I have been struggling to understand how to prove the cardinality of various sets. I know the definition of two sets 'having the same cardinality' as being the fact that there exists a bijection between the two (or using Cantor-Bernstein Theorem, an injection in both directions).
I however, am unable to follow some of the proofs in lectures. Since I believe them to be quite short, and very similar, I have included them both in this question. And although there may be similar questions already on SE I am asking particular questions about two proofs given in my real analysis lectures.
The proof given (possibly incomplete) as presented in the lecture. I struggle with following the proof and have included various comments explaining my questions in more detail.
1. Let $A$ be infinite and $B$ be countable or finite. Then #$(A \cup B)=$#$A.$
Proof: (Possibly incomplete)
Assume union is disjoint. What precisely does this mean and why may we assume this?
Then $(A\smallsetminus C )\ \cup C$ is countable. What is being done here?
So $(A\smallsetminus C) \ \cup (C\cup B)$.
Then C is countable so we may replace it with $\mathbb N$ WLOG and replace $C$ with even natural numbers and $B$ with odd natural numbers WLOG, and thus there is a bijection from $N$ to $C\cup B$.
I do not understand this proof at all, unlike most other proofs we have covered. I cannot see how it has proved anything so would be grateful for explanation of this proof, or a statement of a similar proof. I have searched online but have had difficulties finding one.
2. An example following on from (1): #$[0,1] =$ #$[0,1)$
$[0,1] = [0,1) \cup \{1\}$
$[0,1] = ([0,1) \smallsetminus C) \cup C$
$C = \{ \frac{1}{n}\}^{\infty}_{n=1}$
$[0,1] = ([0,1) \smallsetminus C ) \cup \{ \frac{1}{n}\}^{\infty}_{n=2}$
Then $\varphi(x) = {x, x\in[0,1]} \smallsetminus \{ x_{n}\}^{\infty}_{n=1} $.
$\varphi(x_{n}) = x_{n+1}$ if $x_{n} = \frac{1}{n}.$ This is equivalent to the bijection $f: \mathbb N \to \mathbb N$, with $f(x) = x+1$. This is the same as the Hilbert Hotel.
I do not understand this proof at all; namely since I do not understand (1) and since so little detail is given. Any exposition in more detail of (2) and (1) would be most helpful.
Unfortunately both proofs are so muddled in their given state, be it due to your lecturer's poor exposition or your not copying down every detail he presented, that I'm not sure which proof was intended. So instead of filling in the gaps, I'll start over.
For theorem 1 $A\cup B=A\cup (B\backslash A)$, so without loss of generality $A\cap B=\emptyset$. If $A\backslash B$ is finite, say of size $n$, $A\cup B=(A\backslash B)\cup B$ is countable; just add $n$ elements to the front of an enumeration of $B$. If on the other hand $A\backslash B$ is infinite, it has a countable subset, say $D$. Then $A\cup B=(A\backslash D)\cup (D\cup B)$. Replacing the $n$th element of $D$ in $A\backslash D$ with the $n$th element of $D\cup B$ (for some enumeration of each), $A\cup B$ has an immediate bijection with $A$. So the theorem reduces to $D\cup B$ being countable, which follows from alternating elements of $D$ and $B$.
Theorem 2 is the special case of theorem 1 with $A=[0,\,1),\,B=\{ 1\}$.