Real Analysis
Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au
Last update: 19 July 2014
Lecture 27
Let be a set.
A relation on is a subset of
A partial order on is a relation on such that
(a) |
if and
and
then
|
(b) |
if and
and then
|
A total order on is a relation on such that
(a) |
if and
and
then
|
(b) |
if and
and
then
|
(c) |
if then
or
|
If is a partial order on write
if in
(a partial order that is not a total order.) Let
and let
Let be the relation on given by
In other words, inclusion is a partial order on since
(a) |
if and
and then
and
|
(b) |
if and
and then
|
Inclusion is
not a total order since
and
are in
and
and
Let be a set with a partial order
Write if
Let be a subset of
An upper bound of is an element such that
if then
A lower bound of is an element such that
if then
A maximum of is an element such that
there does not exist such that
(i.e. if then
A minimum of is an element such that
if then
A supremum of is an element such that
(a) |
is an upper bound of and
|
(b) |
if is an upper bound of then
|
An infimum of is an element such that
(a) |
is a lower bound of and
|
(b) |
if is a lower bound of then
|
If
then and
are both maximums of and
Prove that
where
|
|
Proof. |
|
Proof by contradiction.
Assume
is a bijection
Let with
Then and does not appear in the
sequence
So is not surjective. This is a contradiction to being bijective. So
|
Let be a set and let be a partial order on
Write
Let Then let
where
means and
These sets are intervals in
Notes and References
These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100512Lect27.pdf and was given on 12 May 2010.
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