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The reverse triangle inequality $|d(x,z) - d(y,z)| \le d(x,y)$ in any metric space

Statement

Let $(X,d)$ be a metric space (Metric space: $d(x,y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) and let $x, y, z \in X$ . Then

$|d(x,z) - d(y,z)| \le d(x,y),$

where $|\cdot|$ is the absolute value of $\mathbb{R}$ (Absolute value in an ordered field).

Facts & Assumptions

Given: A metric space $(X,d)$ and points $x, y, z \in X$ ; write $t := d(x,z) - d(y,z)$ .

[A1]

The triangle inequality (M3): $d(a,c) \le d(a,b) + d(b,c)$ for all $a,b,c \in X$ (Metric space: $d(x,y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[A2]

Symmetry (M2): $d(a,b) = d(b,a)$ for all $a,b \in X$ (Metric space: $d(x,y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L1]

For every real $u$ , the value $|u|$ equals $u$ or $-u$ (Basic properties of the absolute value, Absolute value in an ordered field).

[L2]

Adding a constant to an inequality: if $a \le b$ then $a + c \le b + c$ . Order is preserved by adding a constant and by adding inequalities states the strict form $a < b \Rightarrow a + c < b + c$ ; the nonstrict form used here is that strict form together with the case $a = b$ , in which the two sides are equal, the order being total (Ordered field, Complete ordered field (least-upper-bound property)).

Proof

technique · direct

1.1

By [A1] at $(a,b,c) = (x,y,z)$ : $d(x,z) \le d(x,y) + d(y,z)$ .

1.2

By [A1] at $(a,b,c) = (y,x,z)$ : $d(y,z) \le d(y,x) + d(x,z)$ , and by [A2] $d(y,x) = d(x,y)$ , so $d(y,z) \le d(x,y) + d(x,z)$ .

A1A2

2.1

Adding $-d(y,z)$ to both sides of step 1.1 gives $t = d(x,z) - d(y,z) \le d(x,y)$ .

step 1.1L2

2.2

Adding $-d(x,z)$ to both sides of step 1.2 gives $d(y,z) - d(x,z) \le d(x,y)$ , that is $-t \le d(x,y)$ .

step 1.2L2

3.1

By [L1] the real number $|t|$ is either $t$ or $-t$ , and both of these are at most $d(x,y)$ by steps 2.1 and 2.2, so $|d(x,z) - d(y,z)| = |t| \le d(x,y)$ .

step 2.1step 2.2L1∎

Remarks

Read with $z$ fixed, this says the function $u \mapsto d(u,z)$ does not increase distances: its values at $x$ and at $y$ differ by at most $d(x,y)$ . That is the model for $|d(x,A) - d(y,A)| \le d(x,y)$ , so the distance to a fixed nonempty set is $1$ -Lipschitz, which proves the same estimate with the point $z$ replaced by a nonempty set.
The inequality specialises, on $\mathbb{R}$ with $d(u,v) = |u-v|$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$ , and it is unbounded), to the familiar $\big||x - z| - |y - z|\big| \le |x - y|$ .

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 19 results over 6 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Triangle inequality (Wikipedia) (standard reference, not scraped)
Metric space (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 2 (standard reference, not scraped)

The reverse triangle inequality ∣d(x,z)−d(y,z)∣≤d(x,y)|d(x,z) - d(y,z)| \le d(x,y)∣d(x,z)−d(y,z)∣≤d(x,y) in any metric space