Reverse Triangle Inequality

Theorem

Let $M = \struct {X, d}$ be a metric space.

Then:

$\forall x, y, z \in X: \size {\map d {x, z} - \map d {y, z} } \le \map d {x, y}$

Normed Division Ring

Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring.

Then:

$\forall x, y \in R: \norm {x - y} \ge \bigsize {\norm x - \norm y}$

Normed Vector Space

Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.

Then:

$\forall x, y \in X: \norm {x - y} \ge \size {\norm x - \norm y}$

Real and Complex Numbers

Let $x$ and $y$ be elements of either the real numbers $\R$ or the complex numbers $\C$.

Then:

$\cmod {x - y} \ge \size {\cmod x - \cmod y}$

Seminormed Vector Space

Let $\struct {K, +, \circ}$ be a division ring with norm $\norm {\,\cdot\,}_K$.

Let $X$ be a vector space over $K$.

Let $p$ be a seminorm on $X$.

Then:

$\forall x, y \in X : \size {\map p x - \map p y} \le \map p {x - y}$

This article, or a section of it, needs explaining. In particular: $\size {\map p x - \map p y}$ You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

Proof

Let $M = \struct {X, d}$ be a metric space.

By Metric Space Axiom $(\text M 2)$: Triangle Inequality, we have:

$\forall x, y, z \in X: \map d {x, y} + \map d {y, z} \ge \map d {x, z}$

By subtracting $\map d {y, z}$ from both sides:

$\map d {x, y} \ge \map d {x, z} - \map d {y, z}$

Now we consider 2 cases.

Case $1$

Suppose $\map d {x, z} - \map d {y, z} \ge 0$.

Then:

$\map d {x, z} - \map d {y, z} = \size {\map d {x, z} - \map d {y, z} }$

and so:

$\map d {x, y} \ge \size {\map d {x, z} - \map d {y, z} }$

Case 2

Suppose $\map d {x, z} - \map d {y, z} < 0$.

Applying Metric Space Axiom $(\text M 2)$: Triangle Inequality again, we have:

$\forall x, y, z \in X: \map d {y, x} + \map d {x, z} \ge \map d {y, z}$

Hence:

$\map d {x, y} \ge \map d {y, z} - \map d {x, z}$

Since we assumed $\map d {x, z} - \map d {y, z} < 0$, we have that:

$\map d {y, z} - \map d {x, z} > 0$

and so:

$\map d {y, z} - \map d {x, z} = \size {\map d {y, z} - \map d {x, z} }$

Thus we obtain:

$\map d {x, y} \ge \size {\map d {x, z} - \map d {y, z} }$

Since these cases are exhaustive, we have shown that:

$\forall x, y, z \in X: \map d {x, y} \ge \size {\map d {x, z} - \map d {y, z} }$

$\blacksquare$

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: Exercise $2.6: 2$