Euclidean Plus Metric is Metric

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\R$ be the set of real numbers.

Let $d: \R \times \R \to \R$ be the Euclidean plus metric:

$\map d {x, y} := \size {x - y} + \ds \sum_{i \mathop = 1}^\infty 2^{-i} \map \inf {1, \size {\max_{j \mathop \le i} \frac 1 {\size {x - r_j}} - \max_{j \mathop \le i} \frac 1 {\size {y - r_j} } } }$


Then $d$ is indeed a metric.


Proof

Recall that $\set {r_j}_{j \mathop \in \N}$ is an enumeration of the rational numbers $\Q$.


Also, we note that:

\(\ds \) \(\) \(\ds \sum_{i \mathop = 1}^\infty 2^{-i} \map \inf {1, \size {\max_{j \mathop \le i} \frac 1 {\size {x - r_j} } - \max_{j \mathop \le i} \frac 1 {\size {x - r_j} } } }\)
\(\ds \) \(\le\) \(\ds \sum_{i \mathop = 1}^\infty 2^{-i} 1\) Definition of Infimum of Subset of Real Numbers
\(\ds \) \(=\) \(\ds 1\) Sum of Infinite Geometric Sequence: Corollary 1 with $z = \dfrac 1 2$

meaning that this is a convergent series and so the definition is meaningful.


Next the axioms for a metric are checked in turn.


Proof of Metric Space Axiom $(\text M 1)$

\(\ds \map d {x, x}\) \(=\) \(\ds \size {x - x} + \sum_{i \mathop = 1}^\infty 2^{-i} \map \inf {1, \size {\max_{j \mathop \le i} \frac 1 {\size {x - r_j} } - \max_{j \mathop \le i} \frac 1 {\size {x - r_j} } } }\)
\(\ds \) \(=\) \(\ds 0 + \sum_{i \mathop = 1}^\infty 2^{-i} \map \inf {1, 0}\)
\(\ds \) \(=\) \(\ds 0\)

$\Box$


Proof of Metric Space Axiom $(\text M 2)$: Triangle Inequality

Let $i \in \N$ be fixed.

Define:

$\map {f_i} x := \ds \max_{j \mathop \le i} \frac 1 {\size {x - r_j} }$


Then:

\(\ds \) \(\) \(\ds \inf \set {1, \size {\map {f_i} x - \map {f_i} y} } + \inf \set {1, \size {\map {f_i} y - \map {f_i} z} }\)
\(\ds \) \(=\) \(\ds \inf \set {1 + 1, 1 + \size {\map {f_i} y - \map {f_i} z}, \size {\map {f_i} x - \map {f_i} y} + 1, \size {\map {f_i} x - \map {f_i} y} + \size {\map {f_i} y - \map {f_i} z} }\) Sum of Infima is Infimum of Sums


Now since $\inf \set {1, \size {\map {f_i} x - \map {f_i} z} } \le 1$, it follows that:

\(\ds \inf \set {1, \size {\map {f_i} x - \map {f_i} z} }\) \(\le\) \(\ds 1 + 1\)
\(\ds \inf \set {1, \size {\map {f_i} x - \map {f_i} z} }\) \(\le\) \(\ds 1 + \size {\map {f_i} y - \map {f_i} z}\)
\(\ds \inf \set {1, \size {\map {f_i} x - \map {f_i} z} }\) \(\le\) \(\ds \size {\map {f_i} x - \map {f_i} y} + 1\)


Suppose now that $\size {\map {f_i} x - \map {f_i} z} \le 1$.

Then:

\(\ds \inf \set {1, \size {\map {f_i} x - \map {f_i} z} }\) \(=\) \(\ds \size {\map {f_i} x - \map {f_i} z}\)
\(\ds \) \(\le\) \(\ds \size {\map {f_i} x - \map {f_i} y} + \size {\map {f_i} y - \map {f_i} z}\) Triangle Inequality for Real Numbers


On the other hand, if:

$\size {\map {f_i} x - \map {f_i} z} > 1$

then also:

$\size {\map {f_i} x - \map {f_i} y} + \size {\map {f_i} y - \map {f_i} z} > 1$

and:

\(\ds \inf \set {1, \size {\map {f_i} x - \map {f_i} z} }\) \(=\) \(\ds 1\)
\(\ds \) \(<\) \(\ds \size {\map {f_i} x - \map {f_i} y} + \size {\map {f_i} y - \map {f_i} z}\)


Combining both cases with the estimates above, we conclude:

$\ds \inf \set {1, \size {\map {f_i} x - \map {f_i} z} } \le \inf \set {1, \size {\map {f_i} x - \map {f_i} y} } + \inf \set {1, \size {\map {f_i} y - \map {f_i} z} }$


Finally, now, we have:

\(\ds \map d {x, z}\) \(=\) \(\ds \size {x - z} + \sum_{i \mathop = 1}^\infty 2^{-i} \map \inf {1, \size {\map {f_i} x - \map {f_i} z} }\)
\(\ds \) \(\le\) \(\ds \paren {\size {x - y} + \size {y - z} } + \sum_{i \mathop = 1}^\infty 2^{-i} \le \inf \set {1, \size {\map {f_i} x - \map {f_i} y} } + \inf \set {1, \size {\map {f_i} y - \map {f_i} z} }\)
\(\ds \) \(=\) \(\ds \paren {\size {x - y} + \sum_{i \mathop = 1}^\infty 2^{-i} \le \inf \set {1, \size {\map {f_i} x - \map {f_i} y} } } + \paren {\size {y - z} + \sum_{i \mathop = 1}^\infty 2^{-i} \le \inf \set {1, \size {\map {f_i} y - \map {f_i} z} } }\)
\(\ds \) \(=\) \(\ds \map d {x, y} + \map d {y, z}\)

$\Box$


Proof of Metric Space Axiom $(\text M 3)$

We have that:

\(\ds \sum_{i \mathop = 1}^\infty 2^{-i} \map \inf {1, \size {\max_{j \mathop \le i} \frac 1 {\size {x - r_j} } - \max_{j \mathop \le i} \frac 1 {\size {y - r_j} } } }\) \(=\) \(\ds \sum_{i \mathop = 1}^\infty 2^{-i} \map \inf {1, \size {-\paren {\max_{j \mathop \le i} \frac 1 {\size {x - r_j} } - \max_{j \mathop \le i} \frac 1 {\size {y - r_j} } } } }\) Absolute Value of Negative
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^\infty 2^{-i} \map \inf {1, \size {\max_{j \mathop \le i} \frac 1 {\size {y - r_j} } - \max_{j \mathop \le i} \frac 1 {\size {x - r_j} } } }\)


Hence:

\(\ds \map d {x, y}\) \(=\) \(\ds \size {x - y} + \sum_{i \mathop = 1}^\infty 2^{-i} \map \inf {1, \size {\max_{j \mathop \le i} \frac 1 {\size {x - r_j} } - \max_{j \mathop \le i} \frac 1 {\size {y - r_j} } } }\)
\(\ds \) \(=\) \(\ds \size {y - x} + \sum_{i \mathop = 1}^\infty 2^{-i} \map \inf {1, \size {\max_{j \mathop \le i} \frac 1 {\size {x - r_j} } - \max_{j \mathop \le i} \frac 1 {\size {y - r_j} } } }\) $\size {\, \cdot \,}$ is a metric
\(\ds \) \(=\) \(\ds \size {y - x} + \sum_{i \mathop = 1}^\infty 2^{-i} \map \inf {1, \size {\max_{j \mathop \le i} \frac 1 {\size {y - r_j} } - \max_{j \mathop \le i} \frac 1 {\size {x - r_j} } } }\) from above
\(\ds \) \(=\) \(\ds \map d {y, x}\)

$\Box$


Proof of Metric Space Axiom $(\text M 4)$

Suppose that $x \ne y$.

Then:

\(\ds \size {x - y} + \sum_{i \mathop = 1}^\infty 2^{-i} \map \inf {1, \size {\max_{j \mathop \le i} \frac 1 {\size {x - r_j} } - \max_{j \mathop \le i} \frac 1 {\size {y - r_j} } } }\) \(\ge\) \(\ds \size {x - y}\)
\(\ds \) \(>\) \(\ds 0\) $\size {\, \cdot \,}$ is a metric

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Euclidean_Plus_Metric_is_Metric&oldid=534078"