Simplicity is directed union-closed

Verbal statement
A directed union of simple subgroups is simple.

Statement with symbols
Suppose $$G$$ is a group, $$I$$ is a nonempty directed set, and $$H_i, i \in I$$ is a collection of subgroups such that $$i < j \implies H_i \le H_j$$. Then, if all the $$H_i$$s are simple, so is their union.

Directed set
A directed set is a partially ordered set $$(I, \le)$$ such that if $$i,j \in I$$, there exists $$k \in I$$, such that $$i \le k, j \le k$$.

Simple group
A nontrivial group is simple if it has no proper nontrivial normal subgroup.

Facts used

 * 1) Directed union of subgroups is subgroup
 * 2) Normality satisfies transfer condition: The intersection of a normal subgroup with another subgroup is normal in that subgroup.

Proof
Given: A group $$G$$, a collection of simple subgroups $$H_i$$ indexed by a directed set $$I$$, such that $$i < j \implies H_i \le H_j$$.

To prove: The union of $$H_i$$ is a simple subgroup.

Proof: Suppose $$H$$ is the union of the $$H_i$$s. By fact (1), $$H$$ is a subgroup, so it suffices to show that $$H$$ is simple. We do this by showing that any normal subgroup $$N$$ of $$H$$ is either trivial or equal to $$H$$.

For each $$i$$, consider $$N \cap H_i$$. By fact (2), this is a normal subgroup of $$H_i$$, hence is either trivial or equals the whole of $$H_i$$. Suppose $$N \cap H_i = H_i$$. Then, $$N$$ is a normal subgroup of $$H$$ containing $$H_i$$.

Now, consider any subgroup $$H_j, j \ne i$$. By the definition of directed set, there exists $$k \in I$$ such that $$i,j < k$$. So, $$H_k$$ contains both $$H_i$$ and $$H_j$$. Thus, the intersection $$N \cap H_k$$ is nontrivial (since the intersection contains $$H_i$$). By the simplicity of $$H_k$$, and fact (2) again, $$N \cap H_k = H_k$$, so $$H_k \le N$$. In particular, $$H_j \le N$$. Thus, every $$H_j, j \in I$$, is contained in $$N$$, Thus, $$N = H$$.

Thus, if $$N \cap H_i = H_i$$ for any $$i \in I$$, $$N = H$$. This leaves the case that $$N \cap H_i$$ is trivial for every $$i \in I$$, forcing $$N \cap (\bigcup_i) H_i$$ to be trivial, and thus forcing $$N \cap H = N$$ to be trivial.