February 24th, 2016, 01:15 PM  #1 
Newbie Joined: Mar 2013 Posts: 24 Thanks: 0  Question on the countability of the power set of natural numbers (Cantor's theorem)
I've been trying, on and off, for weeks now, to wrap my head around this theorem, but I keep running into issues with it. I think I finally have some definitive question(s) to ask, though. Questions I haven't found any documentation of anyone asking. So we are meant to assume that $\mathcal{P}(\mathbb{N})$ is countable, and that $\mathcal{f}: \mathbb{N} \rightarrow \mathcal{P}(\mathbb{N})$ in order to arrive at a contradiction. So we construct the set $\mathcal{X} = \{ x  x \in \mathbb{N} \space and \space x \notin \mathcal{f}(x) \}$ and establish that $\mathcal{X} \in \mathcal{P}(\mathbb{N})$. Okay, so far so good. Here is where I'm having a disconnect of following the logic. The proof claims that since $\mathcal{f}$ is a bijection, we must have $\mathcal{X} = \mathcal{f}(x)$ for some $x$. Why? The punch line: $x \in \mathcal{X} \iff x \notin \mathcal{f}(x) = \mathcal{X}$ Lets say $\mathcal{Q}$ is the set of all such $\mathcal{X}$ as defined above. Now lets define $\mathcal{R}$ as the set of all sets $\mathcal{Y} = \{ y  y \in \mathbb{N} \space and \space y \in \mathcal{f}(y) \}$ Then $\mathcal{Q} \subset \mathcal{P}(\mathbb{N}) \space and \space \mathcal{R} \subset \mathcal{P}(\mathbb{N})$. What I see, is the implication that $\mathcal{Q} \cap \mathcal{R} \neq \emptyset$, but clearly, these two sets need not intersect. What am I missing in the properties of a bijection that means there must be $\mathcal{X} = \mathcal{f}(x)$ for some $x$? The definition of $\mathcal{X}$ says it all, and from my definitions, $\mathcal{Q}$ does not contain any such members where $\mathcal{X} = \mathcal{f}(x) \space and \space x \in \mathcal{X}$. Since $\mathbb{N}$ is an infinite set, and so is $\mathcal{P}(\mathbb{N})$ I don't see where the intersection $\mathcal{Q} \cap \mathcal{R}$ must occur. Is this a condition of reflection? 
February 24th, 2016, 01:40 PM  #2  
Senior Member Joined: Aug 2012 Posts: 2,323 Thanks: 715  Quote:
By the way this proof is far more general. It applies to any set, not just a countable one. So that aspect is best omitted, you don't need that assumption. This proof shows that there is never a bijection between a set and its powerset. Last edited by Maschke; February 24th, 2016 at 01:43 PM.  
February 24th, 2016, 02:10 PM  #3  
Newbie Joined: Mar 2013 Posts: 24 Thanks: 0  Quote:
We never simply "run out of" natural numbers we can use as preimages where we can't simply remap f(x) to some other preimage. Last edited by Polaris84; February 24th, 2016 at 02:16 PM.  
February 24th, 2016, 03:30 PM  #4 
Newbie Joined: Mar 2013 Posts: 24 Thanks: 0 
I guess what I'm trying to say is that for every possible $\mathcal{X} \in \mathcal{P}(\mathbb{N})$, there is some $z \in \mathbb{N}$ where $z \neq x$ and $z \notin \mathcal{X}$ and $\mathcal{f}(z) = \mathcal{X}$. For the contradiction to exist, then my above statement can't be true. 
February 24th, 2016, 04:45 PM  #5  
Senior Member Joined: Aug 2012 Posts: 2,323 Thanks: 715  I like your professor Quote:
Quote:
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This is what in math is called a statement of existence. In math when we say something exists, it exists. We don't need to know any more about it. I can see that this is probably a stumbling block for modern students who are exposed to computer programming long before they ever see formal math. And in general, learning to reason about infinite sets is always a hurdle to be crossed. If f is a surjection and y is some element in the target, then there exists some x in the domain such that f(x) = y. And that's all we care about: that x must exist. Last edited by Maschke; February 24th, 2016 at 04:49 PM.  
February 24th, 2016, 06:23 PM  #6  
Newbie Joined: Mar 2013 Posts: 24 Thanks: 0  Quote:
If the proof said that all members of P(N) are sets defined as X, then I could understand where you would run into a contradiction. But strictly speaking, X is the set of all elements in N which, when mapped, do not produce a set that contains itself as a member. X is a single member P(N). Saying that there must exist an $x \in \mathcal{X}$ such that $x \in \mathcal{f}(x) = \mathcal{X}$ literally just ignores the definition of what we set $\mathcal{X}$ to be. Its forcing a contradiction to exist where there isn't naturally a contradiction, at least from every angle I've looked at it (and I know I'm wrong, but I can't see how). If $x \in \mathcal{f}(x)$ were true, it would never qualify to be in $\mathcal{X}$, by its own definition. EDIT: x is in the domain, but if f(x) > X, then X $\neq$ f(x) by the very definition of X. Last edited by Polaris84; February 24th, 2016 at 06:26 PM.  
February 24th, 2016, 06:27 PM  #7  
Senior Member Joined: Aug 2012 Posts: 2,323 Thanks: 715  Quote:
I didn't address the rest of your post yet but this point is the heart of the matter. If we assume f is a bijection we derive a contradiction. $\displaystyle x \in \mathcal{X}$ if and only if $\displaystyle x \notin \mathcal{X}$. Therefore f is not a bijection. And since f was any bijection, that shows there isn't one. Last edited by Maschke; February 24th, 2016 at 06:31 PM.  
February 24th, 2016, 06:39 PM  #8 
Newbie Joined: Mar 2013 Posts: 24 Thanks: 0  Okay, if I understand this correctly, if I can show that there is some bijection, or arrangement of the mapping between N and P(N) that doesn't hold, then that is all that matters?

February 24th, 2016, 06:56 PM  #9  
Senior Member Joined: Aug 2012 Posts: 2,323 Thanks: 715  Quote:
f is an arbitrary function from N to P(N). The assumption that f is a bijection leads to a contradiction. Therefore f is not a bijection. ps  Have you walked through the details of the argument for a small set, say one with 3 elements and 8 subsets? It might be helpful. Last edited by Maschke; February 24th, 2016 at 07:21 PM.  
February 24th, 2016, 07:40 PM  #10  
Newbie Joined: Mar 2013 Posts: 24 Thanks: 0  Quote:
I may just be reaching the limits of my understanding for now. I can't explain myself any clearer, so I guess I'll either have to wait until I have better tools to form questions, or just give up, because in the end, I'm not a math major, or even a minor, lol, I just like being able to understand. I just can't understand how there obviously is a set X in P(N), that is a fact, that has to exist. But then we say that there has to be an x that is both a member of X and f(x). X is defined so that this doesn't happen, and we can still map some n in N to f(n) = X, where n $\neq$ x. In those cases, n can be in f(n), because n isn't in X. And all x's map to a single set that is a member of P(N) that does not contain the same x. You will never run out of numbers that aren't in X that can map to sets containing themselves. Just the same, you will never run out of numbers that are in X which don't map to sets that contain themselves. I just can't explain my misunderstanding any better. $x \in \mathcal{X}$ so that means there is some $n$ where $\mathcal{f}(n) = \mathcal{X}$ and $n \notin X$. This means that there is no $x = n$. The statement that $x \in X \iff x \notin \mathcal{f}(x) = \mathcal{X}$ appears contrived from no where. There is no obvious $\mathcal{X}$ where $x \in \mathcal{f}(x)$ must be true. In the bijection, $\mathcal{X}$ only requires 1 preimage to map to it that isn't a member of $\mathcal{X}$, which can easily be produced. ***EDIT*** Perhaps this is what I have been miss interpreting: $\mathcal{X}$ must contain every member of $\mathbb{N}$ but some how simultaneously excludes $x \notin \mathcal{f}(x)$? Because I've been reading it as $\mathcal{X}$ must contain every member of $\mathbb{N}$ except $x \notin \mathcal{f}(x)$... Last edited by Polaris84; February 24th, 2016 at 08:32 PM.  

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cantor, countability, natural, numbers, power, question, set, theorem 
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