Unformatted text preview: me up with an example in which the space X is connected. In fact, if
X 0, 1 then X is connected and, if we define
f x x, 0
for all x X, then f fails to send open sets to open subsets of R 2 .
7. Suppose that X is a metric space, that k is a positive integer and that f j : X 220 R for each j 1, 2, , k. Suppose that
f x f 1 x , f 2 x , , f k x
X. Prove that the function f is continuous on X if and only if each of the functions f j is for every point x
continuous on X.
This exercise follows almost at once from the corresponding fact about limits. 8. Suppose that a and b are real numbers, that a b and that f is a function from the metric space a, b to a
metric space Y. Prove that the following conditions are equivalent:
a. The function f is continuous at the number a.
b. For every number 0 there exists a number 0 such that for every number x in the interval a, b
that satisfies the inequality x a we have d f x , f a .
This exercise is obvious because of the fact that x a  x a whenever x
a, b . You can
convert this exercise into one that is a shade more interesting by removing the words “in the
interval” from condition b and replace them by the condition x a. This change would require a
slightly more careful choice of to ensure that it does not exceed b a so that any number
x a that lies within a distance of a would automatically belong to the interval a, b .
9. Prove that if a metric space X has no limit point and f is a function from X to a metric space Y then f must be
continuous on X.
This result follows at once from the fact that a function is always continuous at a point in its domain
if that point is not a limit point of the domain.
10. Suppose that f is a continuous function from a closed subset H of a metric space X into a metric space Y and
suppose that x n is a convergent sequence in the set H. Prove that the sequence f x n is convergent in the
space Y. Is this conclusion still valid if we don’t assume that H is closed? Solution: Suppose that f is continuous on a closed set H and that x n is a convergent sequence in H. We define
x n Ý xn.
lim
Since H is closed we must have x H and therefore f is continuous at the number x. Therefore the fact
f x as n Ý.
that x n x as n Ý guarantees that f x n
To see why the assertion does not remain true without the assumption that H be closed we look at the
example in which
H 1
n Z
n
and we define
f1
n 0 if n is even
1 if n is odd Since no member of the set H is a limit point of H we know that f is continuous at every member of H.
However, in spite of the fact that the sequence 1/n converges (to 0) in the metric space R, the sequence
f 1/n fails to converge.
Actually, much more can be proved: If H is any subset of a metric space X and H fails to be closed then
there exists a convergent sequence x n in H and a continuous function f on H such that the sequence
f x n fails to converge. This stronger assertion is harder to prove and we omit the proof at this point.
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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

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