Q11. Basic Topology (Rudin)
For \(x \in R^1 \) and \(y \in R^1 \) define
(a) \(d_1 (x, y) = (x - y)^2 \)
Determine whether it is a metric or not.
Solution: \(d_1 (x, y) = (x - y)^2 \) satisfies the condition that distance function is positive for all x, y and 0 when x = y since square is always non negative.
Also \(d_1 (x, y) = d_1 (y, x) \) as \((x - y)^2 = (y-x)^2 \)
Finally we check the triangular inequality. We have to show that \( (x-y)^2 + (y-z)^2 \ge (x-z)^2 \) . If this is true then we will have \( 2y^2 - 2xy - 2yz + 2zx \ge 0 \implies (y-x)(y-z) \ge 0 \) . But this is not true for all real numbers. For example if we have x < y < z then (y - x)(y - z) becomes negative. Thus x = 2, y = 4 and z = 7 violates the triangular inequality as \((2-4)^2 + (4-7)^2 \le (2-7)^2 \).
Hence \(d_1 \) is not a metric.
(b) \( d_2 (x, y) = \sqrt {|x -y| } \)
Determine whether it is a metric or not.
Solution: \( d_2 (x, y) = \sqrt {|x -y| } \) satisfies the condition that distance function is positive for all x, y and 0 when x = y since square root of |x-y| is always non negative and equals 0 only when x=y.
Also \(d_2 (x, y) = d_2 (y, x) \) as \(\sqrt {|x -y|} = \sqrt {|y-x|} \)
Finally we check the triangular inequality. We have to show that \(\sqrt {|x -y|} + \sqrt {|y-z|} \ge \sqrt {|x-z|} \)
For \(x \in R^1 \) and \(y \in R^1 \) define
(a) \(d_1 (x, y) = (x - y)^2 \)
Determine whether it is a metric or not.
Solution: \(d_1 (x, y) = (x - y)^2 \) satisfies the condition that distance function is positive for all x, y and 0 when x = y since square is always non negative.
Also \(d_1 (x, y) = d_1 (y, x) \) as \((x - y)^2 = (y-x)^2 \)
Finally we check the triangular inequality. We have to show that \( (x-y)^2 + (y-z)^2 \ge (x-z)^2 \) . If this is true then we will have \( 2y^2 - 2xy - 2yz + 2zx \ge 0 \implies (y-x)(y-z) \ge 0 \) . But this is not true for all real numbers. For example if we have x < y < z then (y - x)(y - z) becomes negative. Thus x = 2, y = 4 and z = 7 violates the triangular inequality as \((2-4)^2 + (4-7)^2 \le (2-7)^2 \).
Hence \(d_1 \) is not a metric.
(b) \( d_2 (x, y) = \sqrt {|x -y| } \)
Determine whether it is a metric or not.
Solution: \( d_2 (x, y) = \sqrt {|x -y| } \) satisfies the condition that distance function is positive for all x, y and 0 when x = y since square root of |x-y| is always non negative and equals 0 only when x=y.
Also \(d_2 (x, y) = d_2 (y, x) \) as \(\sqrt {|x -y|} = \sqrt {|y-x|} \)
Finally we check the triangular inequality. We have to show that \(\sqrt {|x -y|} + \sqrt {|y-z|} \ge \sqrt {|x-z|} \)
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