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What my thought was that the bijection was a piecewise function:

$f(x) = 1/n+1$ if $x = 1/n$ for some $n \in \Bbb N$ and $x=1/n$ if $x \neq1/n$ for some $n \in \Bbb N$.

However, the textbook says this is incorrect. I don't actually see why. Was it a matter of brackets or the open variable $(0,1]$, which is what I tend to think.

Aggrawal Puja
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  • HINT: divide each set in two disjoint subsets, one countable and the other uncountable. Now try to define two bijections, one from countable to countable and the other from uncountable to uncountable. Then from these two bijections you define a new picewise function that is the bijection that you want. The trick is that bijections from countable to countable are easy to manipulate. –  Oct 11 '16 at 15:50

2 Answers2

1

There are several problems with what you propose.

First, note that $f(x)\not\in [0, 1]$ if $x={1\over n}$! So this function doesn't even have the right codomain.

Second, you haven't actually defined it all the way: what exactly is $f(x)$ if $x\not={1\over n}$ for any $n\in\mathbb{N}$?

Finally, you seem to be assuming that there are as many numbers in $[0, 1]$ that are of the form ${1\over n}$ as numbers that are not of that form. However, this is false: the former is countable, while the latter is uncountable.

So this is definitely not the right direction to go.

Noah Schweber
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1

They already pointed out what's wrong with your proposed function so I'll show a working one.

Pick a sequence $\{S_n\}_{n\in\mathbb{N}}$ such that $S_0=0$, $S_n\in(0,1]$ for every $n>0$ and the terms are all distinct.

Consider now the function $f:[0,1]\to(0,1]$ that sends $S_n$ in $S_{n+1}$ and $x$ in itself if it is not one of the $S_n$, this is a bijection between $[0,1]$ and $(0,1]$

Alessandro Codenotti
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