Unformatted text preview: nterval x y , x y is included in
V. Using the fact that U is a neighborhood of y, we choose 0 such that
y , y
Now given any number t in the interval x y , x y we deduce from the fact that
xy t xy
y t x y
and therefore we know that t x U. Therefore, since
t xt x
we know that t V and we have shown that
x y , x y
V. Exercises on Open Sets and Closed Sets
1. Explain why if U is open and H is closed then the set U Solution: The set U H can be expressed as U H must be open.
R H which is the intersection of two open sets. 2. Explain why if U is open and H is closed then the set H U must be closed.
We observe that
which is the intersection of two closed sets. As we know, the intersection of closed sets is always
3. Give an example of an infinite family of open sets whose intersection fails to be open.
We saw earlier that
Ý 1, 1
nn n1 97 0 and, of course, the set 0 isn’t open.
4. Give an example of an infinite family of closed sets whose union fails to be closed.
Ý 0, 1 n1 1
n 0, 1 which is not closed.
5. Give an example of two sets A and B neither of which is open but for which the set A Þ B is open. Solution: Look at the union 0, 1 Þ 1, 2 . 6. Given a set H of real numbers, prove that the following conditions are equivalent:
a. The set H is closed.
b. For every number x R H it is possible to find a number 0 such that x Condition b can be expressed by saying for every number x R
0 such that
x , x
and this is exactly the condition that the set R H is open.
7. , x H . H it is possible to find a number a. Given any number x, prove that the singleton x is closed.
The fact that x is closed follows from the fact that the set R
x is the union of the two open
intervals Ý, x and x, Ý , the fact that an open interval is an open set and the fact that the
union of open sets is always open.
b. Use part a. and the fact that every finite set is a finite union of singletons to deduce that every finite set is
There isn’t really anything to add. The union of finitely many closed sets is always closed. 8. Given that Q H and that H is closed, prove that H R. Solution: The set R H is open and contains no rational number. To prove that R H must be
empty we shall observe that every nonempty open set must contain a rational number. Suppose that U is
open and nonempty. Choose x U and choose 0 such that x , x
U. Since the interval
x , x must contain rational numbers, so must the set U. 9. Given that H is closed, nonempty and bounded below, prove that H must have a least member.
We need to show that inf H belongs to H. Since H is closed we can show that inf H belongs to H by
showing that inf H is close to the set H. Suppose that 0. Since inf H inf H and since inf H is
the greatest lower bound of H we know that inf H must fail to be a lower bound of H. Using this
fact we choose a member x of H such that x inf H . We ob...
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