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I came across this problem while solving MIT 18.100 assignments. The question is to prove that $2^{\mathbb{N}}$ is uncountable.

I have not taken any formal courses so I do not know if what I did is correct or not. Here is my argument. Every possible set of natural numbers is well-ordered. I define a function $f:2^{\mathbb{N}}\to[0.1,1]$. The function concatenates the elements of a well-ordered set in a string after a decimal. For example, $f(\{1\})=0.1$.
The function is not bijective but as far as I can tell, it is surjective. Does it imply that the power set $2^{\mathbb{N}}$ is uncountable?

I am not sure if I am not doing something fundamentally wrong. So please point it out if there is anything wrong with my argument. Also, if this approach if essentially incorrect, can you please push me in the right direction?

PythonSage
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  • No, you've shown that $|2^\Bbb N|\geq | [0.1,1] |$, but you would still have to prove that $[0.1,1]$ is uncountable. – Vsotvep Jan 10 '20 at 22:15
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    @Vsotvep I can show that using the diagonal argument right? I didn't know how to use the diagonal argument in the case of power set. So, using diagonal argument on $[0.1,1]$ in combination with what I have done is correct? – PythonSage Jan 10 '20 at 22:17
  • Sure, that works. – Vsotvep Jan 10 '20 at 22:18
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    I’m happy with OP using the uncountability of a proper closed real interval, @Vsotvep. – Lubin Jan 10 '20 at 22:18
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    @Lubin Well, it's the same thing, really, showing $2^\Bbb N$ is uncountable or that the interval is uncountable, so only showing a relation between $2^\Bbb N$ and the interval would not really be a proof of uncountability. – Vsotvep Jan 10 '20 at 22:20
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    @PythonSage the usual way to show $|\mathcal P(X)|>|X|$ is also by a diagonal argument: you assume that there is a surjective map $f:X\to \mathcal P(X)$ and then consider the diagonal set $Y={x\in X\mid x\notin f(x)}$. Similarly to the proof of uncountability of the interval, we get that $Y$ cannot be an element of $\mathcal P(X)$. – Vsotvep Jan 10 '20 at 22:22
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    Compare https://math.stackexchange.com/q/2629072/403337 –  Jan 10 '20 at 22:24

2 Answers2

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If your function from $2^{\Bbb N}$ to $[\frac1{10},1]$ is onto, then the domain is indeed uncountable.

I’m a little unsure about how your function is defined, though. Could you be more explicit in its definition?

Lubin
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  • I tried writing the exact definition but it got too tedious. So I'll try to demonstrate through few examples. Say $f({1,2,3,4,5})=0.12345$. Also, $f({1,23,45})=0.12345$ as well as $f({1,23,450})$ and so on and so forth.

    It will also cover non-terminating decimal representations because we can always pick a larger number in the set to continue the decimal representation. For example say $0.198198819888.....$ can be derived from $f({198,1988,19888,...})$

     – PythonSage Jan 10 '20 at 22:24
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    So to show surjectivity, you take your real number $\lambda$, and break its decimal representation up into finite blocks ${b_1,b_2,\cdots}$. Then each block $b_i$ is taken as an integer $I_i$. This will give you a set of integers $S$. But how do you do this so that when $i<j$, you always get $I_i<I_j$ ? If this doesn't happen, it seems to me that when the numbers of $S$ are written out in order, the PS function will not necessarily deliver your real number $\lambda$ back to you. – Lubin Jan 10 '20 at 22:45
  • Here is what I thought. Please correct me if I am wrong. For the case you mentioned, there will be other way of breaking it in to another way, say $C_1,...C_q$ such that $C_1\le C_2\le ....C_q$. The idea behind this thought is this. I can make $C_q$ arbitrarily large by adding as many zeroes at the end as I need. Say an apparent decreasing sequence of integers $0.10090807060504030201001000100001...$ can be carved in to a sequence as ${10090,807060,50403020,100100100001,.....}$ and so on and so forth. For a terminating decimal, I can even just take the chunk and make one singleton set – PythonSage Jan 10 '20 at 23:01
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    Of course. But seems to me that you need the application of wisdom for that to work. And an algorithm or any kind of definite construction may not apply wisdom, unless you have that built into the routine. I think the real problem is that a given real number has too many ways of forming the sequence of its decimal digits into blocks. Good try, though. As @Vsotvep says above, and as Henno says in his answer, the diagonal argument works smooth and quick, much simpler in its construction than what you seem to have had in mind. I repeat: good try. – Lubin Jan 11 '20 at 07:26
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Suppose $2^{\Bbb N}$ (I assume the set of all subsets of $\Bbb N$ is meant, as you seem to) is countable, so there is a surjective $$f: \Bbb N \to 2^\Bbb N$$

Now define $A=\{n \in \Bbb N: n \notin f(n)\}$.

Suppose that $A=f(m)$ for some $m \in \Bbb N$. Then $m \in f(m)=A$ iff $m \notin f(m)$ (by definition of $A$). This is absurd, so no such $m$ exists and so $A$ is not in the image of $f$ and $f$ is not surjective. QED.

This is the classic Cantor proof. If you want to use your function to the reals idea, try

$f(A)= \sum_{n \in A} \frac{1}{2^n}$ to assign to each subset a different real number in $[0,1]$ and try to argue it's onto. But that's more indirect as you also need a proof that $[0,1]$ is uncountable. The power set argument directly is cleaner.

Henno Brandsma
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