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I need to the detriment the cardinality of a group and prove why.

The group:

A = {The group of all real numbers in the range between 0 and 1, where the digits can only be ones and zeroes, and after each zero comes a one}

At first I thought this is א, and was planning to prove it using the good old technique of picking one where there is zero, and zero where there is one.

But then I came up with the following function:

f:A ---> Z
f(x) = {
 if the first digit after the point is zero, conversion of the binary digits after the point in x (excluding the first zero) into a decimal number

 if the first digit after the point is one, conversion of the binary digits after the point in x into a decimal number, multiplied by minus one
}

This function seems to be reversible, and since |Z| is א0, |A| should be א0 as well. on the other hand, if proving using the method I intended to use, the result is that |A| is א.

Which one is it, and why?

Noah Schweber
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avivgood2
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    What is "$\aleph$"? Do you mean $2^{\aleph_0}$? (Incidentally, $\mathbb{Z}$ has cardinality $\aleph_0$, not $0_\alpha$ as you've written, and the set in question is not a group.) Also, in general you should use mathjax to include symbols in the question. – Noah Schweber Apr 27 '20 at 16:26
  • @NoahSchweber by א0 I mean listable infininitys – avivgood2 Apr 27 '20 at 16:29
  • As to the question itself, the error is that your process does not in fact yield an integer: e.g. consider $0.10101010...$. – Noah Schweber Apr 27 '20 at 16:29
  • " by א0 I mean listable infininitys" The notation is "$\aleph_0$" (in LaTeX: "$\aleph_0$"). – Noah Schweber Apr 27 '20 at 16:31
  • @NoahSchweber Why is this a problem? the function will just yield a number with infinitely many digits, just as 1010101 is...... – avivgood2 Apr 27 '20 at 16:38

1 Answers1

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The cardinality of your set is that of the continuum, because it's basically the set of all infinite sequences composed of the strings $1$ and $01$.

As mentioned in the comments, the problem with your proof idea is that $f(x)$ is not always an integer. If you would like to interpret $f(x)$ as simply an infinite binary sequence, then that's fine, but then the injectivity of $f$ gives you no grounds on which to claim $A$ is countable.

Jack M
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  • And how do I prove this is a continuum? what reversible function there is from this group to R? – avivgood2 Apr 29 '20 at 07:08
  • @avivgood2 It's well known that the set of binary sequences has the cardinality of the continuum, because each real number has a binary representation. The point is that by treating "01" as a single element, you can see that your set is in bijection with the set of all binary sequences. By the way, you should really stop using the word "group" (which means something else) and say "set". – Jack M Apr 29 '20 at 07:14
  • I understand. using 01 as 0 and 1 as 1 can create binary number that can represent every decimal number. but how do I represent non integers? and negative numbers? – avivgood2 Apr 29 '20 at 09:29
  • @avivgood2 This answer basically addresses how to show that $A$ is equinumerous to the set of all binary sequences $B$. To show that $B$ is equinumerous to $\mathbb R$ is a little bit delicate for the reasons you mentioned, but it's a standard result so you can probably find the details online, probably on this site. – Jack M Apr 29 '20 at 09:44