# How to make this function injective?

I have a question about the following passage:

How is this $F$ injective? For example, $0.0111\sim = 0.1000\sim$. I don't understand how "discarding the nonstandard representations" is defined. The domain of $F$ clearly contains $g$ and $g'$ such that $F(g) = 0.0111\sim = 0.1000\sim = F(g')$. Thanks for help.

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I think he is writing base $10$, so that $0.0111...$ is not $0.1000$ – Thomas Andrews Oct 26 '12 at 15:16
@ThomasAndrews You are right. Your comment answers my question. Thank you! – Rudy the Reindeer Oct 26 '12 at 19:17

## 2 Answers

Discarding the non-standard representations simply means that we don’t allow them: we work only with the standard representations. The real numbers $+.1\overline{0}$ and $+.0\overline{1}$ are not equal: the interpretation is decimal, not binary, so these numbers are $\frac1{10}$ and $\frac1{90}$, respectively.

The function $F$ produces decimals whose digits are all $0$ or $1$, so it cannot produce a non-standard representation of a positive real, and it clearly does not produce $-.\overline{0}$; thus, it produces only standard representations and so must be one-to-one (though obviously not surjective!).

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Noooo -- I thought these were binary representations! Duh! – Rudy the Reindeer Oct 26 '12 at 19:17

I think that when you define $F$ you work in base 10 (or 3)... That issue only occurs in base 2...

Without reading the more I have to guess, it is easy to construct one two one functions in both directions by using base 2 in one direction and another base in the other...

If you want to construct a bijection directly, and define $F$ using the binary representation, then $F$ is onto on $[0,1]$, and the set of pairs where is not one to one is countable, so it is easy to fix.

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