The set obtained by a countable union of countable sets

Question

I tried to proof that the set obtained by a countable union of countable sets is countable by using cardinal arithmetic:

First, notice that if $A_1\subseteq A_2$ then $A_1\cup A_2=A_2$ thus the claim holds trivially.

Therefore, I will assume that the following sets are disjoint and will use induction:

Base: $\left | A_1 \cup A_2 \right | = \left | A_1 \right |+\left | A_2 \right |\underset{a}{\leq} \aleph_0$

Assumption: assume that the claim hold for $\bigcup_{i=1}^{n}A_i$ where each $A_i$ is countable and the sets are disjoint.

Step: $\left | \bigcup_{i=1}^{n+1}A_i \right |\underset{b}{=}\left | \bigcup_{i=1}^{n}A_i \right |+\left | A_{i+1} \right |\underset{c}{\leq} \aleph_0$

a. By cardinal arithmetic we know that the sum of two cardinal numbers that are less or equal to $\aleph_0$ is less or equal to $\aleph_0$.

b. We assume that the sets are disjoint.

c. By the induction assumption and a.

what I'm asking

The mission was to show that the set obtained by a countable union of countable sets is countable.

I showed (hopefully) that the set obtained by a finite union of countable sets, is countable.

Did I show also the claim for union of $\aleph_0$ sets?

You showed that for any finite set, the union is countable because that matches the inductive hypothesis. This doesn’t prove the infinite union case. — Michael Burr, Jul 28 '24 at 16:49
@MichaelBurr Thank you! Will you be able to elaborate why this doesn't proof the infinite union case ? Is there something I could add to the answer which could proof the infinite union case ? — Daniel, Jul 28 '24 at 16:53
@copper.hat It is, $$\aleph_0\cdot\aleph_0=\aleph_0$$
It seems that I just needed the following line:

for $$\bigcup_{i=1}^{\aleph_0}A_i : \forall i, \left | A_i \right |\leq\aleph_0\ \ and\ \ \forall (k\neq j) ,A_k\cap A_j =\emptyset$$

we get $$\left | \bigcup_{i=1}^{\aleph_0} A_i\right |\leq\aleph_0\cdot\aleph_0=\aleph_0$$

Is this what you meant? — Daniel, Jul 28 '24 at 17:12
Cardinals are not my strong suit, but I would argue $A_k \sim {k} \times A_k$ (bijection), $|{k} \times A_k| \le | \mathbb{N}|$ (injection), so $|\cup_k A_k | \le | \cup_k {k} \times A_k | $ (injection, need to be careful here, but everything is countable), $| \cup_k {k} \times A_k | \le | \cup_k {k} \times \mathbb{N} |$ (injection), and finally $| \cup_k {k} \times \mathbb{N} | = | \mathbb{N} \times \mathbb{N}|$ (bijection). — copper.hat, Jul 28 '24 at 17:33
By "countable", do you mean "at most countable", or "infinite countable"? — Anne Bauval, Jul 28 '24 at 18:08

score 3 · Accepted Answer · answered Jul 28 '24 at 18:13

3

To "assume that the following sets are disjoints", you need a little work: set $$B_1:=A_1,\;B_2:=A_2\setminus A_1,\dots,B_n:=A_n\setminus(A_1\cup\dots\cup A_{n-1}).$$ This way, $\cup_{i=1}^nA_i=\sqcup_{i=1}^nB_i.$
Apart from this, your proof (by induction on $n$) for a union of $n$ sets is correct, but you didn't prove anything more.
Obviously, no such proof, i.e. no proof that $\forall n\in\Bbb N\quad P(n)$, for some property $P$, suffices to imply $P(\aleph_0)$. Think for instance of: $P(n)=$ "every union of $n$ finite sets is finite".

answered Jul 28 '24 at 18:13

Anne Bauval

49,005

I got a downvote, while clearly your answer show that the post different from the one you linked - the question asks also if the proof is correct (which you stated in your answer is not complete) . Can you please change it? – Daniel Jul 29 '24 at 12:52
The downvote was not from me, and I discovered only after answering that your post and my answer were true duplicates (I disagree with you on this point). The best I can do is delete my (therefore redundant) answer, but for this I need you to unaccept it. – Anne Bauval Jul 29 '24 at 13:01
The answer you gave is great. Will it change the downvote if I will unaccept it ? – Daniel Jul 29 '24 at 13:10
Again: the downvote has nothing to do with me, and my answer is essentially a duplicate of those in the linked post. – Anne Bauval Jul 29 '24 at 13:12

score 1 · Answer 2 · answered Jul 28 '24 at 17:47

Proving a statement is true for all finite unions (which you've done) doesn't prove the statement for infinite unions.

False claim: All subsets of the natural numbers have an upper bound.

Fake proof: Let $A \subset \Bbb N$ wiith $\vert A \vert = n$ where $n$ is finite and work by induction on $n$.

If $n=1$, then $A= \{k \}$ for some $k \in \Bbb N$, so $k+1$ is an upper bound for $A$.

If $n=m+1$ and we have proven the statement for $m$, then $A=\{k_1, k_2, \ldots, k_m \} \cup \{ k_{m+1} \}$ for some choices of $k_i$. Let $M$ be an upper bound for $\{k_1, k_2, \ldots, k_m \}$. Then $\max \{M, k_{m+1}+1 \}$ is an upper bound for $A$, proving the inductive step, and therefore the (false) claim.

I actually have proven above that all finite subsets of $\Bbb N$ have an upper bound, but that proof does not (and cannot) establish that all infinite subsets of $\Bbb N$ have an upper bound. In fact, my false claim is a false as it can possibly be -- no infinite subsets of $\Bbb N$ have an upper bound.

When you want to use induction to prove a statement about infinite sets, you need to use transfinite induction, which has the additional requirement of proving your statement also is true for limit ordinals (such as $\omega$). That's what's missing from your proof (and my fake proof).

The set obtained by a countable union of countable sets

what I'm asking

2 Answers2