So I have got the following sets $$ \bigcup_{i \in [0, 1]} X_i \ \ \ \text{and} \ \ \ [0, 1] \times [0, 1] $$ where each $X_i$ has cardinality $c$ of the continuum and each pair of $X_i$ where $i \in [0, 1]$ is disjoint.

I am basically trying to show that such a union of sets with cardinality $c$ has cardinality $c$. I have separately been able to prove that the cardinality of $[0, 1] \times [0, 1]$ is $c$ and just need this bijection to arrive at the result. Is there such a bijection and if so, what is it?


Yes, There is a bijection:

Each $X_i$ has the same cardinality as $[0,1]$. So let $f_i:[0,1]\rightarrow X_i$ be a bijection for all $i\in[0,1]$.

Let $f:[0,1]\times [0,1]\rightarrow \bigcup_{j\in[0,1]} X_j$ be the map $$f(j,x) = f_j(x)$$ for all $j\in [0,1]$.

This map is well defined. It is one to one because each of the $X_j$ are disjoint and onto because each of the $f_j$ are onto.


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