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Is there a one-to-one function from $(-∞,0]$ onto $(0,1/2)$?

So I was thinking of a function of the form $$\frac{1}{x+n}+m$$ I'm used to problems in which you have to map an interval into another (where infinity is not involved), which might be solved with a system of linear equalities.

Thanks in advance

Dr. Mathva
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    Yes, such function exists. What have you tried? – Robert Z Aug 30 '18 at 10:32
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    Yes, both are of the same cardinality. Do you mean something like "continuous function"? – MigMit Aug 30 '18 at 10:34
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    My problem here is with the interval $(-∞,0]$ because I've always worked with 'finite' intervals... I think that the function might be something like $$\frac{1}{x+n}+m$$ – Dr. Mathva Aug 30 '18 at 10:37
  • Yes MigMit, that would be in the best case – Dr. Mathva Aug 30 '18 at 10:38
  • It cannot be... It is not defined for $-n$ – dmtri Aug 30 '18 at 10:42
  • There is no continuous one-to-one function like that. There exist both continuous injection and surjection, but no continuous bijection. – MigMit Aug 30 '18 at 10:46
  • @Dr.Valdi: Welcome to Math.SE! You should add your thoughts and specific issues to the question (by editing), rather than leaving them in the comments. Your questions are likely to be better received when you do. – Cameron Buie Aug 30 '18 at 10:51
  • See https://math.stackexchange.com/questions/160738/how-to-define-a-bijection-between-0-1-and-0-1. Then it remains to find a bijection between $(−\infty,0]$ and $(0,1]$ and another one between $(0,1/2)$ and $(0,1)$. – Robert Z Aug 30 '18 at 10:54

2 Answers2

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There exist one-to-one functions from $(-\infty,0]$ onto $(0,\frac12)$. For a proof, consider first $f(x)=\frac1{2(1-x)}$. That maps $(-\infty,0)$ onto $(0,\frac12)$. We now need to change $f$ so $0$ can get mapped somewhere. The standard method is Hilbert's hotel: We let $g$ be function given by $g(x)=f(x)$ for $x\not\in\mathbb{Z}$, $g(0)=f(-1)$, $g(-1)=f(-2)$ and so on.

However, no continuous one-to-one function from $(-\infty,0]$ onto $(0,\frac12)$ exists. A continuous one-to-one function is either strictly monotone increasing or strictly monotone decreasing. It follows that $\lim_{x\to 0} f(x)$ must be $0$ or $\frac12$. As $f$ is continuous, $f(0)=0$ or $f(0)=\frac12$, so the function does not map into $(0,\frac12)$.

Kusma
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You suggested in comments something of the form $\frac{1}{x+n}+m$

  • If you used $f(x) = \frac{1}{x-2}+\frac12$ then you would have $f:(-\infty,0] \to [0,\frac12)$ which is a bijection but not exactly onto $(0,\frac12)$

  • If you used $g(x) = \frac{1}{x-3}+\frac12$ then you would have $g:(-\infty,0] \to [\frac16,\frac12) \subset (0,\frac12)$ which is a one-to-one injection but not a bijection

  • If you combine these and used $h(x) = \frac{1}{x-2}+\frac12$ when $x$ is not an integer and $h(x)= \frac{1}{x-3}+\frac12$ when $x$ is an integer then you would have $h:(-\infty,0] \to (0,\frac12)$ which is a bijection but not a continuous function

You cannot have a continuous bijection because of the problem that it must put $0$ inside the open interval $(0,\frac12)$

Henry
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