Last time we saw the proof of the Erdos-Rado theorem, that \(\left(2^{\aleph_0}\right)^+ \rightarrow (\aleph_1)_2^2\), modulo a few points where the proof requires some facts about cardinal arithmetic.

Addition, multiplication and exponentiation of cardinals is defined at the bottom of page 68. Everyone should look at this, and check that cardinal arithmetic is well-defined. Everyone should think about Exercise 2.29, as well, and **Sean** discuss it.

Exercise 2.30 (in full generality) is more difficult, but we only need two special cases:

- \(2^{\aleph_0} \cdot 2^{\aleph_0} = 2^{\aleph_0}\)
- \(2^{\aleph_0} + 2^{\aleph_0} = 2^{\aleph_0}\)

I challenge everyone to come up with direct proofs of both of these statements (remember that \(2^{\aleph_0}\) counts subsets of \({\mathbb N}\), and that there are easy injections from \(2^{\aleph_0}\) into both \(2^{\aleph_0} \cdot 2^{\aleph_0}\) and \(2^{\aleph_0} + 2^{\aleph_0}\), so by C-S-B it is enough to find injections from \(2^{\aleph_0} \cdot 2^{\aleph_0}\) and \(2^{\aleph_0} + 2^{\aleph_0}\) into \(2^{\aleph_0}\)). We can discuss both of these as a group.

The last tool that was needed in the proof of Erdos-Rado, but that we skipped over, was Lemma 2.34. **Anthony** can present this.