PROVING COMPLETENESS OF THE HAUSDORFF INDUCED
METRIC SPACE
KATIE BARICH
WHITMAN COLLEGE
Acknowledgements
I would like to acknowledge Professor Russ Gordon and Professor Pat Keef for
their advice and g
Math421
2008 Spring
Midterm Exam
May 5, 7:00 pm
Room 404
Caution: You must show all works. Otherwise you will not receive a full credit. Without
proofs, you can use any theorems in the textbook and an
Homework 6
Choi SeongJin
December 15, 2016
1. Determine the components and path components of R .
pf. Let ~x = (xi )iN , ~y = (yi )iN be elements of R . Then : [0, 1] R
defined by
(t) := (1 t)~x + t~y
Introduction
Tietze Extension Theorem
Proof of Tietze Extension Theorem
Appendix
Lecture : Tietze Extension Theorem
Dr. Sanjay Mishra
Department of Mathematics
Lovely Professional University
Punjab, I
Math 421 2011 Fall
General Topology
Mldterm Exam November 4
Caution: You must show all works. Otherwise you may not receive full eredttsv Without
proofs, you can use any theorems m the tertbook and
Homework 4 Solutions
Problem 13.4.
Proposition. If cfw_T is a family of topologies on X, then
T
T is a topology on X.
Proof. Since and X are in each T , they must also be elements of
then C T for all
Homework 8
Choi SeongJin
December 8, 2016
1. Show that a compact metric space X is 2nd countable.
pf. For given q Q+ , there exists finite number of open balls of radius
q which covers X. Denote Bq by
Math 421 F. l E 2011 Fall
General Topology Ina xam December 21
Caution: You must show all works. Otherwise you may not receive full credits. Without
proofs, you can use any theorems in the textbook an
Homework 5
Choi SeongJin
November 30, 2016
2. Suppose A1 , A2 , . . . are connected subspaces of a space X such that
An An+1 6= for each n. Show that X = An is connected.
pf. Suppose X = C D is a sepa
Homework 7
Choi SeongJin
December 8, 2016
3. Suppose (X, d) is a metric space. If f : X X satisfies d(x, y) =
d(f (x), f (y) for all x, y X, then f is called an isometry.
a. Show that an isometry is i
Chapter 1
7. If (E F)c occurs, then E F does not occur, and so
E does not occur (and so Ec does); F does not occur
(and so Fc does) and thus Ec and Fc both occur.
Hence,
1. S = cfw_(R, R), (R, G), (R,
23
Answers and Solutions
Solving for E[N] yields
E[N] =
2 + 2q + q2 /p
1 q2
24. In all parts, let X denote the random variable whose
expectation is desired, and start by conditioning on
the result of
34
Answers and Solutions
(c) If we fix a set of k of the xi and require them
to be the only zeros, thenthere are by (b)
n1
such
(with m replaced by m k)
mk1
m
n1
solutions. Hence, there are
k
mk1
27
Answers and Solutions
1
43. E[T|2n ] =
2n /n
E[T
2
|2n ]
E[Z|2n ] =
1
2n /n
E[Z] = 0
n
n
n
= 2 E[Z2 |2n ] = 2 E[Z2 ] = 2
n
n
n
Hence, E[T] = 0, and
n
Var(T) = E[T 2 ] = E 2
n
n
1 x /2
(x/2) 2
31
Answers and Solutions
(b) Conditioning on the types and using that the
sum of independent Poissons is Poisson gives
the solution
Pcfw_5 = (.18)e4 45 /5! + (.54)e5 55 /5!
+ (.28)e6 65 /5!
67. A run
26
Answers and Solutions
(g) Yes, knowing for instance that i + 1 is the last
of all the cards 1, , i + 1 to be seen tells us
nothing about whether i is the last of 1, , i.
Hence,
E[X] = nE[U]
= E[X 2
30
Answers and Solutions
to the conditional probability that a player about
to compete against the person who won the last
round is the overall winner.
Hence,
P(A) = (1/2)P(A|W) + (1/2)P(A|L)
Also,
=
32
Answers and Solutions
Pcfw_system works
= p3 (p1 + p2 p1 p2 )(p4 + p5 p4 p5 )
+ (1 p3 )(p1 p4 + p2 p5 p1 p4 p2 p5 )
75. (a) Since A receives more votes than B (since a > a)
it follows that if A is
23
Answers and Solutions
Solving for E[N] yields
E[N] =
2 + 2q + q2 /p
1 q2
24. In all parts, let X denote the random variable whose
expectation is desired, and start by conditioning on
the result of
12.4.).
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Ld~ +N EN MI
=> .
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19 3%, :Vg'xio r=r=_
=r/q 30
1.3% we! 91 e/v,
-'-S(o(+N)=o 1m M/N -> (NV t: a f