1.3: Metric spaces
( \newcommand{\kernel}{\mathrm{null}\,}\)
The notion of metric space provides a rigorous way to say: “we can mea- sure distances between points”. That is, instead of (i) on Section 1.1, we can say “Euclidean plane is a metric space”.
Let X be a nonempty set and d be a function which returns a real number d(A,B) for any pair A,B∈X. Then d is called metric on X if for any A,B,C∈X, the following conditions are satisfied:
(a) Positiveness:
d(A,B)≥0.
(b) A=B if and only if
d(A,B)=0.
(c) Symmetry:
d(A,B)=d(B,A)
(d) Triangle inequality:
d(A,C)≤d(A,B)+d(B,C).
A metric space is a set with a metric on it. More formally, a metric space is a pair (X,d) where X is a set and d is a metric on X.
The elements of X are called points of the metric space. Given two points A,B∈X, the value d(A,B) is called distance from A to B.
Let X be an arbitrary set. For any A,B∈X, set d(A,B)=0 if A=B and d(A,B)=1 otherwise. The metric d is called discrete metric on X.
Set of all real numbers (R) with metric d defined by
d(A,B):=|A−B|.
Exercise 1.3.1
Show that d(A,B)=|A−B|2 is not a metric on R.
Metrics on the plane. Suppose that R2 denotes the set of all pairs (x,y) of real numbers. Assume A=(xA,yA) and B=(xB,yB). Consider the following metrics on R2:
- Euclidean metric, denoted by d2, and defined as
d2(A,B)=√(xA−xB)2+(yA−yB)2. - Manhattan metric, denoted by d1 and defined as
d1(A,B)=|xA−xB|+|yA−yB|. - Maximum metric, denoted by d∞ and defined as
d∞(A,B)=max
- Hint
-
Check the triangle inequality for A = 0, B = 1 and C = 2.
Exercise \PageIndex{2}
Prove that the following functions are metrics on \mathbb{R}^2:
(a) d_1;
(b) d_2;
(c) d_{\infty}.
- Answer
-
Only the triangle inequality requires a proof — the rest of conditions in Definition 1.1 are evident. Let A = x_A, y_A), B = (x_B, y_B), and C = (x_C, y_C). Set
x_1 = x_B - x_A, y_1 = y_B - y_A,
x_2 = x_C - x_B, y_2 = y_C - y_B.(a). The inequality
d_1(A, C) \le d_1(A, B) + d_1(B, C)
can be written as
|x_1 + x_2| + |y_1 + y_2| \le |x_1| + |y_1| + |x_2| + |y_2|.
The latter follows since |x_1 + x_2| \le |x_1| + |x_2| and |y_1 + y_2| \le |y_1| + |y_2|.
(b). The inequality
d_2(A, C) \le d_2 (A, B) + d_2(B, C)
can be written as
\sqrt{(x_1 + x_2)^2 + (y_1 + y_2)^2} \le \sqrt{x_1^2 + y_1^2} + \sqrt{x_2^2 + y_2^2}.
Take the square of the left and the right hand sides, simplify, take the square again and simplify again. You should get the following inequality
0 \le (x_1 \cdot y_2 - x_2 \cdot y_1)^2,
which is equivalent to 1.3.9 and evidently true.
(c). The inequality
d_{\infty} (A, C) \le d_{\infty} (A, B) + d_{\infty} (B, C)
can be written as
\max \{|x_1 + x_2|, |y_1 + y_2|\} \le \max \{|x_1|, |y_1|\} + \max \{|x_2|, |y_2|\}.
Without loss of generality, we may assume that
\max \{|x_1 + x_2|, |y_1 + y_2|\} = |x_1 + x_2|.
Further,
|x_1 + x_2| \le |x_1| + |x_2| \le \max \{|x_1|, |y_1|\} + \max \{|x_2|, |y_2|\}.
Hence 1.3.13 follows.