This question is almost a duplicate to the question Q, but I would like to prove it in a more personalised manner.

The fact, that the individual sets, say $A$ and $B$ are countable, as well as infinite, ensures the existence of some bijective function $f: \mathbb{N} \rightarrow A$ and $g: \mathbb{N} \rightarrow B$ . Again, $A$ and $B$ being equipotent, there is some bijective function $h: A \rightarrow B$.

Now, consider their union (keeping in mind that they are disjoint). Define a map $\phi: \mathbb{N} \rightarrow A \cup B $ such that,

$\phi(n)= a_{n/2} $ when $n$ is even; $b_{n+1/ 2}$ when $n$ is odd [ $A=\{a_n\}$ and $B=\{b_n\}$ (enumerability permits this) ] Evidently, this mapping is a bijection.

Now, we consider the composition $f^{-1} \phi: A \rightarrow A \cup B$ and $g^{-1} \phi: B \rightarrow A \cup B$. Both of them are bijective. Furthermore, the equipotency between the sets $A$ and $B$ implies that their union is equipotent to either one of them.

We now use this to prove that $\mathbb{Q}$ and $\mathbb{Q}^+$ are of same cardinality.

We now split $\mathbb{Q}^*$ into $\mathbb{Q}^+$ and $\mathbb{Q}^-$ . Their elements are disjoint and both of them are equipotent. We use the previously proved proposition to deduce that $\mathbb{Q}^*$ is equipotent to $\mathbb{Q}^+$.

  1. Is the above proof correct?

  2. How do I extend the deduction to $\mathbb{Q}$ ?

  • $\begingroup$ Well, we can define an unconventional bijection from $\{0/1, 0/2, 0/3....\}$ to $\mathbb{N}$, but this doesn't seem to be valid. $\endgroup$ – Subhasis Biswas Aug 2 '18 at 10:03
  • $\begingroup$ Other workaround might be splitting the set $\mathbb{Q}$ into $\mathbb{Q}^-$ and $\mathbb{Q}^+ \cup \{0\}$. Setting $b_1=0$ in the second set solves the problem, I guess. $\endgroup$ – Subhasis Biswas Aug 2 '18 at 10:56

I first want to note, that in your first comment where you suggest a bijection from $\{0/1,0/2,0/3,\dots\}$ is only correct if you regard $0/x$ as a syntactical object. If you refer to a classical fractional representation of a rational number, then $\{0/1,0/2,0/3,\dots\}=\{0\}$ which is most certainly not bijective with $\mathbb{N}$.

The proof that for countable disjoint $A,B$, there is a bijection between $A\cup B$ and $A,B$ respectively is fine. Let me add, that I think that the countable case is not really rich in terms of awaiting revelations, as a countable union of countable sets is again countable (as you showed for the case of a pair-union) and every countably infinite set is bijective with any other countably infinite set.

This is why I think there are many ways to arrive relatively immediate at your desired result of equipotency of $\mathbb{Q}$ and $\mathbb{Q}^+$:

  1. You may show that $\mathbb{Q}$ and $\mathbb{Q}^+$ are both countably infinite (e.g. via a diagonal argument a la Cantor), i.e. yielding bijections $f:\mathbb{N}\to\mathbb{Q}$ and $g:\mathbb{N}\to\mathbb{Q}^+$. Then $h:\mathbb{Q}\to\mathbb{Q}^+$, $q\to g(f^{-1}(q))$ may be checked by you to be bijective.
  2. You can proceed by splitting $\mathbb{Q}=\mathbb{Q}^-\cup(\mathbb{Q}^+\cup\{0\})$ or $\mathbb{Q}=(\mathbb{Q}^-\cup\{0\})\cup\mathbb{Q}^+$ and observe that they are both countably infinite and disjoint. Then you can apply your insight for pair-unions of disjoint countably infinite sets.

As I think these results are relatively immediate, to get a greater deal of new knowledge out of this question, maybe try as an exercise to generalize your results and the related insights as far as possible.

  • $\begingroup$ Finally someone read it. Such a nice and insightful answer!!! $\endgroup$ – Subhasis Biswas Aug 2 '18 at 14:19

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