Let be a well ordered set. If
is nonempty,
has a minimal element, which is its maximal lower bound or infimum (inf
). If
is bounded above, it also has a minimal upper bound or supremum (sup
). If
, we define the initial segment of
to be
.
The elements of are called predecessors of
.
The Principle of Transfinite Induction. Let be a well ordered set. If
is a subset of
such that
whenever
, then
.
Proof. If , let
. Then
but $x\not\in A$.
Proposition. If is well ordered and
, then $\cup_{x\in A}I_x$ is either an initial segment or
itself.
Proof. Let $J=\cup_{x\in A}I_x$. If , let
. If $\exists y\in J$ with
, we would have
for some
and hence
, contrary to construction. Hence $J\subset I_b$, and it is obvious that
.
Proposition. If and
are well ordered, then either
is order isomorphic to
, or
is order isomorphic to an initial segment in
, or
is order isomorphic to an initial segment in
.
Proof. Consider the set $\mathcal F$ of order isomorphisms whose domains are initial segments in or
itself and whose ranges are initial segments in
or
itself. $\mathcal F$ is nonempty since the unique
belongs to
, and
is partially ordered by inclusion (its members being regarded as subsets of
). An application of Zorn’s lemma shows that $\mathcal F$ has a maximal element
, with (say) domain
and range
. If
and
, then
and
are again initial segments of
and
, and
could be extended by setting
, contradicting maximality. Hence either
or
(or both), and the result follows.
Proposition. There is an uncountable well ordered set such that
is countable for each
. If
is another set with the same properties, then
and
are order isomorphic.
Proof. Uncountable well ordered sets exist by the well ordering principle; let be one. Either
has the desired property or there is a minimal element
such that
is uncountable, in which case we can take
. If
is another such set,
cannot be order isomorphic to an initial segment of
or vice versa, because
and
are uncountable while their initial segments are countable, so
and
are order isomorphic.
Remark. is called the set of countable ordinals.
Proposition. Every countable subset of has an upper bound.
Proof. If is countable,
is countable and hence is not all of
. Then
such that
, and
is thus an upper bound for
.
The set of positive integers may be identified with a subset of
as follows. Set
, and proceeding inductively, set
. Then
is an order isomorphism from
to
, where
is the minimal element of
such that
is infinite.
It is sometimes convenient to add an extra element to
to form a set
and to extend the ordering on
to
by declaring that
for all
.
is called the first uncountable ordinal.
“Gratitude is the key to happiness. When gratitude is practiced regularly and from the heart, it leads to a richer, fuller and more complete life… It is impossible to bring more abundance into your life if you are feeling ungrateful about what you already have. Why? Because the thoughts and feelings you emit as you feel ungrateful are negative emotions and they will attract more of those feelings and events into your life.” ~ Vishen Lakhiani
To be continued tomorrow 🙂
References
[1] Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, 2ed, page 9-10.
[2] Purpose Fairy’s 21-Day Happiness Challenge, http://www.jrmstart.com/wordpress/wp-content/uploads/2014/10/Free+eBook+-+PurposeFairys+21-Day+Happiness+Challenge.pdf.