Normed Vector Space of Rational Numbers is not Banach Space

Theorem
Let $\struct {\Q, \size {\, \cdot \,}}$ be the normed vector space of rational numbers.

Then $\struct {\Q, \size {\, \cdot \,}}$ is not a Banach space.

Proof
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $\Q$ defined recursively in the following way:


 * $\ds x_0 = \frac 3 2$


 * $\ds \forall n \in \N_{> 0} : x_{n \mathop + 1} = \frac {4 + 3 x_n} {3 + 2 x_n}$

We have that:


 * $\forall n \in \N : x_n \ge 0$

Note that:

Hence:


 * $\paren {x_n \ge \sqrt 2} \implies \paren {x_{n \mathop + 1} \ge \sqrt 2}$

For $n = 0$ we have $\ds \frac 9 4 \ge 2$, or $\ds \frac 3 2 \ge \sqrt 2$.

Therefore:


 * $\forall n \in \N : x_n \ge \sqrt 2$

Furthermore:

Hence:


 * $\forall n \in \N : x_n \ge x_{n \mathop + 1}$

By monotone convergence theorem, $\sequence {x_n}_{n \mathop \in \N}$ converges in $\struct {\R, \size {\, \cdot \,}}$.

By Convergent Sequence in Normed Vector Space is Cauchy Sequence, $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy in $\struct {\R, \size {\, \cdot \,}}$.

On the other hand:


 * $\forall n \in \N : x_n \in \Q$

Thus, $\sequence {x_n}_{n \mathop \in \N}$ is also Cauchy in $\struct {\Q, \size {\, \cdot \,}}$.

Suppose $\sequence {x_n}_{n \mathop \in \N}$ converges to $L \in \Q$.

By combination theorem for sequences:


 * $\ds L = \frac {4 + 3 L} {3 + 2 L}$

or $L^2 = 2$.

$L$ has to be positive, so $L = \sqrt 2$.

However, $\sqrt 2 \notin \Q$.

Hence, $\sequence {x_n}_{n \mathop \in \N}$ does not converge in $\struct {\Q, \size {\, \cdot \,} }$.

Altogether, we have that $\sequence {x_n}_{n \mathop \in \N} \in \Q$ is a Cauchy sequence which is not convergent in $\struct {\Q, \size {\, \cdot \,} }$.

By definition, $\struct {\Q, \size {\, \cdot \,}}$ is not a Banach space.

Also see

 * Rational Number Space is not Complete Metric Space