# Union of Uncountably Many Uncountable Sets

I know that the union of countably many countable sets is countable. Is there an equivalent statement for uncountable sets, such as the union of uncountably many uncountable sets is uncountable? Furthermore, how does this generalize to other cardinals?

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Since a union of uncountable sets contains an uncountable set, it is definitely uncountable. The disjoint union of uncountably many non-empty sets of any size is likewise uncountable. – Thomas Andrews Dec 7 '11 at 2:27
Sure, union of sets of cardinality at least $\kappa$ has cardinality at least $\kappa$. – André Nicolas Dec 7 '11 at 2:40
Every set is either countable (which includes empty or finite sets as well as countably infinite sets) or uncountable. By definition, uncountable means the set is not countable. There are no other choices. So, if you take 1 or more uncountable sets, it will stay in the biggest class, uncountable. Even if you take uncountably many sets that are uncountable, there's no where above uncountable to go. Uncountable isn't a cardinal. If you want to talk about cardinals, then there are infinitely many choices, unlike what I just described. – Graphth Dec 7 '11 at 3:36
Your statement is true, but you should not think of countable and uncountable as being somehow analogous. In the statement "a countable union of countable sets is countable", the restriction to a countable union is necessary, because countable sets aren't supposed to be very big. Uncountable sets are very big, so any union of uncountable sets will be uncountable. – Jim Belk Dec 7 '11 at 4:52

It generalizes very naturally.

If $\kappa$ and $\lambda$ are cardinal numbers such that $\kappa$ is infinite and $0<\lambda\le\kappa$, the union of $\lambda$ sets of cardinality $\kappa$ has cardinality $\kappa$. In other words, the union of at least one and at most $\kappa$ sets of cardinality $\kappa$ has cardinality $\kappa$.

An alternative generalization is that if $\kappa$ is an infinite cardinal number, the union of at most $\kappa$ sets, each of cardinality at most $\kappa$, also has cardinality at most $\kappa$. In symbols, if $\lambda<\kappa$, and $\{A_\xi:\xi<\lambda\}$ is a family of sets such that $|A_\xi|\le\kappa$ for each $\xi<\lambda$, then $$\left|\bigcup_{\xi<\lambda}A_\xi\right|\le\kappa\;.$$

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What about an uncountable union of countable sets? Is that countable as well? – user12279 Dec 6 '13 at 13:43
@user12279: The union of uncountably many pairwise disjoint sets is always uncountable. – Brian M. Scott Dec 6 '13 at 18:29
Even if the pairwise disjoint sets are countable? I'm having a little trouble understanding why. For example, $\mathbb{N}$ is countable, so if I have an interval $[n,n+1)$ and I take an infinite union, wouldn't that still be countable? – user12279 Dec 6 '13 at 21:52
@user12279: It’s true even if the pairwise disjoint sets are singletons, i.e., sets with just one member. In that it’s especially easy to see: their union contains one point for each set, and there are uncountably many sets, so their union must be uncountable. – Brian M. Scott Dec 6 '13 at 22:05
@BrianM.Scott, do you know where can I find a proof for this ? – brick Mar 3 '15 at 21:00

To understand this property correctly we need to realize that "The union of finitely many finite sets is finite" first. If we replace "finite" by the word "small" then it would say $$\text{The cardinality of a union of a small amount of small sets is small.}$$

When jumping from finite numbers to infinite numbers this is also true, for example countable sets are very small in this fashion, and the countable union of countable sets is indeed countable.

Uncountable set is just a set which is not countable, it is like saying $x\neq 0$ for a real number. We have no way of knowing what $x$ is, just that it is nonzero. So talking about "an uncountable union of uncountable sets" makes very little sense.

However, we can "extend" the notion of being small. We can say that sets are small if they have cardinality $\aleph_1$ or less, e.g. if we work in a context of much larger cardinalities. In this case the union of $\aleph_1$ many sets of cardinality $\aleph_1$ is still with cardinality $\aleph_1$.

The same can be said about $\aleph_2$ as small, and so on. However it does not always hold. Having gone through $\aleph_1,\aleph_2,\ldots,\aleph_n,\ldots$ we reach $\aleph_\omega$ which is the least cardinal not being an $\aleph_n$.

If we have $|A_n|=\aleph_n$ then $\bigcup_{n} A_n$ has a cardinality larger than $A_n$ for all $n$, in particular it is larger than $\aleph_n$ for all $n$. In fact it is a countable union of uncountable sets, but the result is not with bijection with any $A_n$. This shows that if we say that a set is small if it has cardinality of $\aleph_n$ for some $n<\omega$ then a small family can be unified into a large set.

Definition: Let $\kappa$ be a cardinal, we say that $\kappa$ is regular if for every $\lambda<\kappa$ and $\lambda_i<\kappa$ for $i<\lambda$ we have: $$|\bigcup_{i<\lambda}\lambda_i|<\kappa$$

The first paragraph is to say that $\aleph_0$ is regular, the second is to say that $\aleph_1$ is indeed regular as well. The later part shows that $\aleph_\omega$ is not regular.

Theorem: If $\kappa$ is an infinite cardinal number then $\kappa^+$ (the successor cardinal of $\kappa$) is regular.

The result of this theorem is that if $\kappa$ is a regular cardinal we can extend the notion of "small" to "being of cardinality less than $\kappa$", while if $\kappa$ is singular we need to change some things in the definition of "small".

Of course that all of the above requires and relies heavily on the axiom of choice, without it we have that a countable union of finite sets can be uncountable; the countable union of countable sets can be of size $\aleph_1$; and even that every uncountable $\aleph$ cardinal is singular, which says that the notion of "small" behaves very strangely in the universe.

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