- #1
Streltsy
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Technically, this is not a homework question, since I solely seek an answer for self-indulgence.
Example 1.1.4. Suppose f and g are functions in a space X = {f : [0, 1] → R}. Does
d(f, g) =max|f − g| define a metric?
(1) d(x, y) ≥ 0 for all x, y ∈ X
(2) d(x, y) = 0 if and only if x = y
(3) d(x, y)=d(y, x)
(4) d(x, z) ≤ d(x, y) + d(y, z)
So, from my understanding: for d(f, g) to define a metric on X, it has to satisfy all the given properties of a metric.
Well, my question is not necessarily whether d(f, g) defines a metric (though I wouldn't mind a proof of it); I was wondering if property (2) is satisfied.
Because in my pursuit of an understanding in topology, I stumbled across a compilation of notes, in which the note-taker mentions that the second property is not satisfied.
The reasoning is: that, "by considering two arbitrary functions at any point within the interval [0, 1]. If |f(x) − g(x)| = 0, this does
not imply that f = g because f and g could intersect at one, and only one, point."
However, I was wondering if that could also be said about d(f, g) =max|f − g|, which is the function being originally considered; since if d(f, g) = 0, then max|f − g|= 0, which means that for all points in [0,1], 0 ≤|f − g| ≤ max|f − g| = 0, or |f − g|= 0; which would further imply f = g.
Homework Statement
Example 1.1.4. Suppose f and g are functions in a space X = {f : [0, 1] → R}. Does
d(f, g) =max|f − g| define a metric?
Homework Equations
(1) d(x, y) ≥ 0 for all x, y ∈ X
(2) d(x, y) = 0 if and only if x = y
(3) d(x, y)=d(y, x)
(4) d(x, z) ≤ d(x, y) + d(y, z)
The Attempt at a Solution
So, from my understanding: for d(f, g) to define a metric on X, it has to satisfy all the given properties of a metric.
Well, my question is not necessarily whether d(f, g) defines a metric (though I wouldn't mind a proof of it); I was wondering if property (2) is satisfied.
Because in my pursuit of an understanding in topology, I stumbled across a compilation of notes, in which the note-taker mentions that the second property is not satisfied.
The reasoning is: that, "by considering two arbitrary functions at any point within the interval [0, 1]. If |f(x) − g(x)| = 0, this does
not imply that f = g because f and g could intersect at one, and only one, point."
However, I was wondering if that could also be said about d(f, g) =max|f − g|, which is the function being originally considered; since if d(f, g) = 0, then max|f − g|= 0, which means that for all points in [0,1], 0 ≤|f − g| ≤ max|f − g| = 0, or |f − g|= 0; which would further imply f = g.