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.
18. All bounded real sequences.
Recall from Theorem 10.1 (Volume 1, Section 10.3) that a sequence $\{f(n)\}$ is called $bounded$ if there exists a positive number $M$ such that $|f(n)| \leq M$ for all $n.$
Let $\{f(n)\}$ and $\{g(n)\}$ be bounded real sequences. That means that there are positive real numbers $M$ and $N$ such that $|f(n)| \leq M$ and $|g(n)| \leq N$ for all $n.$ But this means that $|f(n)| + |g(n)| \leq M + N$ for all $n.$ Using the triangle inequality for real numbers (Theorem I.39 - Volume 1, Section I4.8) we find that $|f(n) + g(n)| \leq M + N$ for all $n,$ showing that the sequence $\{f(n) + g(n)\}$ is bounded, hence verifying closure under addition.
If we multiply both sides of the inequality by $|f(n)| \leq M$ by $|a|,$ with $a$ being a real scalar, then we can see that $\{af(n)\}$ is also bounded, thus verifying closure under multiplication by real numbers. The existence of negatives can be deduced by the definition of a bounded sequence because if $|f(n)| \leq M$ then $ |-f(n)| \leq M$ for all $n.$ Setting $f(n) = 0$ for all $n$ verifies the existence of a zero element since $0$ is bounded by all $M > 0.$ Then, for the remaining axioms, we can verify them by noting that if $\{f(n)\}$ is bounded by some $M > 0$, then for each $n$, $f(n)$ is some real number with $|f(n)| \leq M$.
As such, we have shown that the set of all bounded sequences is a real linear space. $\,\blacksquare$
