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Assignment 1; Due Friday, September 30
1.2: The triangle inequality must hold for every choice of a, b, and c. For instance, it must hold if a = b, so d(a, b) + d(a, c) ≥ d(b, c)
becomes d(b, b) + d(b, c) ≥ d(b, c)
Now d(b, b) = 0 by axiom one, so this gives d(b, c) ≥ d(b, c), which is obvious. You cannot win every time.
The remaining special cases give something interesting. Suppose we set b equal to c. Then the triangle inequality becomes
d(a, c) + d(a, c) ≥ d(c, c)
Since d(c, c) = 0, this gives 2d(a, c) ≥ 0 and so d(a, c) ≥ 0. Since a and c are arbitrary, we have proved that the distance between any two points is ≥ 0, as required.
Finally set a = c. The triangle inequality becomes
d(c, b) + d(c, c) ≥ d(b, c)
Since d(c, c) = 0, we have d(c, b) ≥ d(b, c). This holds for any b and c, so interchanging b and c gives t d(b, c) ≥ d(c, b) and thus d(b, c) = d(c, b) as required.
1.3a: Let d(x, y) = ||x−y||. Then d(x, x) =
(xi − xi)^2
= 0, as required. Conversely,
if d(x, y) = 0, then
(xi − yi)^2
= 0. Squaring,
(xi − yi)^2 = 0. Each term in this expression is non-negative, so the expression can only be zero if each xi − yi = 0 and thus only if x = y.
We must prove the triangle inequality. Since d(a, b) = d(b, a) by the first exercise, we can write the triangle inequality in its more standard form d(x, z) ≤ d(x, y) + d(y, z). Thus we must prove that ||x − z|| ≤ ||x − y|| + ||y − z||
But it is known that ||a + b|| ≤ ||a|| + ||b||. Apply this when a = x − y and b = y − z to get ||x − z|| ≤ ||x − y|| + ||y − z||.
Finally, the graduate students need to prove that ||a + b|| ≤ ||a|| + ||b||. First we prove the Schwarz inequality | < x, y > | ≤ ||x||||y||.
Indeed (^) 〈 x −
< x, y > ||y||^2
y, x −
< x, y > ||y||^2
y
because the length of any vector is non-negative. Expanding
〈 x, x
< x, y >^2 ||y||^2
and so (^) 〈
x, x
y, y
x, y
Since < x, x >= ||x||^2 , the Schwarz inequality follows by taking square roots.
But then ||a + b||^2 =< a + b, a + b >=< a, a > +2 < a, b > + < b, b >
and by the Schwarz inequality this is less than or equal to
< a, a > +2||a|| ||b||+ < b, b >= ||a||^2 + 2||a|| ||b|| + ||b||^2 =
||a|| + ||b||
The required inequality follows by taking square roots.
1.3a continued: Similar arguments hold for d(x, y) =
|xi − yi|. This expression is clearly zero if x = y. Conversely if it is zero, then each term of the sum must be zero, so xi = yi for all i, so x = y.
According to the triangle inequality for ordinary real numbers, |a + b| ≤ |a| + |b|. Set a = xi − yi and b = yi − zi to obtain |xi − zi| ≤ |xi − yi| + |yi − zi|. Summing
∑ |xi − zi| ≤
|xi − yi| +
|yi − zi|
so d(x, z) ≤ d(x, y) + d(y, z).
1.3b: If d(x, y) = (x − y)^2 , the triangle inequality fails. For example, let x = 0, y = 1, z = 2. Then d(x, y) + d(y, z) = 1^2 + 1^2 = 2, but d(x, z) = 2^2 = 4, so it is not true that d(x, z) ≤ d(x, y) + d(y, z).
1.3e: In the exercise set for next week, we will show that the open sets using d and the open sets using d′^ are the same. So the metric spaces using d and using d′^ are homeomorphic. Notice that every subset in the metric space using d′^ is bounded because d′^ < 1. So the condition that a metric space is bounded is not topologically interesting; we can replace any metric space by a homeomorphic one in which every subset is bounded.
Clearly d′(x, y) = 0 if and only if d(x, y) = 0, and so if and only if x = y.
