Question #71393

Is the real line a metric space?
1

Expert's answer

2017-12-05T13:20:06-0500

Answer on Question #71393, Math / Functional Analysis

Is the real line a metric space?

**Solution.** Yes, the real line is a metric space.

Recall that a set XX is a metric space if there is a function d:X2Pd: X^2 \to \mathbf{P} such that for each x,y,zXx, y, z \in X:

(M1) d(x,y)0d(x,y)\geq 0;

(M2) d(x,y)=d(y,x)d(x,y) = d(y,x);

(M3) d(x,y)d(x,z)+d(z,y)d(x,y)\leq d(x,z) + d(z,y);

(M4) d(x,y)=0x=yd(x,y) = 0 \Leftrightarrow x = y.

Let d:P2Pd: \mathbf{P}^2 \to \mathbf{P} be defined as:


d(x,y)=xy for each x,yP.d (x, y) = | x - y | \text{ for each } x, y \in \mathbf {P}.


(M1) d(x,y)0d(x,y)\geq 0 because for each x,yPx,y\in \mathbf{P};

(M2) d(x,y)=xy=(yx)=yx=d(y,x)d(x,y) = |x - y| = |-(y - x)| = |y - x| = d(y,x);

(M3) d(x,y)=xy=(xz)+(zy)xz+zy=d(x,z)+d(z,y)d(x,y) = |x - y| = |(x - z) + (z - y)| \leq |x - z| + |z - y| = d(x,z) + d(z,y);

(M4) d(x,y)=0xy=0xy=0x=yd(x,y) = 0 \Leftrightarrow |x - y| = 0 \Leftrightarrow x - y = 0 \Leftrightarrow x = y.

Hence, the real line is a metric space.

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