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I am working myself into analysis 1 and i came across countable and uncountable sets.

Problem 1: I am very confused about the two terms and I would be very thankful if somebody could explain the meaning of those two terms in a different approach than a text book does.

Problem 2: Further I am interested in showing that the set of finite subsets of $\mathbb{N} $ is countable. Since countable sets have only countable subsets it seems apparent but i want to have a proper proof. Thank you for your advice and help.

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marked as duplicate by MJD, drhab, Najib Idrissi, vadim123, Namaste elementary-set-theory Oct 10 '14 at 14:50

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    $\begingroup$ For each $n$, the set of $n$-element subsets of $\Bbb{N}$ is countable. It is easy to prove. You may know that the countable union of countable sets are countable. $\endgroup$ – Hanul Jeon Oct 10 '14 at 14:11
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    $\begingroup$ Problem 1: We don't know what approach the textbook takes. Problem 2: This was asked several times before on the site, for example, one, two, and three. And there are more. Many more. $\endgroup$ – Asaf Karagila Oct 10 '14 at 14:17
  • $\begingroup$ Note that the set of all subsets of $\Bbb N$ is not countable, so your proposed reasoning, “since countable sets have only countable subsets it seems apparent”, is lacking something, and this should also suggest to you that your judgment of what is apparent cannot be trusted in this area. $\endgroup$ – MJD Oct 10 '14 at 14:28
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First thing first, countable sets.

There are two conventions, one which separates finite sets from countable sets, and another which includes them. Each has its merits, just like there are good reasons to include $0$ in the natural numbers and there are good reasons to exclude it.

Let me take here the approach where a countable set is finite set or a countably infinite set.

Definition. We say that a set $A$ is countable if there is a bijection of $A$ with an initial segment of $\Bbb N$.

This segment can be proper, therefore some finite set, or improper, meaning the entire set of natural numbers.

If $A$ has a bijection with a proper initial segment of $\Bbb N$, we say that $A$ is finite; if it has a bijection with the entire set then we say it's countably infinite; and otherwise we say that it is uncountable.

Next we can prove the following theorems:

  1. If $A\subseteq\Bbb N$ then $A$ is countable.

    We prove this by "collapsing" $A$ to an initial segment. If $A$ is empty, then we're done; otherwise we proceed by recursion and move the minimal element of $A$ to $0$; the next element of $A$ to $1$; and so on. If the induction cannot proceed, then $A$ is finite and we defined a bijection with a proper initial segment of $\Bbb N$; otherwise this is a bijection of $A$ with $\Bbb N$.

  2. $A$ is a countable set if and only if $A$ is empty or there is a surjection of $\Bbb N$ onto $A$ if and only if there is an injection from $A$ into $\Bbb N$.

    This is not a very difficult proof. Note that the key difference from the definition is that here we don't require the injection from $A$ into $\Bbb N$ to have an initial segment as its range.

  3. The union of two countable sets is countable. The Cartesian product of two countable sets is countable.

  4. Assume that $A_i$ are countable sets, and $f_i$ are injections from $A_i$ into $\Bbb N$, then $\bigcup A_i$ is a countable set.

    Using the axiom of choice we can conclude from this theorem that the union of a countable family of countable sets is countable. Since we can choose for each countable set an injection like that. Without the axiom of choice there might be some countable family of countable sets whose union is not countable.


As for the second part, there are more than a handful of different proofs. The easiest, perhaps, is to remember that a finite subset of $\Bbb N$ has a natural order defined on it, therefore we can map a set $\{n_1,\ldots,n_k\}$ to the product of $p_{n_1}^{n_1}\cdot\ldots\cdot p_{n_k}^{n_k}$, where $p_i$ is the $i$-th prime number.

Then the fundamental theorem of arithmetic tells us that this is an injection, so the set is countable.

One additional point which might be worth making is that despite the fact that every subset of $\Bbb N$ is countable, the collection of all subsets of $\Bbb N$ is not countable. We can prove this with Cantor's theorem which states that there is no surjection from a set onto the collection of all its subsets, so this collection is not countable.

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For your first problem: A countable set is a set which you can count its elements in infinite time. For example, the set of natural numbers $\mathbb{N} = \{ 1,2,3, \dots \}$ is countable because you can count it as $1,2,3,4,5, \dots$.

The set of integers $\mathbb{Z}= \{ \dots,-3,-2,-1,0,1,2,3,\dots \}$ is also countable because there is a way you can count all of its elements in infinite time. If you count it as $0,1,2,3,4,5,\dots$, you will never count $-1$. However, if you count the elements of $\mathbb Z$ as $0,1,-1,2,-2,3,-3,4,-4, \dots$ then you will be able to count all elements of $\mathbb Z$ in infinite time.

An uncountable set is a set such that whatever counting way you try, there are uncounted elements left. Remember: up above, we tried to count $\mathbb Z$ in a way, but the element $-1$ was uncounted. But we did not conclude that $\mathbb Z$ is uncoubtable, because there is a way to count it.

An example of an uncountable set is $\mathbb R$ and it is easy to prove using Cantor's diagonal argument.

For your second problem: You are asked to show that the number of finite subsets of $\mathbb N$ is countable. There are many many subsets of $\mathbb N$ and if you try to count all of these subsets (not the elements) (i.e. if you try to count the elements of $\mathcal P ( \mathbb N )$ ) you will see that it is uncountable. However, if you want to count only finite subsets of $\mathbb N$, then you will be able to count all of them in infinite time, and this is what you are asked to prove.

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For the second question there is an obvious bijection $f$ from $\Bbb N$ to the finite subsets of $\Bbb N$: write $n\in \Bbb N$ in binary notation $n\sum_{i\in\Bbb N}b_i2^i$ with $b_i\in\{0,1\}$, and put $f(n)=\{\,i\in\Bbb N\mid b_i=1\,\}$, which is finite.

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