Birkhoff-Kakutani Theorem/Topological Vector Space/Lemma

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Lemma

Let $\GF \in \set {\R, \C}$.

Let $\struct {X, \tau}$ be a first-countable Hausdorff topological vector space over $\GF$.

Let $D$ be the set of real numbers with a terminating binary notation.

Let $\sequence {V_n}_{n \mathop \in \N}$ be a local basis of ${\mathbf 0}_X$ such that:

$V_{n + 1} + V_{n + 1} \subseteq V_n$ for each $n \in \N$.

For $r \ge 1$, set $\map A r = X$.

For $r \in D$, set:

$\ds \map A r = \sum_{j \mathop = 1}^\infty \map {c_j} r V_j$

where $\ds \sum_{j \mathop = 1}^\infty$ denotes linear combination.

Let $r, s \in D$ be such that $r + s < 1$.


Then:

$\map A r + \map A s \subseteq \map A {r + s}$


Proof

Proof by induction:

For all $m \in \N$, let $\map P m$ be the proposition:

if $\map {c_n} r = \map {c_n} s = 0$ for $n > m$, then:
$\map A r + \map A s \subseteq \map A {r + s}$


Basis for the Induction

We prove $\map P 1$.

If $\map {c_n} r = \map {c_n} s = 0$ for $n > 1$, then:

$r \in \set {0, 2^{-1} }$ and $s \in \set {0, 2^{-1} }$.

Since we have also imposed $r + s < 1$, we have $\set {r, s} = \set {0, 2^{-1} }$.

Then, we have $r + s = 2^{-1}$ and so:

$\map A r + \map A s = 2^{-1} V_1 = \map A {2^{-1} } = \map A {r + s}$

So we have shown that $\map P 1$ holds.

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P N$ is true, where $N \ge 1$, then it logically follows that $\map P {N + 1}$ is true.

So this is our induction hypothesis:

if $\map {c_n} r = \map {c_n} s = 0$ for $n > N$, we have:
$\map A r + \map A s = \map A {r + s}$


Then we need to show:

if $\map {c_n} r = \map {c_n} s = 0$ for $n > N + 1$, we have:
$\map A r + \map A s = \map A {r + s}$


Induction Step

This is our induction step.

Let $r, s \in D$ be such that $r + s < 1$ and $\map {c_n} r = \map {c_n} s = 0$ for $n > N + 1$.

Let:

$r' = r - \map {c_{N + 1} } r 2^{-\paren {N + 1} }$
$s' = s - \map {c_{N + 1} } s 2^{-\paren {N + 1} }$

Then we have $\map {c_n} {r'} = \map {c_n} {s'} = 0$ for $n > N$, so that:

$\map A {r'} + \map A {s'} \subseteq \map A {r' + s'}$

and:

$\map {c_n} {r' + s'} = 0$ for $n > N$.

We also have:

$\map A r = \map A {r'} + \map {c_{N + 1} } r V_{N + 1}$

and:

$\map A s = \map A {s'} + \map {c_{N + 1} } s V_{N + 1}$

Then, we have:

$\map A r + \map A s \subseteq \map A {r' + s'} + \map {c_{N + 1} } r V_{N + 1} + \map {c_{N + 1} } s V_{N + 1}$

If $\map {c_{N + 1} } r = \map {c_{N + 1} } s = 0$, then $r = r'$ and $s = s'$ and we obtain $\map A r + \map A s \subseteq \map A {r' + s'} = \map A {r + s}$.


Now consider the case that $\map {c_{N + 1} } r = 0$ and $\map {c_{N + 1} } s = 1$.

We have:

$\ds \map A r + \map A s \subseteq \map A {r' + s'} + V_{N + 1}$

We have:

\(\ds \map A {r' + s'} + V_{N + 1}\) \(=\) \(\ds \sum_{n \mathop = 1}^N \map {c_n} {r' + s'} V_n + V_{N + 1}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^N \map {c_n} {r' + s'} V_n + V_{N + 1}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^{N + 1} \map {c_n} {r' + s' + 2^{-\paren {N + 1} } } V_n\) since $\map {c_n} {r' + s'} = 0$ for $n > N$
\(\ds \) \(=\) \(\ds \map A {r' + s' + 2^{-\paren {N + 1} } }\)
\(\ds \) \(=\) \(\ds \map A {r + s}\)

So we obtain $\map A r + \map A s \subseteq \map A {r + s}$ for $\map {c_{N + 1} } r = 0$ and $\map {c_{N + 1} } s = 1$.

By swapping $r$ and $s$, we obtain $\map A r + \map A s \subseteq \map A {r + s}$ for $\map {c_{N + 1} } r = 1$ and $\map {c_{N + 1} } s = 0$.


Now consider the case $\map {c_{N + 1} } r = 1$ and $\map {c_{N + 1} } s = 1$.

We have:

$\map A r + \map A s \subseteq \map A {r' + s'} + V_{N + 1} + V_{N + 1} \subseteq \map A {r' + s'} + V_N = \map A {r' + s'} + \map A {2^{-N} }$

Since $\map {c_n} {r' + s'} = 0$ for $n > N$ and $\map {c_n} {2^{-N} } = 0$ for $n > N$, we have:

\(\ds \map A {r' + s'} + \map A {2^{-N} }\) \(\subseteq\) \(\ds \map A {r' + s' + 2^{-N} }\)
\(\ds \) \(=\) \(\ds \map A {\paren {r' + 2^{-\paren {N + 1} } } + \paren {s' + 2^{-\paren {N + 1} } } }\)
\(\ds \) \(=\) \(\ds \map A {r + s}\)

Hence we obtain $\map A r + \map A s \subseteq \map A {r + s}$ in the case $\map {c_{N + 1} } r = 1$ and $\map {c_{N + 1} } s = 1$.

Hence we have proven that $\map A r + \map A s \subseteq \map A {r + s}$ if $\map {c_n} r = \map {c_n} s = 0$ for $n > N + 1$, completing the proof.

$\blacksquare$

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