Fall 2013 Math 451
Homework Solutions (2)
12. (Exercise 1.2) Let d be a metric on the set A. Prove: For all a, b A,
d(a, b) = d(b, a).
Proof. Let a, b A. We know that for all c,
d(a, b) + d(a, c) d(b, c)
Replacing c with a, we have
d(a, b) + d(a, a) d(b,
Fall 2013 Math 451
Exam One
In Class Portion
You are only expected to set up the skeleton structure for each of the following
proofs. If you cannot set up the skeleton of a proof, at the very least, provide
the denition of the phrase in bold italics.
1. L
Fall 2013 Math 451
Homework Solutions (3)
15. (Exercise 1.5) Let d0 be the discrete metric and let Y be any other set
with metric d. Prove: every function f : X Y is continuous.
Proof. Let f : X Y be any function, let a X, and let
we choose = 1/2 and we h
Fall 2013 Math 451
Homework Solutions (1)
For Problems 1 to 6, assume f : X Y with W, W X and Zj , Z
Y.
1. Prove: f (W W ) = f (W ) f (W ).
Proof. First, we will show f (W W ) f (W ) f (W ). Let y
f (W W ), then there exists w W W with f (w) = y.
Case 1
Fall 2013 Math 451
Exam One
In Class Portion
You are only expected to set up the skeleton structure for each of the following
proofs. If you cannot set up the skeleton of a proof, at the very least, provide
the denition of the phrase in bold italics.
1. L
Fall 2013 Math 451
Exam Two
1. Prove that the following denition of U denes a topology on R.
U U x U, n Z+ , x + n U
Proof. U is vacuously true and R U because for all x R and
n Z+ , x + n R.
If U, V U, and x U V , then for all n Z+ , x U forces x + n U
a
Fall 2013 Math 451
Homework Solutions (4)
20. 2.3(a) Before doing the proof, here are a couple of things to note about
the sets in U. First, they are nested. This means that if U1 , U2 U,
then either U1 U2 or U2 U1 . Second, if x < y U U, then
x U.
(i) U
Fall 2013 Math 451
Homework Solutions (7)
33. 6.2(a) If X1 X2 and Y1 Y2 , then X1 X2 Y1 Y2 .
=
=
=
Proof. Because X1 X2 , there exists a one-to-one, onto, bicontinuous
=
function fX : X1 X2 and because Y1 Y2 , there exists a one-to-one,
=
onto, bicontinuo
Fall 2013 Math 451
Homework Solutions (9)
39. 9.8(a) The largest connected subset of Q is a point and so Q is disconnected.
Proof. Assume that A Q and A contains two distinct points a, b A,
without loss of generality, assume a < b. Then there exists an ir
Fall 2013 Math 451
Homework Solutions (6)
28. 4.2
1
Fall 2013 Math 451
Homework Solutions (6)
29. 4.5(a)
Let UX be the topology on X. The subspace topology on Y X
is UY = cfw_Y U : U UX . The subspace topology on Z Y is
UZY = cfw_Z U : U UY . The subspace