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I recently learned of Cantor's diagonal argument, and was thinking about why there can't be a bijection between any infinite set of integers and any infinite set of real numbers. I understood the basic idea behind the proof, but I was thinking of a particular transformation, for which I don't see why it doesn't form a bijection.

Let's say we want to map all of the integers, to the real numbers between $[-1, 1]$. My transformation is created in such a way, that for every integer, I transform it into a mirrored number that is placed after a decimal.

For example:

$$0 \rightarrow 0.0$$ $$1 \rightarrow 0.1$$ $$2 \rightarrow 0.2$$ $$\vdots$$ $$100 \rightarrow 0.001$$ $$\vdots$$

It would seem to me that this is one to one, and every number between $[-1, 1]$ is hit. Even in the example of an irrational number, say $(\sqrt2 - 1)$, there is sum integer of infinite size that maps exactly to $(\sqrt2 - 1)$, because $(\sqrt2 - 1)$ can be written as a decimal made of up an infinite number of integer sitting behind a decimal point. It could be mapped as:

$$...73265312414 \rightarrow 0.41421356237...$$

Now, I'm assuming I was probably not the first person to think of this, so, why does this not work as a bijcetive mapping?

Asaf Karagila
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Ephraim
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  • There can be, and in fact there are, injections between the integers and lots (in fact, an infinite number) of infinite sets of real numbers. For example, $;|\Bbb N|=|\Bbb Q|=|{-1,-2,-3,....}|=...;$ – Timbuc Apr 22 '15 at 08:14

2 Answers2

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There are no integers with infinite length (in decimal system). The image of your map is contained in the rational numbers, hence can not be all of the real interval.

Note that the image is not even all of the rational numbers between $0$ and $1$. For instance $\frac{1}{3}$ does not appear.

MooS
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For example $1/3=0.333\ldots$ is not in the image, and so the function is not onto.(This is already contained in the answer by MooS).

P Vanchinathan
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