Unformatted text preview: e conclude that
L AÞB L A ÞL B .
7. Is it true that if A and B are subsets of a metric space X then
LA B LA
What if A and B are closed? What if A and B are open? What if A and B are intervals in R?
The answers are no, no, no and no. Look at the following example: 129 A 0, 1 and B 1, 2
These two sets are closed in R and
LA B L 1 while
L B 0, 1
Now look at the following example:
A 0, 1
In this case
LA B L
L B 0, 1 1, 2 1 .
B 1, 2 .
1, 2 1 . 8. Is it true that if D is a dense subset of R then L D R?
The assertion is true. Suppose that D R. We know that whenever a and b are real numbers and
a b there must be members of D lying between a and b. Now suppose that x is a real number. To
show that x is a limit point of D, suppose that 0. Since there must be members of D in the
interval x, x we conclude that the set x , x
x is nonempty.
9. Is it true that if D is a dense subset of a metric space X then L D X?
No! Consider, for example, a metric space like 1, 2, 3 that has no limit points at all. The set
1, 2, 3 is dense in this space.
10. Is it true that if D is a dense subset of a connected metric space X then L D X?
Yes, as long as the space X contains at least two different points then if X connected and D is
dense in X we must have L D X. As a matter of fact, we can do better than this. The condition
that X be connected says that X has no open closed subsets other than and X. All we actually
need to know is that no singleton in X can be an open set.
Suppose that X is a metric space in which no singleton is open. Suppose that D is a dense subset
of X. Suppose that x X and, to show that x is a limit point of D, suppose that 0. Since the ball
B x, is open and x is not open we know that B x,
x , in other words, B x,
x B x,
which is open, the set B x,
x , being a nonempty open subset of X, must intersect with the set
11. Is it true that if X is a metric space and L X X then for every dense subset D of X we have L D X?
The answer is yes. See the solution to Exercise 10.
12. Given that a set S of real numbers is nonempty and bounded above but that S does not have a largest
member, prove that sup S must be a limit point of S. State and prove a similar result about inf S.
To show that sup S is a limit point of S, suppose that 0. Since sup S sup S and since sup S is
the least upper bound of S the number sup S fails to be an upper bound of S. Choose a member
x of S such that sup S x. Since x sup S and since sup S does not belong to S we have x sup S.
We conclude that
sup S , sup S
13. Given that S is a closed subset of a metric space X and that every infinite subset of X has at least one limit
point, prove that every infinite subset of S must have at least one limit point that belongs to S.
Every infinite subset of S, being an infinite subset of X, must have a limit point somewhere in X....
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