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So the set can be rewritten as:

$$\{0,1\}^{\mathbb{N}}=\{(a,b,c,d,....) : a,b,c,d,... \in (0,1) \}$$

And so we can write those sequences as:

$$111111111...$$ $$010101010...$$

etc.

And then we pick the $1$st number of the first sequence and change it to $0$, we pick the second of the second etc. etc. and so we will end up with a sequence that is not in any of the sequences in the set, and so $\{0,1\}^{\mathbb{N}}$ is uncountable.

And then a follow up a question. I do not necessarily have to pick the "diagonal" numbers, do I? I can just pick every first number of every sequence and I would still end up with a sequence that is different to every sequence from which it is constructed, correct?

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    What does "pick every first number of every sequence" mean? – 2017-02-25
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    @bof I interpreted that as meaning that their new sequence is $(1-S_1(1), 1-S_2(1), 1-S_3(1), . . .)$ - that is, make the $i$th bit of the new sequence different from the first bit of the $i$th sequence. – 2017-02-25
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    As posted answers show, the first answer is "yes". It's standard to think of those sequences of $0$s and $1$s as expressing a number in the unit interval as a binary decimal. That shows $\{0,1\}^\mathbb{N}$ counts the real numbers in $[0,1]$. (You have to fuss a little bit with decimals that end with all $1$s.) – 2017-02-25
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    @NoahSchweber correct. sorry for not making it more clear – 2017-02-25

3 Answers 3

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The answer to your first question is yes.

The answer to your second question, however, is no: in order to conclude that some sequence $D$ is not on your list of sequences $S_1 S_2, . . . $, you need to argue:

For each $i$, $D$ disagrees with the $i$th sequence somewhere: that is, for some $n$, $S_i(n)\not=D(n)$.

(This is just because this is what it means for two sequences to be different! Two sequences are different iff they disagree somewhere; so if I want to conclude that $D\not=S_i$ for any $i$, I have to argue that - for each $i$ - there is some "point of disagreement".)

The diagonal argument picks $n=i$: the $i$th bit of $D$ is different from the $i$th bit of $S_i$, for each $i$. We could pick differently: e.g. make the $i+17$th bit of $D$ different from the $i+17$th bit of $S_i$.

By contrast, what you've described - making the $i$th bit of $D$ different from the $1$st bit of $S_i$ - isn't enough! There's no reason to believe that this actually produces a sequence not on the list.

For a concrete example of how this can go wrong, suppose your list looks like:

  • $S_1=0111111111...$

  • $S_2=1000000000...$

  • $S_2=1100000000...$

  • $S_3=1110000000...$

  • And so on.

Then if we do the "change the first bit" trick you describe, the sequence we'll produce is $$10000000000...,$$ which is on the list as $S_2$.

By contrast, the classically-constructed (anti)diagonal sequence is $$11111111111...,$$ which isn't on the list (we know it can't be $S_i$ since its $i$th bit isn't the $i$th bit of $S_i$). Of course, this sequence is in a sense the "limit" of the list, but that's not relevant - it's not on the list.

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    I see.. and obviously, in my case, they are not different because the only $n$ that I am picking from original sequences is $1$ and so all the other indices may not and will not be different. Thanks – 2017-02-25
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To your first question, yes, you don't have to pick the diagonals, but to your second, no, the new sequence should differ from all those listed in at least one entry.

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Associate a sequence $s\in\left\{ 0,\,1\right\}^\mathbb{N}$ with $S_s:=\left\{ n\in\mathbb{N}|s_n=1\right\}$. A list $\left( s^j\right) _{j\in\mathbb{N}}$ of such sequences is now associated with $S_{s^j}$ (we use superscripts to number sequences and subscripts to number entries within them). If $\left\{ n\in\mathbb{N}|s^n_n=0\right\}=S_{s^k}$, $s^k_k=0$ iff $k\in S$, i.e. iff $s^k_k=1$.

Depending on how you read this proof by contradiction, you can consider it either the "diagonal argument" on sequences or a special case of the proof of Cantor's theorem (i.e. the result that taking the power set obtains a greater cardinality). Just as one needs to construct a certain set to prove Cantor's theorem, one needs to construct a certain sequence to prove the $s^j$ aren't exhaustive. This relation between subsets and sequences on $\left\{ 0,\,1\right\}$ motivates the description of the proof of Cantor's theorem as a "diagonal argument".