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$X$ is defined as the collection of sets $X = \{X_i: i \in [0, 1] \}$ where each set has cardinality $c$ of the continuum and each pair is disjoint.

How do I prove that for each $i \in [0, 1]$, the set $[0, 1] \times \{i\}$ has the same cardinality as $X_i$(which can therefore be replaced by any cardinality)?

I know that the general idea is that the union clearly has a subset of cardinality $c$ and so the union must have a cardinality greater than or equal to $c$. So if I am able to show that this union also has cardinality less than or equal to $c$, I can probably use the Schröder–Bernstein Theorem to deduce the result.

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The cardinality of X$_i$ is c.
The cardinality of [0,1] is c.
The cardinality of {i} is 1.
Clearly #([0,1]×{i}) = c×1 = c = #X$_i$.

As all the X's are pairwise disjoint and have cardinality c and there are c of them the cardinality of the union of the X's is c×c = c.

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