# Let [ ] be a countable number of finite sets. Prove [ ]

## Homework Statement

Problem:
Let A_1 , A_2 , . . . be a countable number of finite sets. Prove that the union S = ⋃_i A_i is countable.

Solution:
Included in the TheProblemAndSolution.jpg file.

## Homework Equations

Set-theoretic algebra.

## The Attempt at a Solution

Unless I missed something, it seems that I understand everything about that solution except why the author chose to use ##f(b_{ij}) = m_1 + m_2 + . . . + m_{i - 1} + j##.

Could someone please explain to me why that specific function was constructed? I get that the author is trying to show that there is a one-to-one correspondence between the set S and the set of natural numbers, but I don't understand that particular sum was chosen. What's the logic behind the decision?

Any input would be GREATLY appreciated!

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Dick
Homework Helper

## Homework Statement

Problem:
Let A_1 , A_2 , . . . be a countable number of finite sets. Prove that the union S = ⋃_i A_i is countable.

Solution:
Included in the TheProblemAndSolution.jpg file.

## Homework Equations

Set-theoretic algebra.

## The Attempt at a Solution

Unless I missed something, it seems that I understand everything about that solution except why the author chose to use ##f(b_{ij}) = m_1 + m_2 + . . . + m_{i - 1} + j##.

Could someone please explain to me why that specific function was constructed? I get that the author is trying to show that there is a one-to-one correspondence between the set S and the set of natural numbers, but I don't understand that particular sum was chosen. What's the logic behind the decision?

Any input would be GREATLY appreciated!
You didn't link you jpg correctly. So I can't see it. But it appears the author is just writing a function that counts total number of elements in the finite union of all of the sets before the i'th set and then includes j elements of the i'th set.

Oops!

Here is the image I meant to attach (just in case you misinterpreted the situation, which I don't think you did, but just in case).

Is ##f(b_{ij}) = m_1 + m_2 + . . . + m_{i - 1} + m_i## equivalent to ##f(b_{ij}) = m_1 + m_2 + . . . + m_{i - 1} + j##? I don't know if it's because it's late as I write this, but I still don't see why the authors chose the specific letters they did.

Also you said "finite union", was that a mistake? I say this, because there can be a countably infinite amount of sets (where each set has a finite number of elements).

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Dick
Homework Helper
Oops!

Here is the image I meant to attach (just in case you misinterpreted the situation, which I don't think you did, but just in case).

Is ##f(b_{ij}) = m_1 + m_2 + . . . + m_{i - 1} + m_i## equivalent to ##f(b_{ij}) = m_1 + m_2 + . . . + m_{i - 1} + j##? I don't know if it's because it's late as I write this, but I still don't see why the authors chose the specific letters they did.

Also you said "finite union", was that a mistake? I say this, because there can be a countably infinite amount of sets (where each set has a finite number of elements).
The author is just counting the elements in ##S## to get a 1-1 correspondence with the natural numbers. If there are three elements in ##B_1## then assign them to 1,2,3. Then if there are four elements in ##B_2## assign them to 4,5,6,7. Etc, etc. That formula just expresses that way of counting. So, for example, for the second element of ##B_2##, ##b_{22}=3+2=5##.

So, would that be a yes to the question

"Is ##f(b_{ij}) = m_1 + m_2 + . . . + m_{i - 1} + m_i## equivalent to ##f(b_{ij}) = m_1 + m_2 + . . . + m_{i - 1} + j##?"

?

haruspex