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"Subset of countable set is countable", keeping this fact aside for a moment, consider even numbers.

let $A$ be the set of all even numbers and E be a countable subset of $A$, and similar to cantor's diagonal process, one can form a even number by writing the elements of $E$ continuously.

and since this number will not be an element of $E$, one can form a new even number every time. every countable subset of $A$ becomes proper subset of $A$. by this argument even numbers form an uncountable set correct?

But i know even numbers are not uncountable due to bijection from even numbers to natural numbers

What is wrong here?

Andrés E. Caicedo
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jnyan
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  • How exactly do you go about this diagonal process? An even number is finite 'in both directions' (before/after the decimal point). – Fimpellizzeri May 23 '17 at 05:28
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    Even numbers don't have infinitely many digits. What exactly are doing with these even numbers to get a new one? That is not at all clear in your argument. – fleablood May 23 '17 at 05:28
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    If E is infinite then writing the elements of E results in an infinite string. That is not an even number. But you can use this to prove the even numbers are infinite. – fleablood May 23 '17 at 05:30
  • " and since this number" it's not a number " will not be an element of E" that's because it's not a number, " one can form a new even number" it's not a number " every time." Only if E is finite. Then you have successfully proven the even numbers are not finite. – fleablood May 23 '17 at 05:37
  • i start writing even number sequences 2,10,2164,.... i can write a new even number using the all the previous even numbers, 2102164. this even number appears nowhere in the sequence before and does it matter to have infinite digits in cantor adigonal argument? all one does is get a number which is not in the sequence. correct? – jnyan May 23 '17 at 05:40
  • But that just shows it doesn't appear before. It certainly can appear later. If E is all evens, you get the number 2468101214. That number does appear in E. – fleablood May 23 '17 at 05:45
  • one doesnt prove that in cantor argument, i mean reapperance of a number in the sequence. – jnyan May 23 '17 at 05:46
  • please help me out. I have this doubt since a long time – jnyan May 23 '17 at 05:47
  • The Cantor argument takes something from every element in E to create the new real number. If you take something from every element it has infinite representation. An integer does not have infinite representation. What you get is not an integer. – fleablood May 23 '17 at 05:51
  • Let us continue this discussion in chat. – jnyan May 23 '17 at 05:56
  • @fleablood why cant i have infininite representation of even number for example i can represent 2 by 0000.....2 appending 0 many times in front – Aayush Neupane Jan 05 '21 at 09:41
  • "why cant i have infininite representation of even number for example i can represent 2 by 0000.....2 appending 0 many times in front" An infinite number of zeros before is negligible becuase it is adding an infinite number of zeros. But this is the only way we can "represent" an integer with infinite digits. We can't have an infinite number of any other digit and we can't have any infinite number of zeros after any digit. Anyway that is not important. This procedure creates an infinite sum with an infinite number of non-zero digits and that is not resolvable. It doesn't make an integer. – fleablood Jan 06 '21 at 05:40
  • Okay, I reread the argument. to get $00000000.....02$ means you must have an infinitely countable list of even numbers of which all of them but one or only zeros, and the one that isn't only zeros must be the last in the list. But if the list is countably infinite you can't have a last term. – fleablood Jan 06 '21 at 05:48

3 Answers3

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I realize two answer is bad form but:

Let's go over Cantor's diagonal argument in detail and your argument in detail.

First we must define what we think a real number and an even number is.

Cantor: A real number between $0$ and $1$ is a value $\sum_{k=1}^{\infty} a_i*10^{-i}; a_i\in \mathbb N_{10}=\{0...9\}$. There is a bijection from $j:[0,1] \leftarrow\rightarrow X =$ {the set of infinite sequences of $\mathbb N_{10}$} via $j(\sum_{k=1}^{\infty} a_i*10^{-i}) = \{a_i\}$.

We are going to do our Cantor diagonal argument on $X = \{\text{infinite sequences of } \mathbb N_{10}\}$ rather than on $[0,1]$ but that will be fine as $[0,1] \approx X$.

Your: An even integer is a value $\sum_{k=0}^na_k*10^k; a_i \in \mathbb N_{10}; a_0 \in \{0,2,4,6,8\}$. There is a bijection $h: \{\text{even integers}\} \leftarrow \rightarrow Y = \{\text{finite sequences of }\mathbb N_{10}\text{ with first term even}\}$ via $h(\sum_{k=0}^na_k*10^k) = \{a_i\}$.

We will do our Cantor argument on $Y$ rather than the set of even integers but this is fine because $Y$ and the set of even integers or equivalent.

But NOTE: $X$ is a set of infinite sequences and $Y$ is a set of finite sequences. That is going to make all the difference.

Cantor: Let $E$ be a countable infinite subset of $X$. That is $E = \{v_i| v_i = \{a_{i,1}, a_{i,2},......\}\}$.

Your: Let $F$ be a countable infinite subset of $Y$. That is $F = \{w_i|w_i = \{a_{i,0}, a_{i,1} ....... a_{i, m}\}\}$

Cantor: We are going to find an $f: \{\text{countable infinite subsets of } X\} \rightarrow X$. so that $f(E) \in X$ but $f(E) \not \in E \subset X$.

Thus no countable subset of $X$ is all of $X$ and thus $X$ is uncountable.

Your: We are going to attempt (but ultimately fail) to find a $g: \{\text{countable infinite subsets of }Y\}\rightarrow Y$. What we are going to define instead is a $g:\{\text{countable infinite subsets of }Y\}\rightarrow X$. We will fail to show that $f(F) \in Y$. We will find instead that $f(F) \not \in Y$. (Although $f(F) \in X$).

As $f:\{\text{countable infinite subsets of}Y\}\not\rightarrow Y$ the fact that $f(F) \not \in F$ does not demonstrate anything.

Cantor: Let $f(E) = \{b_i\}$ so that $b_i = a_{i,i} + 1 \mod 10$ we $\{a_{i,k} = v_i \in E$. Thus $\{b_i\}$ is an infinite sequence and thus in $X$. So $f: \{\text{countable infinite subsets of } X\}\rightarrow X$. And $b_i \ne a_{i,i}$ for any $i$, $\{b_i\} \ne v_i$ for any $v_i \in E$ so $f(E) \not \in E$.

We are done.

Your: Let $g(F) = \{b_i\}$ were $b_{(\sum_{i=0}^k i)+j} = a_{k+1, j}$; that is ... we simply write every sequence one after another.

However as $F$ is an infinite set this $g(F)$ is an infinite sequence. $Y$ is a set of finite sequences. So $g: \{\text{countable infinite subsets of }Y\}\not \rightarrow Y$. Instead $g: \{\text{countable infinite subsets of }Y\}\rightarrow X$. So $g(F) \not \in Y$. As $g(F) \not \in Y$, the fact that $g(F) \not \in F \subset Y$ is not useful.

fleablood
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  • Thank you for all the effort. – jnyan May 23 '17 at 16:32
  • I got some more insight myself, coming up with new even number works fine, but I should be able to do that even after countably infinite times. Which is possible only if the number which I am trying to form has infinite digits. This is where infinite digits come in to picture and after infinite times, I can't get an even number the way I could after finite times. – jnyan May 23 '17 at 16:42
  • exactly. What you get is not an integer as it has infinite digits. The closest you can claim is that $a = \sum_{i=0}^{\infty}a_i10^i$ which diverges and is not a meaningful value and is not* an even integer. – fleablood May 23 '17 at 16:51
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similar to cantor's diagonal process, one can form a even number by writing the elements of $E$ continuously

This is mistaken. For simplicity, let $E$ be the positive even numbers. The notation $2468101214161820\ldots$ doesn't name a number. Nor does $\ldots2018161412108642$. The decimal notation $0.2468101214161820\ldots$ does name a real number, but it's not an integer, let alone an even integer.

Chris Culter
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  • given any even number sequence, say, 10,14,216,....i can form an even number, an integer, 1014216, and this even number is nowhere in the sequence, correct? – jnyan May 23 '17 at 05:37
  • That's a finite sequence. So that proves the evens are not finite. – fleablood May 23 '17 at 05:38
  • it also proves that every subset of E is proper subset of E, meaning E is uncountable. – jnyan May 23 '17 at 05:43
  • " given any even number sequence, say, 10,14,216,....i can form an even number, an integer, 1014216, and this even number is nowhere in the sequence, correct?" Incorrect, why wouldn't 1014216 appear later in the sequences? You need some way to guarantee that the even number formed by some of terms never appears later. The Cantor way of doing that is to take some ing from every term. But that results in something infinite that is not a number at all. – fleablood May 23 '17 at 05:43
  • by that logic, whatever number is obtained,in cantor argument, which is not in the sequence till then, could also be somewhere in the sequence later – jnyan May 23 '17 at 05:45
  • It only proves that if E is finite. If E is infinite it proves nothing of the sort. – fleablood May 23 '17 at 05:47
  • Let us continue this discussion in chat. – jnyan May 23 '17 at 05:47
  • The Cantor argument is that you take a digit from every element,* all* of them. Your argument is you create a new number from all the even numbers in E. If E is infinite the result will be an infinite string. That is NOT an integer. – fleablood May 23 '17 at 05:55
  • If you stopped doing the Cantor diagonal after only a finite number of term, that would be true. But in the Cantor diagonal you must do it for every term. If you copied all the even numbers what you get simply is not an integer at all. That's the difference and it is essential. A real number can have infinite decimals. Therefore there are $2^{N} $ of them. That's uncountable. An integer can not have infinite digits. So there only $\sum N=N^2$ of them, that's countable. You method creates something that is NOT an integer. – fleablood May 23 '17 at 06:06
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If E is finite, the result will be an even number not in E. Ergo There are infinite even numbers.

That's fine and valid.

If E is infinite the result is an infinite string of digits. That is not a number at all. Nothing relevant can be concluded.

fleablood
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