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I am trying to prove the below statement.

Let $A$ be a countable set, and let $B_n$ be the set of all $n$-tuples $(a_1,\ldots, a_n)$, where $n\in \mathbb{N}$ and $a_k\in A$ ($k=1,\ldots, n$), and the elements $a_1,\ldots, a_n$ need not be distinct. Then $B_n$ is countable.

My approach is to have a bijection for $B_n$ with natural numbers/subset of natural numbers.

Since $A$ is countable, every $a_k\in A$ ($k=1,\ldots, n$) is assigned a natural number. So I will count the number of digits in the natural number assigned to $a_k$ write those many $1$'s followed by zero for each $a_k$ and after I am done with all the $a_k$ I continue the number with natural number assigned to each $a_k$. this way I get a bijection with every $n$ tuple. I dont even have to know the numbers assigned to original countable set. I just used that fact to prove the statement.

Eg: [$a_1$ $a_2$ $a_3$] is the $3$ tuple, where $a_1$ is assigned 105, $a_2$ is assigned 23, $a_3$ is assigned 1431, in the original bijection. So by my theory the $3$-tuple is assigned to $111011011110105231431$.

I know there are far better methods, but I wanted to solve this on my own and I came up with this. I have thought of every possibility to maintain the bijection. If I have missed something, please help.

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    It's not clear that the map $B_n\to \mathbb N$ you describe is a bijection, although it is injective. For instance it will never map to a number with first digit $2$ – 2017-02-24
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    I am sorry, but I meant bijective with a subset of natural numbers. – 2017-02-24
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    I take that back, it's not clear that it's injective either. What about when the numbers involved are combinations of 1s and zeros. Where do you draw the line? – 2017-02-24
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    one would know what tuple is it before. – 2017-02-24
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    Maybe if you wrote the numbers on the right hand of the string backwards rather than forwards. That way you know where to start and you can read off the left hand side then know how much to read from the right, etc. – 2017-02-24
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    Yeah, but you need to prove it is injective... how do you know that there isn't a string that has multiple interpretations? – 2017-02-24
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    if I have three tuples, i will start after getting three zeros, if n tuple, i start after getting n zeros. For a set of N tuples, there is a unique representation. and nautral numbers dont start with zero. – 2017-02-24
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    I understand that the algorithm works forward... to prove injectivity you need to show it works in reverse. (to prove that it produces a *unique* integer for every tuple) – 2017-02-24
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    For instance, the modification I said earlier with writing the numbers on the RHS backwards would be manifestly injective cause you can recover the tuple from the string. Perhaps I'm missing something, but I don't think it's as clear the way you have it. It may well be injective- I haven't produced an example of a string that has two interpretations-but you would have to prove one doesn't exist. – 2017-02-24
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    I am using distinct natural numbers as each $a_k$ has distinct natural assigned to it. if certain digits are added to distinct natural numbers, overall natural number is distinct. so the final natural number assigned to a tuple is distinct. – 2017-02-24
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    It is certainly injective. 1110110111011011101 has to be 110,11,101. (Because you know it must be a 3 tuple and never a 4 tuple.) But it certainly isn't surjective -- nothing is mapped to 23, for isstance, or 11101101056-- but that isn't a requirement. – 2017-02-24
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    @jnyan Yep, sorry I missed that there was a fixed $n$ here. (Even though you told me so at one point and the problem wouldn't make sense without it.) – 2017-02-24
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    It's injective. The result will begin with a specific string of 1s and 4 zeros. That can only be interpreted as there by the number of digits between 0s. As three uniquely determined numbers, a,b,c. Then it will be followed after the fourth 9 be a+b+c digits that must break into three numbers of a,b,and, c digits. The is only one posible input for each outcome. But it isn'isn't surjective as illegitimate strings far out number the legitimate ones. – 2017-02-24
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    "Where do you draw the line?" You draw the line at the fourth 0. The number can only be a triplet (this is a drawback as it can not be extending easily to multiple B_n s. It can only be used specifically for n-tuples and you *must* know the value of n). – 2017-02-24
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    @fleablood agree, I was thinking about parsing a tuple of arbitrary length for some reason . You're right, since we know $n=3$ we know where the line is. – 2017-02-24
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    It's reasonable assumption. A proof of the countability of all arbitrary length tuples will be asked almost immediately and this will not be very applicable. – 2017-02-24
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    @fleablood well not reasonable cause OP was clear in the question, but you're right. – 2017-02-24
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    i am sorry but arbitary length of tuples is not countable right?? I mean power set of Natural numbers is a set of arbitary tuples of countable set. But power set is not countable. What am I missing? – 2017-02-24

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