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This question is asked many time when I search it. But i didn't find what am I really looking for.

My approach:

Let $A_1,A_2$ be two countable sets. If they have some common elements then redefine it by $B_2=A_1\setminus A_2=\{x\in A_2:x\notin A_1 \}$. The point of this is that the union $A_1 \cup A_2=A_1 \cup B_2$ and the sets $A_1,B_2$ are disjoint. I guess three case be happen for $B_2:$

  1. If $B_2=\emptyset$, then $A_1\cup A_2=A_1 \cup B_2=A_1$ which we already know to be countable.
  2. If $B_2=\{b_1,b_2,\cdots,b_m \}$ has $m$ elements, then how can I define and make sure $h:A_1 \cup B_2\to \mathbb{N}?$ In order to satisfy the statement $A$ is countable iff there is a bijective $(1-1$ and onto$)$ mapping $f: A \to \mathbb{N}$, where $\mathbb{N}$ is the set of all natural numbers.
  3. If $B_2$ is infinite again the same problem faced.

I have no idea how to approach and fix all these things. Besides, I wanted to know Is this approach is enough to proof the statement or not? Any hint or solution will be appreciated.

Thanks in advance.

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    $\begingroup$ You started well by splitting the sets into two disjoint sets and this is often forgotten by new students. Now, to continue, you can think of a bijection to a subset of the natural numbers as a sequence (because that is how a sequence is actually defined). Now... to write the sequence with all elements of your two sets, if $B$ is finite, just write all of them out at the front of the sequence and then write out the elements of $A_1$ after that. If they are both infinite, then just alternate $b_1,a_1,b_2,a_2,b_3,a_3,\dots$. Now, just make those ideas rigorous. $\endgroup$ – JMoravitz Mar 17 at 15:50
  • $\begingroup$ I've fixed your formatting. A couple main pieces of advice. Firstly, in English, we put spaces after punctuation that precedes another word. I.e., instead of "[...] when I search it.But I [...]" use "[...] when I search it. But I [...]". Secondly, paragraph breaks and proper numbered lists greatly improve readability. Other than that, generally speaking a pretty good question, and I appreciate the use of LaTeX. $\endgroup$ – jgon Mar 17 at 15:53
  • $\begingroup$ Thanks @JMoravitz sir for your great advise.I guess you tell me $$ h(n) = \begin{cases} b_n,n\le m \\ a_n\in A_1, n\gt m \end{cases}$$ but I can't think what have to do when they are both infinite? $\endgroup$ – emonHR Mar 17 at 16:06
  • $\begingroup$ Terminology: "Countable" means "finite or countably infinite". That is, $X$ is countable iff there exists an injective $f:X\to \Bbb N$. So that "uncountable" means "not countable". $\endgroup$ – DanielWainfleet Mar 18 at 6:08
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I would proceed like this:

  1. We know that there is a bijection between $A_1$ and $\Bbb{N}$ and $A_2$ and $\Bbb{N}$. Let's call them $f_1: A_1 \longrightarrow \Bbb{N}$ and $f_2: A_2 \longrightarrow \Bbb{N}$.

  2. We need to prove that there is $g: B \longrightarrow \Bbb{N}$ bijective ($B = A_1 \cup A_2$), or equivalently, a $g': \Bbb{N} \longrightarrow B$.

So define $g'$ as follows: $$g'(n) = \begin{cases} f_1^{-1}(n), & \text{if n is even} \\ f_2^{-1}(n), & \text{if n is odd} \end{cases}$$

Note that $f_1$ and $f_2$ are invertible because they are bijective. Now take $g = g'^{-1}$ and you have what you were looking for.

Note: a way to visualize it is to consider the following two sets: $A = \{(+, n) : n \in \Bbb{N}\}, B = \{(-, n) : n \in \Bbb{N}\}$ which union gives $\Bbb{Z} \setminus {0}$ and notice that $h$ defined as $$h(n) = \begin{cases} (+, n/2), & \text{if n is even} \\ (-, (n+1)/2), & \text{if n is odd} \end {cases}$$ is a bijection between $A \cup B$ and $\Bbb{Z} \setminus {0}$. And you know that $\Bbb{Z} \setminus {0}$ is countable.

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Here is hint to show the way when taking the union of two disjoint, countably infinite sets; it examines a concrete example. But the constructions/mechanics of this example can be used to prove the OP's case 3 (infinite/infinite).

Let $A = \{\, (m, 0) \, | \, m \in \mathbb N \, \}$ and $B = \{\, (m, 1) \, | \, m \in \mathbb N \, \}$. Since both of these sets are subsets of $\mathbb N \times \mathbb N$, we can apply the functions $\pi_1$ and $\pi_2$, the projections onto the first and second coordinate:

$\quad \pi_1: \mathbb N \times \mathbb N \to \mathbb N, \; \text{ with } \pi_1\left( (x,y) \right) = x$

$\quad \pi_2: \mathbb N \times \mathbb N \to \mathbb N, \; \text{ with } \pi_2\left( (x,y) \right) = y$

For any $z \in A \cup B$, define

$\tag 1 F(z) = 2 \, \pi_1(z) + \pi_2(z)$

It is easy to demonstrate that $F: A \cup B \to \mathbb N$ is a bijection, sending the set $A$ to the even integers and $B$ to the odd integers.

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  • $\begingroup$ Thanks @CopyPasteIt sir but i can't get what you mean by "the projections onto the first and second coordinate" and Why $2 \, \pi_1(z) + \pi_2(z) \equiv \mathbb N $.It will teach me a totally new thing thanks again $\endgroup$ – emonHR Mar 17 at 16:41
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    $\begingroup$ Oops after a while accidentally again I read your answer and finally understood it.Thanks again for a nice intuitive answer $\endgroup$ – emonHR Apr 3 at 14:04
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Numberphile has a video on this:

summary is:

  • first case is fine
  • your second case, can be done by mapping the previous cases up m.
  • your third case, can be done by noting both evens and odds are countably infinite sets, map n to 2n and place the rest in 2n-1
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Sets are countable if you can enumerate the elements ($=$ assign an index to each), without omission.

Let $C:=A\cup B$. From the enumerations

$$a_1,a_2,a_3,\cdots$$ and $$b_1,b_2,b_3,\cdots,$$

we form the enumeration

$$a_1,b_1,a_2,b_2,a_3,b_3,\cdots$$

also written

$$c_1,c_2,c_3,c_4,c_5,c_6,\cdots$$

In other terms, the elements of $A$ get the odd indexes and those of $B$ the even ones. You can check that this is an enumeration without omission.


If $A$ and $B$ have common elements, they will be cited twice. But this doesn't matter. (If you want, you can just skip the duplicates, but it is not even required.)

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