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I ran into some claims of my own while doing an exercise that I would like to be true for my solution to be correct. Are the following true or false, how to prove or disprove it?

  1. Let $ \varphi : \mathbb{N} \setminus I \rightarrow \mathbb{N} $ with $I \subseteq\mathbb{N}$, is it bijective ?

I know it holds for I finite, because by Dedekind's criterion there exist a bijection between an infinite set and some proper subset. But I am not sure if I can say the same for I infinite, because what if for example I is $\mathbb{N} \setminus \{1\}$,then the domain of $\varphi$ would be just $\{1\}$, wouldn't it? and $\varphi$ wouldn't be a bijection . But it might as well be just an ilusion like when one thinks even numbers are half of natural numbers, and then it turns out they are equinumerous

2) If $ \psi : Z \rightarrow \mathbb{N} $ is a bijection,with $Z$ a subset of $\mathbb{N}$, then $ \psi : Z\setminus I \rightarrow \mathbb{N}\setminus I$ is also a bijection, $I \subseteq\mathbb{N}$

Same kind of doubt here if I is infinite, but also I guess I have to consider two cases: $I \cap Z \neq\phi$ and $I \cap Z =\phi$

Asaf Karagila
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some_math_guy
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  • For 1, you got it. Basically, the sets are bijective iff $\mathbb{N} \backslash I $ is countably infinite. For 2, I think you want $ \psi_I : Z\setminus I \rightarrow \mathbb{N}\setminus \psi(I)$. If so, then the statement is true, and the proof is almost immediate (so give it a try, and explain where you're stuck at) – Calvin Lin Jun 13 '20 at 23:02
  • For 1, Can't I extract a condition on $ I$ directly? I needed 1 to be true for any $I$, finite or countably infinite, looks like for $I$ countably infinite, nothing can be concluded – some_math_guy Jun 13 '20 at 23:11
  • @juancarlosvegaoliver: No. There is a bijection between $\Bbb N\setminus I$ and $\Bbb N$ if and only if $\Bbb N\setminus I$ is infinite. That’s guaranteed if $I$ is finite, but if $I$ is infinite, you really cannot conclude anything without more specific information about $I$. – Brian M. Scott Jun 13 '20 at 23:14
  • @Brian M. Scott I guess the same goes for 2) then? – some_math_guy Jun 13 '20 at 23:17
  • @juancarlosvegaoliver: I’m afraid so, yes. – Brian M. Scott Jun 13 '20 at 23:18

1 Answers1

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Suppose that $Z=\{2n:n\in\Bbb N\}$, and $I=Z\setminus\{2\}$. Then the map

$$\psi:Z\to\Bbb N:2n\mapsto n$$

is a bijection, but $Z\setminus I=\{2\}$ is finite and cannot map bijectively to the infinite set $\Bbb N\setminus I$. (Specifically, $\Bbb N\setminus I=\{2\}\cup\{2n-1:n\in\Bbb N\}$.) This is basically the same difficulty that you already noticed with (1).

Brian M. Scott
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    Just an off-topic question. Do you use any software to write in Latex? I am amazed how fast everyone writes the math answers here. It takes me a lot. – some_math_guy Jun 13 '20 at 23:27
  • @juancarlosvegaoliver: No, I just type directly to the site. It took me a while to get comfortable with it when I started, partly because I’d never used $\LaTeX$ at all, but by now I’m fairly comfortable with it (though I do consider myself slow compared with some folks here!). – Brian M. Scott Jun 13 '20 at 23:29