The Banach Space of the Real Numbers and Complex Numbers

# The Banach Space of the Real Numbers and Complex Numbers

Recall from the Banach Spaces page that a normed linear space $X$ is said to be a Banach space if every Cauchy sequence in $X$ converges in $X$, that is, $X$ is complete.

We now prove a very basic result - the normed linear space of the real numbers is a Banach space.

Theorem 1: $\mathbb{R}$ with the absolute value norm is a Banach space. |

**Proof:**Let $(x_n)_{n=1}^{\infty}$ be a Cauchy sequence in $\mathbb{R}$. Then for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then:

\begin{align} \quad |x_m - x_n| < \epsilon \quad (*) \end{align}

- In particular, for $\epsilon = 1 > 0$ there exists an $N_1 \in \mathbb{N}$ such that if $m, n \geq N_1$ then:

\begin{align} \quad |x_m - x_n| < 1 \quad \Leftrightarrow x_n - 1< \quad x_m < x_n + 1 \quad \Leftrightarrow |x_m| < |x_n| + 1 \end{align}

- We first show that $(x_n)_{n=1}^{\infty}$ is a bounded sequence. Let:

\begin{align} \quad M = \max \{ |x_1|, |x_2|, ..., |x_{N_1}| + 1 \} \end{align}

- Then for all $n \leq N_1$ we have that $|x_n| \leq M$ and for all $n \geq N_1$ we have that $|x_n| \leq M$. So indeed, $(x_n)_{n=1}^{\infty}$ is a bounded sequence. We use the Bolzano-Weierstrass theorem which says that every bounded sequence of real numbers has a convergent subsequence. Let $(x_{n_k})_{k=1}^{\infty}$ a convergent subsequence of $(x_n)_{n=1}^{\infty}$. Then for some $x \in \mathbb{R}$ we have that:

\begin{align} \quad \lim_{k \to \infty} x_{n_k} = x \quad (**) \end{align}

- We claim that $(x_n)_{n=1}^{\infty}$ converges to $x$. Let $\epsilon > 0$ be given. By $(*)$ there exists an $M_1 \in \mathbb{N}$ such that if $k, m \geq M_1$ then:

\begin{align} \quad |x_k - x_m| < \frac{\epsilon}{2} \end{align}

- And by $(**)$ there exists an $M_2 \in \mathbb{N}$ such that if $k \geq M_2$ then:

\begin{align} \quad |x_{n_k} - x| < \frac{\epsilon}{2} \end{align}

- Let $M = \max \{ M_1, M_2 \}$. Then if $k \geq M$ we have that:

\begin{align} \quad |x_k - x| \leq |x_k - x_{n_k}| + |x_{n_k} - x| < \frac{\epsilon}{2} + \frac{\epsilon}{2} \end{align}

- Therefore $(x_n)_{n=1}^{\infty}$ converges to $x \in \mathbb{R}$. So $\mathbb{R}$ is a Banach space. $\blacksquare$

An analogous proof shows that the space of complex numbers is also a Banach space. We state this result below.

Theorem 2: $\mathbb{C}$ with the absolute value norm is a Banach space. |

While $\mathbb{R}$ and $\mathbb{C}$ are Banach spaces, the space of rational numbers $\mathbb{Q}$ is not a Banach space.

Theorem 3: $\mathbb{Q}$ with the absolute value norm is NOT a Banach space. |

**Proof:**We exhibit a Cauchy sequence of rational numbers that converges to an irrational number.

- Consider the irrational number $\sqrt{2}$. For each $n \in \mathbb{N}$ let:

\begin{align} \quad x_n = \frac{\left \lfloor 10^n \sqrt{2} \right \rfloor}{10^n} \end{align}

- The first few terms of the sequence $(x_n)_{n=1}^{\infty}$ are:

\begin{align} \quad x_1 &= 1.4 \\ \quad x_2 &= 1.41 \\ \quad x_3 &= 1.414 \\ \quad x_4 &= 1.4142 \end{align}

- It is easy to see that each term $x_n$ gives us the first $n + 1$ digits in the decimal expansion for $\sqrt{2}$. Furthermore, this sequence is Cauchy. For each $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that:

\begin{align} \quad \frac{1}{10^N} < \epsilon \end{align}

- And for all $m, n \geq N$ we have that:

\begin{align} \quad |x_m - x_n| \leq \frac{1}{10^N} < \epsilon \end{align}

- Lastly, it is clear that $(x_n)_{n=1}^{\infty}$ converges to $\sqrt{2}$. But since $\sqrt{2} \not \in \mathbb{Q}$, we have exhibited a Cauchy sequence of rational numbers that does not converge IN $\mathbb{Q}$. So $\mathbb{Q}$ is not a Banach space. $\blacksquare$