Generalized Hilbert Sequence Space is Metric Space/Metric Space Axiom M2

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Theorem

Let $\alpha$ be an infinite cardinal number.

Let $I$ be an indexed set of cardinality $\alpha$.

Let $A$ be the set of all real-valued functions $x : I \to \R$ such that:

$(1)\quad \set{i \in I: x_i \ne 0}$ is countable

$(2)\quad$ the generalized sum $\ds \sum_{i \mathop \in I} x_i^2$ is a convergent net.

Let $d_2: A \times A \to \R$ be the real-valued function defined as:

$\ds \forall x = \family {x_i}, y = \family {y_i} \in A: \map {d_2} {x, y} := \paren {\sum_{i \mathop \in I} \paren {x_i- y_i}^2}^{\frac 1 2}$

Then $d_2$ satisfies Metric Space Axiom $(\text M 2)$: Triangle Inequality.

Proof

Let $H = \struct{\ell^2, d_{\ell^2}}$ denote the Hilbert sequence space, where:

$\ell^2$ denotes the real $2$-sequence space, that is, the set of all real sequences $\sequence {x_n}$ such that the series $\ds \sum_{n \mathop = 0}^\infty x_n^2$ is convergent

$d_{\ell^2}$ denotes the real $2$-sequence metric, that is, the real-valued function $d_{\ell^2}: \ell^2 \times \ell^2: \to \R$ defined as:

$\ds \forall x = \sequence {x_n}, y = \sequence {y_n} \in \ell^2: \map {d_{\ell^2}} {x, y} := \paren {\sum_{n \mathop = 0}^\infty \paren {x_n - y_n}^2}^{\frac 1 2}$

Lemma

Let $x_1, x_2, \ldots, x_m \in A$.

Then there exists $y_1, y_2, \ldots, y_m \in \ell^2$:

$\forall a,b \in \closedint 1 m : y_a \ne y_b \iff x_a \ne x_b$

$\forall a,b \in \closedint 1 m : \map {d_{\ell^2} } {y_a, y_b} = \map {d_2} {x_a, x_b}$

$\Box$

Let $x_1, x_2, x_3 \in A$.

From Lemma:

$\exists y_1, y_2, y_3 \in \ell^2 : $

$\forall i, j \in \set{1, 2, 3} : \map {d_2} {x_i, x_j} = \map {d_{\ell^2}} {y_i, y_j}$

We have:

\(\ds \map {d_2} {x_1, x_3}\)

\(=\)

\(\ds \map {d_{\ell^2} } {y_1, y_3}\)

Lemma

\(\ds \)

\(\le\)

\(\ds \map {d_{\ell^2} } {y_1, y_2} + \map {d_{\ell^2} } {y_2, y_3}\)

Metric Space Axiom $(\text M 2)$: Triangle Inequality applied to $d_{\ell^2}$

\(\ds \)

\(=\)

\(\ds \map {d_2} {x_1, x_2} + \map {d_2} {x_2, x_3}\)

Lemma

The result follows.

$\blacksquare$