Power sets and Cardinalities.

Cardinality

The cardinality of a set, roughly speaking, is the number of elements it has. For a set like \(A=\{1, 2, 3\}\) the cardinality of \(A\) (written as \(|A|\)) is \(3\) or \(|A|=3\). This concept is fine until you consider an infinite set.

We cannot assign numbers to infinite quantities however we can describe their cardinalities it terms of size. They may all be infinite but some are more "infinite" than others!

For two sets to have the same cardinality $|A|=|B|$ there has to be a bijection (one-to-one and onto function) $f:A\to B$. This means that for every element of $A$ there is exactly one corresponding element of $B$. We do not need to prove the map from $B$ to $A$ as the inverse of a bijective function is itself a bijection.

Example: $A=\{1,2,3\}$ and $B=\{2, 4, 6\}$ now if we find a bijection we have $|A|=|B|$. The bijection $f:x\mapsto 2x$ is such a function.Therefore $|A|=|B|$.

Power sets

The power set of a set $A$ is the set of all possible subsets of $A$. E.g. $A=\{1, 2, 3\}$ then the power set is $\mathscr{P}(A)=\{\{1, 2, 3\}, \{1, 2\}, \{2, 3\}, \{3, 1\}, \{1\},\{2\}, \{3\}, \varnothing\}$

Clearly $A\in \mathscr P(A)$ for all $A$. Now using simple permutations we have the cardinality of a sets power set is as follows $|\mathscr P(A)|=2^{|A|}$.

Back to cardinalities

Let's discuss infinite sets. There are a lot of infinities however we shall start from the smallest. Countable infinity or $\aleph_0$ (pronounced aleph naught) is a type of infinity which is the cardinality of the natural numbers ($\Bbb N$). The name comes from the fact that for a set $A$to have the same cardinality as $\Bbb N$ there has to be a bijection $f:\Bbb N \to A$ as previously established. In a sense, each element in $A$ can be counted using a rule. Sets with the cardinality $\aleph_0$ include $\Bbb{Z, Q}$ and the set of even numbers. (Prove this! The case for $\Bbb Q$ is especially fun!).

Now using the idea of a power set we can construct sets with even larger cardinalities! For example $|\mathscr P(\Bbb N)|=2^{\aleph_0}=\aleph_1$. In fact without hurting your brain we can demonstrate at least countable amount of infinities $|\mathscr P(\Bbb N)|=2^{\aleph_0}=\aleph_1<|\mathscr P(\mathscr P(\Bbb N))|=2^{\aleph_1}=\aleph_2\\<|\mathscr P(\mathscr P(\mathscr P(\Bbb N)))|=2^{\aleph_2}=\aleph_3\cdots$

The next biggest infinity after $\aleph_0$ is $\aleph_1$... or is it? This is a fundamental question in set theory and has been proven to be impossible to prove! It is called the continuum hypothesis: There exists a cardinal number $\mathfrak c$ such that $\aleph_0<\mathfrak c< \aleph_1$.

Moving on we have the cardinality of $\Bbb R$ to consider. Is it countable ($\aleph_0$) or uncountable ($\aleph_1$). We actually find that it is the latter! And we will prove this now! Firstly let us lay some lemmas down.

Lemma 1 The cardinality of the interval $(0, 1)$ is the same as the cardinality of $\Bbb R$. Proof We need to demonstrate the existence of a bijection $f:(0,1)\to\Bbb R$. The function $f:x\mapsto\cot(\pi x)$ suffices.

Lemma 2 We have this equality $|(0, 1)|=|[0, 1]|$. Proof $(0,1)\subset[0, 1]\subseteq\Bbb R$ therefore $|(0, 1)|=|[0, 1]|$.

OK! We are ready for the proof!

First we would like to show that every real number in between $0$ and $1$ can be shown to be equivalent to a subset of $\Bbb N$. This is easier to visualize in binary. We can construct a bijection $\varphi:[0,1]\to\mathscr P(\Bbb N)$ such that $\varphi(0)=\varnothing$ and $\varphi(1\text{ (or) }0.1111\cdots)=\Bbb N$. Now represent $\displaystyle [0,1]\ni x=\sum^\infty_{n=1}\frac{b_n}{2^n}$ where $b_n$ repersents the nth binary digit now we can define $\varphi:0.b_1b_2b_3\cdots\mapsto\{n\in\Bbb N : b_n=1\}$ therefore for every binary number there is a corresponding subset of $\Bbb N$. $\square$

Example values for $\varphi$ in the proof are $\varphi(0.010101\cdots)=\{2, 4, 6\cdots\}$

We have now proved that $|R|=|\mathscr P(\Bbb N)|$.

I hope this post on some set theoretic ideas has inspired and enlightened you!