I saw the proof of cantor proving set of all sequence of zeros and one is uncountable(in Rudin analysis book) and I have a doubt, if we take set of all real number between 0 and 1.1111111...and so on and let me denote all irrational numbers by small letters and rationals as capital letters. Suppose I take two irrational numbers a and b then I can find a rational number between a and b, let the rational number be A which is located between a and b, so if their is another irrational number c greater than A, then I can find another rational number greater than A and less than c. Now the question is, haven't I found a bijective function between rationals and irrationals between 0 and 1.11111111111....

I am pretty sure their is a flaw here, please help me to find this flaw.

  • $\begingroup$ I have tried to edit your question, but I still feel something is not being properly asked. $\endgroup$ – IgotiT Jul 2 '16 at 7:37
  • $\begingroup$ Your premise is that if you extend this process in a countably infinite way, then you will have defined a function on all real numbers (well, on the interval, but that's superfluous). Unfortunately, your premise is exactly what you would need to prove. $\endgroup$ – rnrstopstraffic Jul 3 '16 at 21:38

You haven't even defined a function at all. Let $X$ be your set of rationals and $Y$ be your set of irrationals. To define a bijection $f:X\to Y$, you first of all have to define a function. That is, you have to describe a rule that given an element $x\in X$, gives you a single element $f(x)\in Y$. You haven't done that.

  • $\begingroup$ I to the contradiction feel like i have proved the number of rational and irrational between 0 and 0.1111111..to be same. $\endgroup$ – Mukil Jul 2 '16 at 8:09
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    $\begingroup$ @Mukil What is the image under your map for $1/2,$ and what is the image of $\sqrt{2}/2$? The description you gave does not allow one to find either of these explicitly, so it seems the map is not uniquely defined. $\endgroup$ – coffeemath Jul 2 '16 at 8:14

So, you've sort of described a bijection, but not quite. It sounds like you're saying that to each pair of irrationals you can attach a rational, and vice versa; but you haven't said anything to indicate that the rational is different each time. Certainly there is a rational between $\sqrt{2}/2$ and $\sqrt{2}/2 + 0.1$; certainly there is also a rational between $\sqrt{2}/2 - 0.1$ and $\sqrt{2}/2 + 0.2$. But in both cases, $0.75$ would fit the bill.

Conversely, you're attaching an irrational to each pair of rationals, but you haven't shown you obtain every irrational that way; there's no reason to believe that there aren't many irrationals that simply aren't needed.

  • $\begingroup$ "you've sort of described a bijection": I disagree strongly with this statement. $\endgroup$ – TonyK Jul 3 '16 at 20:13
  • $\begingroup$ The poster has described an association relating every element of one set with an element of the other, and vice versa; I'd call that a "sort of" bijection. As the rest of my response indicated, it's not actually a bijection, but it's reminiscent enough of one that the poster's confusion is understandable. $\endgroup$ – Reese Jul 3 '16 at 20:19
  • $\begingroup$ The poster has described no such thing. $\endgroup$ – TonyK Jul 3 '16 at 20:21
  • $\begingroup$ The poster seems to disagree with you. In any case, do you have a substantive mathematical objection to my response? $\endgroup$ – Reese Jul 3 '16 at 20:35
  • $\begingroup$ @Reese yeah but since playing with infinity causes so much trouble,as you've said 0.75 fits but so will 0.76 and 0.74 so cant we fix this numbers to each number? $\endgroup$ – Mukil Jul 6 '16 at 6:19

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