(Note that ‘well ordering’ is an important general idea, but ‘special chains’
are an ad hoc notion for this particular proof. Note also that if
C
is a special
chain and
x
∈
C
then
C
x
is a special chain.)
The key point is that, if
K
and
L
are special chains, then either
K
=
L
or
K
=
L
x
for some
x
∈
L
or
L
=
K
x
for some
x
∈
K
.
Subproof
If
K
=
L
, we are done. If not, at least one of
K
\
L
and
L
\
K
is
non-empty. Suppose, without loss of generality, that
K
\
L
6
=
∅
. Since
K
is
well ordered,
x
= min
K
\
L
exists. We observe that
K
x
⊆
L
. If
K
x
=
L
, we
are done.
We show that the remaining possibility
K
x
6
=
L
leads to contradiction.
In this case,
L
\
K
x
6
=
∅
so
y
= min
L
\
K
x
exists. By definition of
y
and
the fact that
K
x
⊆
L
, we have
K
y
=
L
y
. But
K
and
L
are special chains so
y
=
κ
(
K
y
)
∈
K
contradicting the definition of
y
.
End subproof
We now take
S
to be the union of all special chains.
Using the key
observation, it is routine to see that:
(i)
S
is a chain.
(If
a, b
∈
S
, then
a
∈
L
and
b
∈
K
for some special
chains. By our key observation, either
L
⊇
K
of
K
⊇
L
. Without loss of
generality,
K
⊇
L
so
a, b
∈
K
and
a
”
b
or
b
”
a
.)
(ii) If
a
∈
S
, then
S
a
is a special chain. (We must have
a
∈
K
for some
special chain
K
.
Since
K
⊆
S
, we have
K
a
⊆
S
a
.
On the other hand, if
b
∈
S
a
then
b
∈
L
for some special chain
L
and each of the three possible
relationships given in our key observation imply
b
∈
K
a
. Thus
S
a
⊆
K
a
, so
S
a
=
K
a
and
S
a
is a special chain.)
(iii)
S
is well ordered. (If
E
is a non empty subset of
S
, pick an
x
∈
E
. If
S
x
∩
E
=
∅
, then
x
is a minimum for
E
. If not, then
S
x
∩
E
is a non-empty
subset of the special, so well ordered chain
S
x
, so min
S
x
∩
E
exists and is a
minimum for
E
.)
(iv)
S
is a special chain. (If
x
∈
S
, we can find a special chain
K
such
that
x
∈
K
. Let
y
=
κ
(
K
). Then
L
=
K
∪ {
y
}
is a special chain. As in (ii),
S
y
=
L
y
, so
S
x
=
L
x
and
κ
(
S
x
) =
κ
(
L
x
) =
x
.)
We can now swiftly obtain a contradiction. Since
S
is well ordered
κ
(
S
)
exists and does not lie in
S
. But
S
is special, so
S
∪
κ
(
S
) is, so
S
∪
κ
(
S
)
⊆
S
,
so
κ
(
s
) lies in
S
. The required result follows by reductio ad absurdum
5
.
Lemma 36 (Hammel basis theorem).
(i) Every vector space has a basis.
(ii) If
U
is an infinite dimensional normed space over
F
(with
F
=
R
or
F
=
C
) then we can find a discontinuous linear map
T
:
U
→
F
.