(0,n) can't map to (0,1) without dividing by some f(n).

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I was trying to map a set x=(0,n) to y=(0,1) by using a single(non-piecewise) function.

For example, you could say:

f(x)={0, x=0 && 1, x \dne 0}

however this is a piecewise function.

I wanted the function to like this f(x) = x/x. However, this leaves x=0 as undefined.

I have tried other ways but it always seems that you must divide by some f(x) which could have the possibility of being 0.

Thus, I am trying to prove that:

(0,n) can't map to (0,1) without dividing by some f(n) where f(n)=0.

Has this been proven? Or, how would I go about proving it?

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Let $n = 1$. Then every element maps to itself.

If $n \neq 1$, then let $f(x) = x/n : x \in (0, n)$.

$0$ is not in the domain of $n$ because $0$ cannot map to $(0,1)$ surjectively.