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 disjointfamily 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 Hypothesisand 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.