Mathematical Immaturity

1.5 Exercises

• 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. 19. All convergent real sequences.
• Recall the definition of a convergent sequence from Volume 1, Section 10.2. Definition.$\quad$ A sequence $\{f(n)\}$ is said to have a limit $L$ if, for every positive number $\epsilon,$ there is another positive number $N$ (which may depend on $\epsilon$) such that $$ |f(n) - L| < \epsilon \quad \textit{for all}\ n \geq N. $$In this case, we say the sequence $\{f(n)\}$ converges to $L$ and we write $$ \lim_{n \to \infty} f(n) = L, \quad \text{or} \quad f(n) \to L \,\ \text{as}\,\ n \to \infty. $$
• Let $\{f(n)\}$ and $\{g(n)\}$ be convergent real sequences with limits $L$ and $M,$ respectively. Then for every positive number $\epsilon,$ there is another positive number $N$ (which may depend on $\epsilon$) such that $$ |f(n) - L| < \epsilon, \quad \text{and} \quad |g(n) - M| < \epsilon $$for all $n \geq N.$ Adding these two inequalities, we find that for all $n \geq N,$ we have $$ |f(n) - L| + |g(n) - M| < 2\epsilon $$But, recalling the triangle inequality for real numbers, we know that $$ |f(n) + g(n) - (L + M)| \leq |f(n) - L| + |g(n) - M| $$Thus, for all positive $\epsilon$ and all $n \geq N,$ we have $$ |f(n) + g(n) - (L + M)| < 2\epsilon $$Which means that $\{f(n) + g(n)\}$ is convergent, and the set of all convergent real sequences satisfies closure under addition. Now, suppose $\{f(n)\}$ is a convergent real sequence with limit $L.$ We know by definition that there exists some positive $N$ such that $|f(n) - L| < \epsilon$ for all $n \geq N$ and for all $\epsilon > 0.$ If we multiply both sides of this inequality by $|a|,$ where $a$ is some real scalar, then we find that for all $n \geq N,$ $$ |af(n) - aL | < |a|\epsilon $$But since this is the case for all $\epsilon > 0,$ this implies that $af(n) \to aL$ as $n \to \infty,$ which implies that $\{af(n)\}$ is convergent, and the set of all convergent sequences satisfies closure under multiplication by real numbers. We can see that the sequence with $f(n) = 0$ for all $n$ is convergent with limit $L = 0,$ satisfying the existence of a zero element. In verifying closure under multiplication by real numbers, we can deduce the existence of negatives by setting $a = -1$ and noting that $$ |aL - af(n)| = |af(n) - aL|. $$ And since $f(n)$ is real-valued for all $n,$ we can easily verify the remaining axioms on addition and multiplication by scalars. Hence, the set of all convergent real sequences is a real linear space. $\,\blacksquare$