Suppose that $A$ and $B$ are two infinite sets and $|A|<|B|$. The question is that how to prove that $|A∪B|=|B|$. The proof is related to the Axiom of Choice.

# [Math] Cardinality of the union of two infinite set

axiom-of-choiceelementary-set-theory

#### Related Solutions

Suppose $A$ and $B$ are two sets of equal cardinalities, without any additional structure there are $2^{|A|}$ many bijections between them.

Using the axiom of choice we can choose $f_i\colon X_i\to Y_i$ which is a bijection, and using these bijections construct one from $\bigcup X_i$ to $\bigcup Y_i$ in the intuitive manner that you'd expect.

However since we are choosing from infinitely many sets *at once* we cannot just say that the sequence of bijections exists. Such assertion needs to be backed up, and the axiom of choice is exactly what allows us to back it up - the collections of bijections are nonempty, and so we may choose one from each one. From these choices we have a sequence of bijections which we can use to construct the bijection between the unions.

Without the axiom of choice it is possible that we cannot choose the bijections. It is possible to have that $A_i$ is countable for $i\in\omega$ but $|\bigcup A_i|=\aleph_1$, which is of course uncountable.

By your argument we could say that $A_i$ has a bijection with $\{i\}\times\omega$, however $\bigcup A_i$ is uncountable while $\bigcup(\{i\}\times\omega)=\omega\times\omega$ is countable. Therefore the unions are no longer in bijection.

Wait, it gets even worse. It is possible to have a *countable* family of disjoint *pairs* (sets of two elements each), and the union of is uncountable. Of course the uncountability of the union does not imply it has cardinality $\aleph_1$, but rather that it cannot be well ordered at all. On the other hand, we can write $\mathbb Z\setminus\{0\}$ as the pairs $\{-n,n\}$. The union of these countably many pairs is indeed countable.

With products the use of choice is even clearer. It is not trivial that the product is nonempty, but it does not imply that *every* product is empty.

In the above example of the pairs, we have countably many disjoint pairs $P_i$ such that $\prod P_i=\varnothing$. However if you take pairs of natural numbers, for example $\{2n,2n+1\}$ then the product is $2^\omega$ and the union is countable. This, once again, should hint you of the example.

So to answer both questions, the point we use the axiom of choice is in our choice of bijections to construct the bijection between the products or the unions. Of course this can be averted if additional structure exists on the sets, if they are all subsets of the same well ordering, for example.

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## Best Answer

$|A\cup B| \leq |A| + |B| \leq |B| + |B| = |B|$ since $|B|$ is of infinite cardinality and $2\cdot \aleph_k = \aleph _k$ for whichever cardinal $|B|$ happens to be.

Then also $|B|\leq |A\cup B|$ by subadditivity.

Hence $|A\cup B| = |B|$ when $B$ is an infinite set and $|A|\leq |B|$

I don't think you need to use the axiom of choice for the proof, though perhaps I have naive notions on how set operations work on higher sized sets.