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Set $S$ is a collection of disjoint sets each having cardinality of that of $\mathbb{R}$.The cardinality of set $S$ is also that of $\mathbb{R}$. $F$ is the set which is union of all the elements of $S$. Is the cardinality of set $F$ is equal to that of $\mathbb{R}$ ?

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Yes. The cardinality of $F$ is $\sum\limits_{x\in S}|x| = \sum\limits_{x\in S}\mathfrak{c} = \mathfrak{c}\mathfrak{c}=\mathfrak{c}$.

In cardinal arithmetic, if $\kappa$ and $\lambda$ are nonzero cardinals, and at least one is infinite, you have $$\kappa+\lambda = \kappa\lambda = \max\{\kappa,\lambda\}.$$

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(Minor technicality: $\kappa\lambda=0$ if $\kappa=0$ or $\lambda=0$.) –  Jonas Meyer Dec 2 '10 at 17:09
    
@Jonas Meyer: Quite right! Thank you. –  Arturo Magidin Dec 2 '10 at 18:35

You are asking whether $\mathbb{R} \times \mathbb{R}$ has the same cardinality as $\mathbb{R}$. The answer is yes, thanks to the fact that $|\mathbb{R}| = 2^{\aleph_0}$ and $2^{\aleph_0} \times 2^{\aleph_0} = 2^{\aleph_0 + \aleph_0} = 2^{\aleph_0}$. More explicitly, $\mathbb{R}$ has the same cardinality as the set $S = \{ 0, 1 \}^{\mathbb{N}}$ of binary sequences, and there is an obvious bijection $S \times S \to S$ given by "interweaving" sequences: that is, sending

$$(a_1, a_2, a_3, ...) \times (b_1, b_2, b_3, ...) \to (a_1, b_1, a_2, b_2, a_3, b_3, ...).$$

In general, the statement that $A \times A$ has the same cardinality as $A$, for $A$ infinite, is true for all alephs. I think I have been told that whether this statement is true for all infinite sets is equivalent to AC.

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Yes: the statement that $A\times A$ is bijectable to $A$ for all infinite $A$ requires AC. –  Arturo Magidin Dec 2 '10 at 15:50
    
@Arturo Magidin, @Qiaochu Yuan : thank you for the answers and explanation. –  Rajesh D Dec 2 '10 at 16:57
    
This is a theorem by Alfred Tarski, when he went to publish it from one end he heard "AC is obviously false, so the theorem is useless" and from the other end he got "This is obviously trivial, and it is pointless to publish that". The proof itself is quite elegant nonetheless. –  Asaf Karagila Dec 2 '10 at 17:09

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