Answer to Question #236909 in Discrete Mathematics for Arshdeep

Question

Check the extension of d given by d(x,y) = d(x,a) + 1 + d(b,y) where x belongs to X and y belongs to Y/ X

Answer 1

Given that

$d(x,y) = d(x,a) + 1 + d(b,y) \\ x \in X, y \in \frac{Y}{X}\\ So, y = x_0+y_0 \space s.t \space x_0 \in x, y_0 \in y\\ \implies d(x,y) - d(x,a) = 1 + d(b,y) \\ d(y,a) ≤ d(x,y) -d(x,a) = 1 + d(b,y) \\ d(y,a) -d(b,y) ≤1 \\ d(y,a) -d(y,b) ≤1 \\ d(a,b) ≤ d(y,a) -d(y,b)≤1\\ d(a,b)≤1 \space for \space all \space a,b$