Space of Bounded Sequences with Pointwise Addition and Pointwise Scalar Multiplication on Ring of Sequences forms Vector Space

Theorem
Let $\ell^\infty$ be the space of bounded sequences.

Let $\struct {\R, +_\R, \times_\R}$ be the field of real numbers.

Let $\paren +$ be the pointwise addition on the ring of sequences.

Let $\paren {\, \cdot \,}$ be the pointwise multiplication on the ring of sequences.

Then $\struct {\ell^\infty, +, \, \cdot \,}_\R$ is a vector space.

Proof
Let $\sequence {a_n}_{n \mathop \in \N}, \sequence {b_n}_{n \mathop \in \N}, \sequence {c_n}_{n \mathop \in \N} \in \ell^\infty$.

Let $\lambda, \mu \in \R$.

Let $\sequence 0 := \tuple {0, 0, 0, \dots}$ be a real-valued function.

Let us use real number addition and multiplication.

Define pointwise addition as:


 * $\sequence {a_n + b_n}_{n \mathop \in \N} := \sequence {a_n}_{n \mathop \in \N} +_\R \sequence {b_n}_{n \mathop \in \N}$.

Define pointwise scalar multiplication as:


 * $\sequence {\lambda \cdot a_n}_{n \mathop \in \N} := \lambda \times_\R \sequence {a_n}_{n \mathop \in \N}$

Let the additive inverse be $\sequence {-a_n} := - \sequence {a_n}$.

Closure Axiom
By assumption, $\sequence {a_n}_{n \mathop \in \N}, \sequence {b_n}_{n \mathop \in \N} \in \ell^\infty$.

By definition:


 * $\displaystyle \sup_{n \mathop \in \N} \size {a_n} < \infty$


 * $\displaystyle \sup_{n \mathop \in \N} \size {b_n} < \infty$

Consider the sequence $\sequence {a_n + b_n}$.

Then:

Hence:


 * $\sequence {a_n + b_n} \in \ell^\infty$

Commutativity Axiom
By Pointwise Addition on Ring of Sequences is Commutative, $\sequence {a_n} + \sequence {b_n} = \sequence {b_n} + \sequence {a_n}$

Associativity Axiom
By Pointwise Addition on Ring of Sequences is Associative, $\paren {\sequence {a_n} + \sequence {b_n} } + \sequence {c_n} = \sequence {a_n} + \paren {\sequence {b_n} + \sequence {c_n} }$.

Convergent Subsequence of Cauchy Sequence
Let $\epsilon > 0$.

Let $N \in \N$ be such that:


 * $\displaystyle \forall n, m > N : \norm {x_n - x_m} < \frac \epsilon 2$

Let $n_K \in \N$ be such that:


 * $\displaystyle n_K > N \implies \norm {x_{n_K} - x} < \frac \epsilon 2$

Then: