Unformatted text preview: S E is open in
the metric space S. Using the fact that E
choose a point x E. Since V is a neighborhood of x
and x S we know that S V
. This shows that S E
.
The same argument applied to the set S E shows that S E is a nonempty open subset of the
metric space S. Therefore the space S fails to be connected, as we promised.
22. True or false? If S is a bounded subset of a metric space X then the set S is also bounded.
This statement is true. Suppose that S is a bounded subset of a metric space X and choose a
number p such that the inequality d x, y
p holds whenever x and y belong to S. We shall now
observe that the ame inequality holds for all members x and y of S. In other words, diam S
p. 125 To obtain a contradiction, suppose that x and y belong to S and that d x, y p. We define
d x, y p.
Using the fact that x and y are close to S we choose members u and v of S such that u B x,
and v B y, . We observe that
2
d x, u d u, v d v, y
d u, v d u, v d u, v d x, y
2
2
which implies that d u, v p, contradicting the way in which p was chosen.
2 d x, y p 23. True or false? If S is a totally bounded subset of a metric space X then the set S is also totally bounded.
This statement is true. Suppose that S is a totally bounded subset of a metric space X and that
0. Using the fact that S is totally bounded we choose a positive integer n and points x 1 , x 2 , , x n
such that
n B S xj, j1 . 2 We shall now show that
n B xj, S . j1 Suppose that y S. Choose a member x of S such that d x, y
x B x j , 2 . Since
d y, x j
we see that y d y, x d x, x j 2 2 . Choose j such that 2 B xj, . 24. Prove that the closure of a convex subset of R k is also convex.
25. Suppose that S is a convex subset of R k , that x is an interior point of S, that y
that the point
1 t xy
is an interior point of S. S and that 0 t 1. Prove 26. This exercise refers to the notion of a subgroup of R that was introduced in an earlier exercise. That
exercise should be completed before you start this one.
a. Given that H and K are subgroups of R, prove that the set H K defined in the sense of
an earlier exercise is also a subgroup of R.
To prove that H K is a subgroup of R we need to show that H K is nonempty and that the
sum and difference of any members of H K must always belong to H K.
To show that H K is nonempty we use the fact that H and K are nonempty to choose x H
and y K. Since x y H K we have H K
.
Now suppose that w 1 and w 2 are any members of the set H K. Choose members x 1 and x 2 of
H and members y 1 and y 2 of K such that w 1 x 1 y 1 and w 2 x 2 y 2 . Since the numbers
x 1 x 1 and x 1 x 2 belong to H and the numbers y 1 y 2 and y 1 y 2 belong to K, and since
w1 w2 x1 x2 y1 y2
and
w1 w2 x1 x2 y1 y2
we see at once that w 1 w 2 and w 1 w 2 belong to H K.
b. Prove that if a, b and c are integers and if
a 2 b 3 c
then a b c 0. 126 Solution: From the equation
a 2 b 3 c
we see...
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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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