# Prove that the countable union of sets of cardinality $c$ again has cardinality $c$. [duplicate]

Where $c = 2^{\aleph_0}$.

I have no idea how to tackle this problem. Any help is greatly appreciated!

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Let $\{ X_n \mid n < \omega \}$ be a countable family of sets of cardinality $2^{\aleph_0}$, and suppose (without loss of generality) that they're pairwise disjoint. For each $n$, you can inject $X_n$ into the interval $(n,n+1) \subseteq \mathbb{R}$. Piecing together these injections yields an injection $\bigcup_n X_n \to \mathbb{R}$.

For each $n < \omega$, the interval $(n,n+1)$ has cardinality $2^{\aleph_0}$. Proving this is an easy exercise: first biject any open interval with $\mathbb{R}$, then use a linear ($\Rightarrow$ bijective) function to scale/translate it to $(n,n+1)$.
Since $|X_n|=2^{\aleph_0}$ for each $n$, there is a bijection $f_n : X_n \to (n,n+1)$, which we can view as an injection $f_n : X_n \to \mathbb{R}$ whose image is equal to $(n,n+1)$. The reason we choose this interval is so that, when we build a function $\bigcup_n X_n \to \mathbb{R}$, distinct sets $X_n$ are embedded into disjoint subsets of $\mathbb{R}$.
Since the sets $X_n$ are pairwise disjoint, there is a unique function $f : \bigcup_n X_n \to \mathbb{R}$ such that $f(x)=f_n(x)$ for each $x \in X_n$. Moreover, $f$ is injective, since its restrictions to each $X_n$ are injective and the images of each $X_n$ are pairwise disjoint.
Since $\bigcup_n X_n$ injects into $\mathbb{R}$, we see that $\left| \bigcup_n X_n \right| \le 2^{\aleph_0}$.