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I've had a look on the website but nothing seems to answer the particular question I have.

(i) Let F be a finite set. Show that any injective (that is, one-to-one) map ψ : F → F is surjective (that is, onto).

(ii) (Paradox of Galileo) Give an example of an injective map ϕ : N → N that is not surjective.

(iii) Now let T be a countably infinite set. Deduce that T can be put in one-to-one correspondence with some proper subset of itself; so that there exists an injective map φ : T → T that is not surjective.

(iv) Let S be an infinite set. Show that there exists a countably infinite subset T of S.

(v) Deduce from (i), (iii) and (iv) the Dedekind-Pierce Theorem: A set S is infinite, if and only if S can be put into one-to-one correspondence with some proper subset of itself.

All the other parts of the question are fine, but i don't understand how to do part V. How do I use the other parts of the question to derive this fact.

Help is much appreciated always.

Thanks

user416936
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  • Check my answer to this question: https://math.stackexchange.com/questions/524029/a-set-is-infinite-iff-there-is-a-one-to-one-correspondent-with-one-of-its-proper/ – Akerbeltz Jun 08 '19 at 22:15

2 Answers2

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Suppose $S$ is infinite, and using (iv) take a countably infinite subset $T = \{s_0, s_1, \dotsc \}$ of $S$. Then build the map $\phi\colon S\to S$ such that $\phi(x) = x$ for $x \notin T$ and $\phi(s_i) = s_{i+1}$ for elements of the subset $T$. I assume this is the map that you came up with in (iii)?

This map $\phi$ is not surjective, so by (i) $S$ is not finite. Can you finish writing up the details from this?

Mike Pierce
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  • Yes that is the map I came up with in (iii) after doing some reading earlier. I also understand that by (i) S is not finite but don't think it would be right to simply say because of this fact S is infinite. Would you be able to elaborate further? Thank you for your help. – user416936 Oct 11 '17 at 15:11
  • @user416936 But that's exactly what it means for a set to be infinite. A set is infinite if it is not finite. Or if you don't agree with this definite of a set being infinite, then what definition do you have in mind? – Mike Pierce Oct 11 '17 at 15:30
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    No I do agree with the definition. I think I've just been looking for detail when this is entirely sufficient. The only bit I've been missing is that (phi) is not surjective and therefore S is not finite. Thanks. – user416936 Oct 11 '17 at 15:36
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$$S\supseteq T = \{t_1,t_2,t_3,\ldots\}$$ $$ \begin{array}{rcc} t_1 & \leftrightarrow & t_2 \\ t_2 & \leftrightarrow & t_4 \\ t_3 & \leftrightarrow & t_6 \\ t_4 & \leftrightarrow & t_8 \\ t_5 & \leftrightarrow & t_{10} \\ & \vdots \\ \text{and for } s\in S\smallsetminus T,\quad s & \leftrightarrow & s \end{array}$$ This is a one-to-one correspondence between $S$ and a proper subset of $S.$ The proper subset contains as a member everything that appears to the right of the arrows above.

Michael Hardy
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