If card$(A_n)\leq$ card$(\mathbb{R}), \forall n \in \mathbb{N}$ show that card$( \bigcup \{A_n : n\in \mathbb{N}\} ) \leq$ card$(\mathbb{R})$

My attemp:

If card$(A_n)\leq$ card$(\mathbb{R})$ for each $n\in\mathbb{N}$ exist $f_n:A_n\to\mathbb{R}$ an injective function.

First case. If $ \bigcup \{A_n : n\in \mathbb{N}\}$ is such that $A_n \cap A_m = \emptyset, n\neq m$

Let $g: \bigcup \{A_n : n\in \mathbb{N}\}\to\mathbb{R}$ defined as

$$ g(x)= \left\{ \begin{array}{lcc} f_1(x) & \textit{if} & x \in A_1 \\ \\ f_2(x) & \textit{if} & x \in A_2\\ \hspace{0.5cm}\vdots \\ f_m(x) & \textit{if} & x \in A_m \\ \hspace{0.5cm}\vdots \\ \end{array} \right. $$

Since each $f_n$ is an injective function and $A_n \cap A_m = \emptyset, n\neq m$ then $g$ is an injective function. Therefore card$( \bigcup \{A_n : n\in \mathbb{N}\} ) \leq$ card$(\mathbb{R})$

But, what happens with the case when $A_n \cap A_m \neq \emptyset$ for some $n, m \in \mathbb{N}$?

  • 1
    $\begingroup$ Even if the sets $A_n$ are pairwise disjoint, the sets $f_n[A_n]$ need not be pairwise disjoint, so $g$ need not be injective. There may be $x\in A_0$ and $y\in A_1$ such that $f_0(x)=f_1(y)$, even if $A_0\cap A_1=\varnothing$. $\endgroup$ – Brian M. Scott Apr 2 at 2:50

I’ll assume that you know that for each $n\in\Bbb Z$ there is a bijection $$h_n:\Bbb R\to(n,n+1)\,.$$


$$\varphi:\bigcup_{n\in\Bbb N}A_n\to\Bbb N:x\mapsto\min\{k\in\Bbb N:x\in A_k\}\,,$$

and let

$$g:\bigcup_{n\in\Bbb N}A_n\to\Bbb R:x\mapsto (h_{\varphi(x)}\circ f_{\varphi(x)})(x)\,.$$

Now show that $g$ is injective.


What you are assuming in your argument is not that $\{A_n : n\in\mathbb{N}\}$ are pairwise disjoint but that $\{ f_n(A_n): n\in \mathbb{N}\}$ are pairwise disjoint. We can reduce the problem to this case.

Choose your favourite bijection $g(x)$ between $\mathbb{R}$ and the interval $(0,1)$. You can find some in this site. Then for any $n$ let $g_n(x)=(g\circ f_n)(x)+n$, where $f_n$ are the injections that you describe. Then $g_n$ is an injection into $(n,n+1)$. Finally you apply your argument with the $g_n$ instead of $f_n$.


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