Q: Number of images from $\mathbb{N}$ to {0, 1}. 
Are the number of images from $\mathbb{N}$ to {0, 1} countably infinite or uncountably infinite?


I was thinking of counting in base 2 to make a bijection between  $\mathbb{N}$ and {0, 1}. So, a table like this one:
 Image  |  0  1  2  3  ...         <- Here is N
-------------------------------
  0     |  0  0  0  0  ...         <- And here is the image in {0,1}
  1     |  1  0  0  0  ...
  2     |  0  1  0  0  ...
  3     |  1  1  0  0  ...
  4     |  0  0  1  0  ...
...

So, right now, in the left column, I can identify every number of $\mathbb{N}$ with an image on the right. Which makes it countably infinite, by definition.
But a co-student, claimed that the prof said it was uncountably infinite. Any idea why? 

A: The table you made doesn't make a bijection between $\mathbb N$ and the set of mappings from $\mathbb N$ to $\{0,1\}$. Consider such a scheme $f:\mathbb N\to\{0,1\}$ such that $f(n)=1-f_n(n)$, where $f_n$ is your $n$-th scheme. You will find that $f$ is different from any existed scheme $f_n$, while $f$ is indeed a scheme. This contradicts the number of schemes is countable.

A: Another attempt at clarifying: Imagine each sequence from $\mathbb N$ to $\{0,1\}$ to represent the binary decimal places of a number between $0$ and $1$.  Then you have a surjection from $\{0,1\}^{\mathbb{N}}$ onto the real interval $[0,1]$ (which is uncountable).  (Almost a bijection, actually, except for minor technical differences.)

A: You can use that if $A\subseteq \mathbb{N}$ then $f(a)=1$ if $a\in A$ and $f(b)=0$ if $b\notin A$ then $\mid \{0,1\}^\mathbb{N}\mid=\mid P(\mathbb{N})\mid$ by Cantor's theorem, $\mid \mathbb{N}\mid<\mid P(\mathbb{N})\mid$.
