I saw this exercise in a book:

Suppose $G$ is a group. Suppose $T$ is an index set and for each $t \in T$, $H_t$ is a subgroup of $G$. Furthermore, if $\{H_t\}$ is a monotonic collection, show that $\bigcup_{t \in T} H_t$ is a subgroup of $G$.

My question is if $T$ is not $\mathbb N$, how to define "monotonic"?

If it is defined by what Jyrki Lahtonen suggested:

For arbitrary $k$, $l \in T$, either $H_k \subseteq H_l$ or $H_l \subseteq H_k$ (or both).

Then here is my proof:

Suppose $a$, $b \in \bigcup_{t \in T}H_t$, then $a \in H_k$ for some $k$ and $b \in H_l$ for some $l$. So either $a$, $b \in H_k$ or $a$, $b \in H_l$ (or both). Then either $a \cdot b \in H_k$ or $a \cdot b \in H_l$ (or both). Hence $a \cdot b \in \bigcup_{t \in T}H_t$. Therefore $\bigcup_{t \in T}H_t$ is a subgroup.

My question is:

  1. Is the above proof correct?

  2. I think it looks too trivial. In my opinion, if you have a monotonic collection of sets, all satisfying certain criterion, then "obviously" the union also satisfies that criterion. Can someone give a counter-example?

  • 2
    $\begingroup$ This doesn't make sense. If $T$ is just a set, then there is no possibility of defining monotonicity. If $T$ is a partial order, then monotonic probably means $s \leq t \Rightarrow H_s \subseteq H_t$. But then the claim is not true. For example, if $T$ is discrete (i.e. $s \leq t$ iff $s=t$), then every collection of subgroups is monotonic, but it is well-known that the union of distinct subgroups is almost never a subgroup. What we really need is that $T$ is a directed partial order. And in that case the proof is a two-liner. $\endgroup$ Jun 21, 2014 at 14:08
  • 2
    $\begingroup$ I guess that monotonic simply means that if $\ell$ and $k$ are two arbitrary elements of $T$, then you have either $H_\ell\subseteq H_k$ or $H_k\subseteq H_\ell$ (or possibly both). The usual subgroup criterion then gives the claim relatively quickly. $\endgroup$ Jun 21, 2014 at 21:08
  • 2
    $\begingroup$ Sensible people call such a thing a chain of subgroups. $\endgroup$ Jun 22, 2014 at 2:59
  • 1
    $\begingroup$ BY the way, it is never a good idea to mention a book saying «a book»: what book is it? Adding more or less complete bibliographic information always helps everybody: you, the OP,because it makes it easier for others to answer your question by giving context, and others because, well, because it does. $\endgroup$ Jun 22, 2014 at 3:03
  • 1
    $\begingroup$ You definition of "monotonic collection" is quite often called chain, see Wikipedia, ProofWiki. $\endgroup$ Jun 22, 2014 at 6:10

3 Answers 3


The definition of monotonic is indeed the one suggested in Jyrki's comment

In the book Elements of Abstract and Linear Algebra by Edwin H. Connell1 we can find the following definition:

Definition A collection of sets is said to be monotonic if, given any two sets of the collection, one is contained in the other.

I would also mention that such system of sets is quite often called chain, see Wikipedia, ProofWiki. (Or we could simply call it totally ordered subset, or linearly ordered subset.)

The above definition is mentioned in the book when discussion Hausdorff maximality principle. The formulation given in the book is that "every monotonic collection is contained in a maximal monotonic collection".

1The OP mentioned in another post that this is the book where the problem comes from.

You also asked this:

I think it looks too trivial. In my opinion, if you have a monotonic collection of sets, all satisfying certain criterion, then "obviously" the union also satisfies that criterion. Can someone give a counter-example?

Well a trivial example would be considering the property "a set is finite". You can easily find a monotonic collection (chain) of finite subsets, such that the union is finite.

But Teichmüller–Tukey lemma say that your conjecture is true for properties of finite character.

A comment to your proof: To show that the union is indeed a subgroup, you should also show that: 1) it is non-empty, 2) it is closed under taking inverses (i.e., $a\in\bigcup H_t$ $\Rightarrow$ $a^{-1}\in\bigcup H_t$).

These parts are rather easy, but you should mention both of them for the proof to be complete.


Does the book define "index set"? I'm guessing that by index set they actually mean what would normally be called a directed set, which is a set with a preorder in which every pair of elements has an upper bound. So, in this case monotonic would mean that if $s \le t$, then $H_s \subseteq H_t$. I don't see how the claim can be true without assuming $T$ is directed.


Your proof is not enough, since not every subset of a group that is closed under products is a subgroup. Consider the positive integers as a subset of the positive rationals, with multiplication.

However, the idea of the proof is there, just recall that to be a subgroup, a set needs to be not just closed under products, but also under inverses, and that can be verified in this setting just as easily.

However, your suggestion in 2. is too naive, and does not work without some care: Consider the case where you have a chain $N_0\subset N_1\subset N_2\subset\dots$ with each $N_i$ order isomorphic to $\mathbb N$, say $N_0=\{a_0<a_1<\dots\}$ and $N_1=\{a_{-1}<a_0<a_1<\dots\}$ and $N_2=\{a_{-2}<a_{-1}<a_0<\dots\}$, etc. Now, you could object that this is not a good example, in that I am changing the first element, but the criterion all models satisfy is simply that there is a least element, not which element it is.

Naturally, this suggests that there may be a version of 2. that works, but special attention has to be paid to either the language or the theory we want to preserve. The right framework to study this question is model theory, and there are indeed theorems giving positive versions of 2.

Theorem. A $\Pi_2$ formula $\varphi$ is preserved by unions.

To understand what this means, assume you have a family of structures in a common language -- in the sense of model theory, but this is very general: These could be sets with a binary relation, or sets $S$ with a unary function $f:S\to S$ and some distinguished "constant" $a\in S$, or ...

A first order formula $\phi$ is $\Pi_2$, or universal-existential, iff it has the form $$\forall x_1\dots\forall x_n\exists y_1\dots\exists y_m \psi(x_1,\dots,x_n,y_1,\dots,y_m,a_1,\dots,a_k)$$ where $\psi$ has no quantifiers. for instance, the field axioms have this form.

The theorem says that if we are given a chain of such structures, and the statement $\phi$ is true in all of them, then it is also true in their union. All that matters is that $\phi$ is "not too complicated" syntactically, in the specific form described above.

In fact, the same holds not just for chains but for direct limits of directed systems, so the result really covers many natural examples.

The topic of what statements or theories are preserved under unions is a classical problem well studied in model theory. Remarkably, we have:

Theorem. A theory $T$ is preserved under unions of chains iff each axiom of $T$ is universal-existential.

This is Theorem 3.2.3 in the classical Chang-Keisler Model theory, but it is treated in detail in many other sources as well.

For instance, partial orders are preserved under unions. For a more interesting example, the theory of Boolean algebras is also preserved under unions: Given a chain of Boolean algebras, their union is again a Boolean algebra. Here we need to be careful on how the language is specified. Boolean algebras are completely described by their natural partial order, but perhaps you find more natural to describe them in terms of a least and largest elements, with conjunction, disjunction and negation, and to say that an algebra $A$ is a substructure of an algebra $B$ means that, as sets, $A\subset B$, and the operations of $A$ are the restrictions of the operations of $B$, plus the requirement that the $0$ of $A$ is the $0$ of $B$, and also the $1$ of $A$ is the $1$ of $B$.

On the other hand, the theory of atomic Boolean algebras is not preserved under unions. Table 3.2.1in Chang-Keisler provides many other examples.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .