I saw this exercise in "Elements of Abstract and Linear Algebra" by E. H. Connell:
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 $G$.
Suppose "monotonic" is defined as follows:
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:
Is the above proof correct?
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?
