# Bijection between $\mathbb{Z} \longmapsto \mathbb{R}$ [duplicate]

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?

## marked as duplicate by Asaf Karagila♦ set-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 22 '15 at 13:35

• 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 8:14
Note that the image is not even all of the rational numbers between $0$ and $1$. For instance $\frac{1}{3}$ does not appear.
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).