First, study todays handouts while the class is still fresh in your memory.
One of these handouts was about onto functions. If you got the quiz-question
about onto functions wrong, then turn it in as HW.

Next, read section 2.3.  It contains the proof of the Bernstein-Schroder theorem
which says that if a,b are cardinals with a <= b and b <= a then a = b.
Remember that the proof of that in todays class involved a tricky idea with
one-sided and two-sided sequences  (called cycles in the book).  Read this
proof (theorem 7,8,9 in the book).  It is not an easy proof, and it is OK
if you don't fully grasp it, but this is something you should see at least once.
Hopefully it will make sense if you read it with todays class still fresh in
your memory.

The proof of Theorem 10,11 is shorter, but this proof is even harder due to the
fact that it is very abstract compared with other math you have encountered thus
far. It is OK if this proof is too hard.
Instead of this proof, I will cover a simplified version of the proof, then run
into a technical issue, and then I'll explain the idea how to fix that issue
without going into the technical details.

For the turn-in HW, read Theorem 14 on page 41 and use it to give yet another
proof for 2.2 Ex 5.

Here are some hints: Let Q be the rational numbers and R be the real numbers.
Let D = Q  and  E = R - Q.
Let d = o(D) and e = o(E).
Now from the definition in the first lines of section 2.4 we get:

	(1)	d + e = c

where c = o(R).  Explain (turn in) why that is so.
Now Theorem 14 says that

	(2)	d + e = max(d, e)

if e is infinite (which it is).
But now we have to ask ourselves.  If we define

	(3)	m := max(d, e)

then m = d or m = e.

Now (also turn in) explain why m can not be equal to d.
(hint: combining (1)+(2) tells you what m is, and looking
at d = o(D) = o(Q) tells you what d is).

Then (also turn in) put this all together to prove that e = c.

At that point we have 3 (!) proofs for 2.2 Ex 5
(we only do the irrational numbers, we'll skip the transcendental numbers in Ex 5).

About the other two proofs that were due today, if you struggled with them but if
you understand them better now, it is OK to turn them in next time.

================================================

PS.  If you are wondering, when d,e are cardinal numbers, why should d,e have
    a maximum?   Why should m in line (3) above be well defined?  Well, if you
    wondered about that, then that is a very good sign, it means you paid attention
    to the details!   The answer to that question is: Theorem 10,11.