A collection of subsets of
is called an independent family iff for any pairwise distinct
in
, we have that the set
is infinite.
- Show that there is an independent family of the same size as the reals. (This means that there is an independent family
for which there is a bijection between
and
. If it is easier to think of it this way, you can use that
has the same size as the set
of infinite sequences of zeros and ones.)
- Given a set
recall that its characteristic function is the map
such that
if
and
if
We can think of each
as a vector in the space
Suppose now that
is an independent family, and show that
is a linearly independent set. (Thus any basis for
over
must have the same size as the reals.)
- Consider
as a vector space over
and show that any basis must have the same size as the reals.
Here is a `hint’ for the first problem: If
is countable, it clearly suffices to find an independent family as required, but consisting of subsets of
(so we replace each
with
of course).
Take
is finite, and
Check that
is countable.
Now, given
let
The claim is that
is as wanted.
Check that the assignment
is injective.
To show that
satisfies the stronger independence requirement, suppose
are infinite, pairwise distinct, subsets of
Write
For each
and
let
Show that for any finite
there is an
such that 
Hmm… Here is an easier way of arguing about the second problem (easier meaning that we do not need problem 1):
An almost disjoint family is a collection of subsets of
that are infinite, but the intersection of any two of them is finite. It is much easier (I think) to come up with an almost disjoint family of the same size as
than it is to come up with an independent family.
For example: For each real
fix a sequence of (pairwise distinct) rationals converging to
and let
be the range of this sequence. Then
is almost disjoint. (And, of course,
is countable.)
To solve problem 2, one could start with an almost disjoint (rather than independent)
For problem 3: a. If you assume the Continuum Hypothesis
and answer the problem under this additional assumption, you get some partial credit, but to get full credit you need to answer the question independently of whether
holds.
b. In case you need this, you may use that all bases have the same size. Again, one can solve the problem without appealing to this fact, but feel free to use it if you need it.
c. Also, in case you need it, feel free to assume the Schröder-Bernstein theorem: If
and
are sets and we have injections from
into
and from
into
then
and
have the same size (i.e., there is a bijection between the two of them). Again, the problem can be solved without using this fact, but feel free to use it if you see the need.
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