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Unformatted text preview: al 3 , 3 which is included in the
interval 1, 3 . Since all other members of this interval are obviously interior points, this interval is
open in the metric space 0, 3 .
10. Given that S is a nonempty subset of a metric space X, d and that U is an open subset of the metric space
X, d and that U S, prove that U must be open in the metric space S, d .
Suppose that x U. Using the fact that U is open in the space X, d , choose a number 0 such
that B x,
U. In the metric space S, d , the ball center x with radius is S B x, and since
S B x,
B x,
U
we have shown that x is an interior point of the set U in the metric space S, d .
11. Suppose that S is a nonempty subset of a metric space X, d and that U
conditions are equivalent: S. Prove that the following two a. The set U is open in the metric space S, d .
b. There exists an open subset V of the metric space X, d such that
U S V. Solution: First we prove that condition b implies condition a. Suppose that condition b holds and choose an open subset V of the metric space X such that U S V. To prove that the set U is
open in the subspace S, suppose that x U. Using the fact that x V and the fact that V is open in the
metric space X we now choose 0 such that
B x,
V
and we observe that
y S d x, y
S V.
Since the set y S d x, y is the ball center x with radius in the metric space S, we have
shown that the set U is open in the metric space S.
Now we want to prove that condition a implies conditio b. Suppose that condition a holds. In other
words, suppose that the set U is open in the metric space S. We know that for every point x U it is
possible to find a positive number such that
B x,
S U.
We now define V to be the union of all the balls of the form B x, for which x U and 0 and
B x,
S U.
The set V, being a union of balls in the space X, must be open in X. Finally, we need to explain why
U V S.
U we can choose 0 such that
B x,
SU
B x, we have x V. Now given any point x
and since x Now suppose that x V S. Using the definition of the set V we choose y 119 U and 0 such that B y,
and x B y, . Thus x B y, S S U U. 12. Given that S is an open nonempty subset of a metric space X, d and that U is an open subset of the metric
space S, d , prove that U must be open in the metric space X, d .
Using Exercise 11 we choose a set V that is open in the metric space X, d such that U V S.
Since both of the sets S and V are open in the space X, d , so is their intersection, which is U.
13. Given that S is a nonempty subset of a metric space X, d and that H is a closed subset of the metric space
X, d and that H S, prove that H must be closed in the metric space S, d .
From Exercise 11 and the fact that the set X H is open in the metric space X, d we conclude that
the set S
X H is open in the metric space S, d . Since
S
X H S H
we conclude that S H is open in the space S, d which tells us that H is closed in the space S, d .
14. Given that S is a closed nonempty subset...
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 Fall '08
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 Math, Calculus

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