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Consider the following statement:

Prove that it is possible to write $\Bbb R$ as a union $\Bbb R= \bigcup_{i\in I} A_{i}$ where $A_{i} \cap A_{j}= \emptyset$ if $i\neq j$, $i,j \in I$,and such that each $A_{i}$ and $I$ are uncountable sets.

There is a same question here (The real numbers as the uncountably infinite union of disjoint uncountably infinite sets). And my question is about one of the answers from that question.

Thanks for Kyle Gannon who gives a constructive proof:

Since $|\mathbb{R}| = |\mathbb{R} \times \mathbb{R}| $, there exists a bijection $f$ from $\mathbb{R} \to \mathbb{R} \times \mathbb{R} $. Then $\mathbb{R} = \bigcup_{a \in \mathbb{R}} f^{-1}(\mathbb{R},a)$ where $f^{-1}(\mathbb{R},a) = \{b \in \mathbb{R}: f(b) = (c,a)$ for some $c \in \mathbb{R} \}$.

This answer is reasonable to me. But I am struggling with proving the the facts that the set $f^{-1}(\mathbb{R},a)$ is uncountable, each $f^{-1}(\mathbb{R},a)$ is disjoint from one another and the union of them is the whole set of real numbers. They all seem intuitively true to me but I just want to know how to formally prove them.I would really appreciate if someone could help me. Thanks so much!

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This just says that the plane is the union of vertical lines each of which contains uncountably many points, and all vertical lines are disjoint. Then you transfer this partition back to $\mathbb R$ using the mysterious bijection $f$.

Instead of working with such a mysterious bijection, one can use a more meaningful embedding of $\mathbb R$ into a hyperreal field ${}^\ast\mathbb R$ (obtained as ultrapower with index set $\mathbb N$) which is known to have the same cardinality. Then each galaxy in ${}^\ast\mathbb R$ is uncountable, and there are uncountably many galaxies.

Here a galaxy is a set of hyperreal numbers at finite distance one from another.

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  • $\begingroup$ Thanks so much for your answer. I fully understand your explanation on "the union of all vertical lines". But as I said, my problem is trying to formally prove why there is an uncountable number of points in every vertical line. I certainly know that every vertical line does not intersect any other line. But again, the problem is how to formally prove that using mathematical theories. $\endgroup$ – PropositionX Mar 8 '17 at 13:51
  • $\begingroup$ Each vertical line is just a copy of $\mathbb R$ so if you believe that $\mathbb R$ has uncountably many points then so will each vertical line. $\endgroup$ – Mikhail Katz Mar 8 '17 at 13:53
  • $\begingroup$ Let me try this: the set $f^{-1}(\mathbb{R},a)$ denotes the points in the real line that are bijectively mapped to some vertical line in the plane. Therefore this set is equivalent to the set of all points in that vertical line. Since the set of all points in that vertical line is equivalent to $\Bbb R$. Therefore the set $f^{-1}(\mathbb{R},a)$ is uncountable. $\endgroup$ – PropositionX Mar 8 '17 at 14:01
  • $\begingroup$ Ok. But here is another problem : why, in formal theory, does the fact that different vertical lines in the plane does not intersect with each other imply that different sets of points (in the real line) that are bijectively mapped to different vertical lines are disjoint? $\endgroup$ – PropositionX Mar 8 '17 at 14:05
  • $\begingroup$ Well, functions are assumed to be single-valued. So if there were a common point to two such sets, it would have to map to a common point on distinct vertical lines. But parallel lines are disjoint :-) $\endgroup$ – Mikhail Katz Mar 8 '17 at 14:07

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