# The set of all infinite binary sequences

Let's suppose that we have the set $S$ of all possible infinite binary sequences $s_i$:

$$S=\{s_1,s_2,s_3,\ldots \}$$

where the sequences $s_i$ are like $\{1,1,1,1,\ldots\}$, $\{0,0,0,0,\ldots\}$, $\{0,1,0,1,\ldots\}$ etc.

However following the Cantor's diagonal argument we can construct an sequence $s_0$ that is not in the set $S$. So there is a contradiction because we had supposed that the set $S$ contains ALL such infinite sequences of $1$'s or $0$'s. If we add the $s_0$ to $S$, then we can again construct another $s_0'$ that will not be in the set $S$, and so on.

So where's the problem? How do we define/construct the set that contains all such sequences?

You were supposed to be assuming, for the sake of contradiction, that $S$ was countably infinite; Cantor's diagonal argument tells you how to construct, for any countably infinite collection of binary sequences, a binary sequence not in that collection. Thus, the set $S$ of all binary sequences (which is a perfectly well-defined object) is uncountable.
The fact that it uses is that you have a map from $\Bbb N\to A$ (the collection of all sequences(sets)), so it assumes that collection to be countable. But, since this argument contradicts the completeness of the collection , therefore, it implies that you can't have any map from $\Bbb N\to A$ and thus, $A$ is uncountable.
You can define $A$ or $S$ as $$\{(s_1,s_2,s_3,\cdots): s_i\in\{0,1\} \forall i\in \Bbb N\}$$
Let $$S$$ denote the set of inﬁnite binary sequences. Here is Cantor’s famous proof that $$S$$ is an uncountable set. Suppose that $$f : S → \mathbb{N}$$ is a bijection. We form a new binary sequence $$A$$ by declaring that the n'th digit of $$A$$ is the opposite of the n'th digit of $$f^{−1} (n)$$. The idea here is that $$f ^{−1} (n)$$ is some binary sequence and we can look at its n'th digit and reverse it. Supposedly, there is some N such that $$f(A) = N$$. But then the $$N$$'th digit of $$A = f ^{−1} (N)$$ is the opposite of the $$N$$'th digit of $$A$$, and this is a contradiction.