Normed Vector Space of Rational Numbers is not Banach Space

From ProofWiki
Jump to navigation Jump to search

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:

\(\ds x_{n \mathop + 1}^2 - 2\) \(=\) \(\ds \frac {\paren {4 + 3 x_n}^2 } {\paren {3 + 2 x_n}^2} - 2\)
\(\ds \) \(=\) \(\ds \frac {16 + 24 x_n + 9 x_n^2 - 18 - 24 x_n - 8 x_n^2} {\paren {3 + 2 x_n}^2}\)
\(\ds \) \(=\) \(\ds \frac {x_n^2 - 2} {\paren {3 + 2 x_n}^2}\)

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:

\(\ds x_n - x_{n \mathop + 1}\) \(=\) \(\ds x_n - \frac {4 + 3 x_n} {3 + 2 x_n}\)
\(\ds \) \(=\) \(\ds \frac {3 x_n + 2 x_n^2 - 4 - 3x_n} {3 + 2 x_n}\)
\(\ds \) \(=\) \(\ds \frac {2 \paren {x_n^2 - 2} } {3 + 2 x_n}\)
\(\ds \) \(\ge\) \(\ds 0\)

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.

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Normed_Vector_Space_of_Rational_Numbers_is_not_Banach_Space&oldid=505565"