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What can be the Cardinality of the union of disjoint sets, each of which have a cardinality of reals? How should this be proved.

I know using Schroder bernstein theorem, it is easy to see that the cardinality of the union must be equal to the cardinality of reals. But without using it, how should this be proved?

I have an idea: [0,1) has the same cardinality as that of reals. Similarly, [1,2) has the same cardinality as reals. Similarly a semi-open interval [n,n+1) has the same cardinality and they are all disjoint. So, can I combine them and simply say that their cardinality is equal to reals? Is this correct. I am looking at a more formal proof along this line.

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  • $\begingroup$ It';s not true for arbitrary unions. If the family of disjoint sets has greater cardinality than the reals, then their union will have greater cardinality than the reals. (Assuming the axiom of choice.) Also, if the family of disjoint sets is empty, then their union will have cardinality zero which is less than the cardinality of the reals. $\endgroup$ – bof Sep 24 '14 at 19:30
  • $\begingroup$ @bof- I have already mentioned that each set has cardinality equal to the reals, so they cannot be empty sets $\endgroup$ – Silver moon Sep 24 '14 at 19:32
  • $\begingroup$ I didn't say they were empty sets. I said the family was empty. $\endgroup$ – bof Sep 24 '14 at 19:38
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Map one of the sets to $(-\infty, 0)$, one to $[0, 1]$, and one to $(1, \infty)$.

(added as requested)

Therefore, each element of each of these sets gets mapped into a unique real, so the cardinality of their union is that of the reals.

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  • $\begingroup$ Can you please elaborate? $\endgroup$ – Silver moon Sep 24 '14 at 19:20
  • $\begingroup$ @ marty- I just gave an example of 3 sets. There can be finitely any number of such sets. $\endgroup$ – Silver moon Sep 24 '14 at 19:22
  • $\begingroup$ @ marty- Exactly. But is there any way to show this obvious fact without using cardinal arithmetic? Just curious $\endgroup$ – Silver moon Sep 24 '14 at 19:26
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If you have $2^{\mathbb{R}}$ disjoint sets with at least 1 element then their union will have a cardinality of at least $2^{\mathbb{R}}$.

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One (extreme) option is to use the Axiom of Choice to prove that $\kappa+\lambda=\max\{\kappa,\lambda\}$ when at least one of $\kappa$ or $\lambda$ are infinite cardinals. From this, it follows that given $A$ and $B$ disjoint sets with $|A|=|B|=|\mathbb{R}|$ that $$|A|+|B|=\max\{|A|,|B|\}=\max\{|\mathbb{R}|,|\mathbb{R}|\}=|\mathbb{R}|$$

More generally, given any family $\mathcal{A}$ whose elements have cardinality of the continuum and where $|\mathcal{A}|\leq 2^{\aleph_0}$, then $$\left|\coprod_{A\in \mathcal{A}}{A}\right|=\sum_{A\in \mathcal{A}}{|A|}=\sum_{A\in \mathcal{A}}{2^{\aleph_0}}=|\mathcal{A}|\cdot 2^{\aleph_0}=2^{\aleph_0}$$ using the related fact that $\kappa\cdot \lambda=\max\{\kappa,\lambda\}$ when at least one of $\kappa$ or $\lambda$ are infinite cardinals.

When $|\mathcal{A}|>2^{\aleph_0}$, the result no longer holds true, for then $$\left|\coprod_{A\in \mathcal{A}}{A}\right|=\sum_{A\in \mathcal{A}}{|A|}=\sum_{A\in \mathcal{A}}{2^{\aleph_0}}=|\mathcal{A}|\cdot 2^{\aleph_0}=|\mathcal{A}|$$

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  • $\begingroup$ @ Hayden- This is not very suitable though it is correct. A more generalized proof should do. $\endgroup$ – Silver moon Sep 24 '14 at 19:06
  • $\begingroup$ @Martin I don't quite understand what you mean. In what ways is this not suitable? This is an incredibly general result; do you mean you want a generalization dealing with arbitrary disjoint union of sets of cardinality of the continuum? $\endgroup$ – Hayden Sep 24 '14 at 19:10
  • $\begingroup$ @Hayden- I have edited the question. Please check. $\endgroup$ – Silver moon Sep 24 '14 at 19:11
  • $\begingroup$ @Hayden- Yes, I am talking about a generalization dealing with arbitrary disjoint union of sets of cardinality of the continuum $\endgroup$ – Silver moon Sep 24 '14 at 19:12
  • $\begingroup$ @Hayden- Can you please suggest if my idea is correct? $\endgroup$ – Silver moon Sep 24 '14 at 19:14

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