y?Y} . According to Exercise 17.13 ? is closed in Y iffY is Hausdorff. Since it is
closed, C is closed.
Therefore, clA?C?clA.
The requirement "if it may be extended" is needed, of course, because not every continuous
function can be extended onto the closure of the domain:
for example, 1/x defined on R+ cannot be extended onto closure of R+ .

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13. Let A subset X; let f A arrow Y be continuous; let Y be Hausdor.pdf

  • 1. 13. Let A subset X; let f : A arrow Y be continuous; let Y be Hausdorff. Show that if f may be extended to a continuous function g : A arrow Y, then g is uniquely determined by f. Solution Let f(x) and g(x) be two continuous functions on clA such that they agree on A . Let C={x?clA:f(x)=g(x)} . Then, in particular, A?C . According to Theorem 18.4 (f,g) is continuous, and C=(f,g)?1(?) , where ?={(y,y)|y?Y} . According to Exercise 17.13 ? is closed in Y iffY is Hausdorff. Since it is closed, C is closed. Therefore, clA?C?clA. The requirement "if it may be extended" is needed, of course, because not every continuous function can be extended onto the closure of the domain: for example, 1/x defined on R+ cannot be extended onto closure of R+ .