# Thread: Help in cardinal numbers of sets question...

1. ## Help in cardinal numbers of sets question...

I need to prove that the cardinal number of {0,1}^N is the cardinality of the continuum, c.
{0,1)^N is that set {f(x) | f:{0,1} --> N}, i.e., all the functions from the set {0,1} to Natural numbers.

The thing is, I have never learned of sets and cardinal numbers, and the lecturer mentioned briefly what a cardinal number is - so I have no idea how to approach this kind of problem.

Thanks!

2. The set $\left\{ {0,1} \right\}^\mathbb{N}$ can be thought of as the set of infinite bit-strings.
Use the Cantor Diagonal argument to show that the set is not countable.
But can be mapped injectively into [0,1].
If you are allowed to accept the Continuum Hypothesis then you are done.

3. Thank you very much.

4. You do not need to use the Continuum Hypothesis. In fact we should stay away from it.
(By construction) $\mathbb{R}$ is equal to $P(\mathbb{N})$ in cardinality.

To show that $|P(\mathbb{N})| = |2^x|$ where $2=\{0,1\}$ we are required to construct a bijection.

Define, $\mu_{X}:\mathbb{N}\mapsto 2$ as $\mu_{X}(a) = \left\{ \begin{array}{cc}1\mbox{ if }a\in X\\0\mbox{ if }a\not \in X \end{array} \right.$ for every $X\subset \mathbb{N}$.

Now define $\phi: P(\mathbb{N})\mapsto 2$ as $\phi (X) = \mu_X$.
This is a bijection.