# $\langle S\rangle = \bigcup \langle T \rangle$ where T $\subset$ S and T is finite.

Let $G$ be an abelian group and $| S |$ is infinite with $S\subset G$.

Then $\langle S\rangle = \bigcup \langle T\rangle$ where $T \subset S$ and $T$ is finite.

My attempt is below.

$\langle S \rangle \supset \bigcup\langle T\rangle$ is clear, since every $T$ is contained in $S$ and then each $\langle T \rangle$ is contained in $\langle S\rangle$.

Remains to show is $\langle S \rangle \subset \bigcup \langle T\rangle$.

Take any $x \in \langle S\rangle$.

Case I. $x \in S$. In this case, there is $T'$ containing $x$. so it is clear.

Case II. $x \notin S$. In this case, there are $a, b \in \langle S \rangle$ such that $ab=x$. and then there is $T''$ containing $a$ and $b$. Then $\langle T'' \rangle$ contains $x$.

But I'm not sure the case II. Could I say there are such $a$ and $b$ ? Does $x \in \langle S \rangle$ guarantee it?

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If $\,x\in\langle S\rangle\,$ , then $\;x\;$ is a finite combination of elements in $\,S\,$ , meaning:
$$\exists\,s_1,...s_k\in S\;\;s.t.\;\;x=s_1^{\epsilon_1}\cdot\ldots\cdot s_k^{\epsilon_k}\;,\;\;\epsilon_i=\pm 1\;\;\forall\,i$$