In Exercises 1 through 28, determine whether each of the given sets is a real linear space, if addition and multiplication by real scalars are defined in the usual way. For those that are not, tell which axioms fail to hold. The functions in Exercises 1 through 17 are real-valued. In Exercises 3, 4, and 5, each function has domain containing 0 and 1. In Exercises 7 through 12, each domain contains all real numbers.
21. All absolutely convergent real series.
Recall the definition of absolute convergence from Volume 1, Section 10.18.
$\text{Definition. $\quad$}$ A series $\sum a_n$ is called absolutely convergent if $\sum |a_n|$ converges. It is called conditionally convergent if $\sum a_n$ converges but $\sum |a_n|$ diverges.
Suppose we have two absolutely convergent series $\sum a_n$ and $\sum b_n,$ and let $\alpha$ and $\beta$ be real scalars. Then, by the triangle inequality for real numbers, we have
\begin{align*}
\sum_{n = 1}^{M} |\alpha a_n + \beta b_n| &\leq |\alpha|\sum_{n = 1}^{M} |a_n| + |\beta|\sum_{n = 1}^{M} |b_n|
\\
\\
&\leq |\alpha|\sum_{n = 1}^{\infty} |a_n| + |\beta|\sum_{n = 1}^{\infty} |b_n|
\end{align*}But since $\sum|\alpha a_n + \beta b_n|$ is bounded by $|\alpha|\sum|a_n| + |\beta|\sum |b_n|,$ we know from Volume 1, Theorem 10.1, that the sequence of the partial sums of $\sum|\alpha a_n + \beta b_n|$ must be convergent. Hence $\sum \alpha a_n + \beta b_n$ is absolutely convergent. Moreover, this shows that the set of absolutely convergent series satisfies closure under addition and closure under multiplication by real numbers. This also satisfies the existence of negatives by setting $\beta = -\alpha$ and $b_n = a_n.$
The existence of the zero element is satisfied by setting $a_n = 0$ for all $n,$ and the remaining axioms can be verified by noting that the parial sums of any absolutely convergent series are real numbers whose absolute value is bounded by some positive real number $S.$
As such, we can see that the set of all absolutely convergent series is a real linear space. $\,\blacksquare$
