1.3: Metric spaces
- Page ID
- 23579
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 \(\mathcal{X}\) be a nonempty set and \(d\) be a function which returns a real number \(d(A, B)\) for any pair \(A, B \in \mathcal{X}\). Then \(d\) is called metric on \(\mathcal{X}\) if for any \(A, B, C \in \mathcal{X}\), the following conditions are satisfied:
(a) Positiveness:
\[d(A, B) \ge 0.\]
(b) \(A = B\) if and only if
\[d(A, B) = 0.\]
(c) Symmetry:
\[d(A, B) = d(B, A)\]
(d) Triangle inequality:
\[d(A, C) \le d(A, B) + d(B, C).\]
A metric space is a set with a metric on it. More formally, a metric space is a pair \((\mathcal{X}, d)\) where \(\mathcal{X}\) is a set and \(d\) is a metric on \(\mathcal{X}\).
The elements of \(\mathcal{X}\) are called points of the metric space. Given two points \(A, B \in \mathcal{X}\), the value \(d(A, B)\) is called distance from \(A\) to \(B\).
Let \(\mathcal{X}\) be an arbitrary set. For any \(A, B \in \mathcal{X}\), set \(d(A, B) = 0\) if \(A = B\) and \(d(A, B) = 1\) otherwise. The metric \(d\) is called discrete metric on \(\mathcal{X}\).
Set of all real numbers (\(\mathbb{R}\)) with metric \(d\) defined by
\[d(A, B) := |A - B|.\]
Exercise \(\PageIndex{1}\)
Show that \(d(A, B) = |A - B|^2\) is not a metric on \(\mathbb{R}\).
Metrics on the plane. Suppose that \(\mathbb{R}^2\) denotes the set of all pairs \((x, y)\) of real numbers. Assume \(A = (x_A, y_A)\) and \(B = (x_B, y_B)\). Consider the following metrics on \(\mathbb{R}^2\):
- Euclidean metric, denoted by \(d_2\), and defined as
\[d_2(A, B) = \sqrt{(x_A - x_B)^2 + (y_A - y_B)^2}.\] - Manhattan metric, denoted by \(d_1\) and defined as
\[d_1(A, B) = |x_A - x_B| + |y_A - y_B|.\] - Maximum metric, denoted by \(d_{\infty}\) and defined as
\[d_{\infty}(A, B) = \max \{|x_A - x_B|, |y_A - y_B|\}.\]
- Hint
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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
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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.