Solutions of Second Homework
Solution of 7.1 Ex. 7:
Let x, y ∈ X. We have
kx + yk2 + kx − yk2 = (x + y) · (x + y) + (x − y) · (x − y)
= x · x + 2x · y + y · y + x · x − 2x · y + y · y
= 2x · x + 2y · y = 2kxk2 + 2kyk2 .
Solution of 7.2 Ex. 6:
Let (xn ) and (yn ) be two sequences in Rd . Assume that the first
sequence converges to 0 and the second is bounded. Therefore, we
have lim kxn k = 0 and kyn k ≤ M for all n for some large M ≥ 0. By
Cauchy-Schwartz inequality we have
|xn · yn | ≤ kxn kkyn k ≤ M kxn k −→ 0
as n tends to infinity. Hence xn · yn −→ 0.
Solution of 7.2 Ex. 12:
Clearly we have δ(x, y) = 0 if and only if x = y. Moreover, from the
definition also follows that δ(x, y) = δ(y, x) for any x and y.
It remains to prove the triangle inequality: δ(x, y) ≤ δ(x, z) + δ(z, y)
for any x, y and z. If x = y, δ(x, y) = 0 and there is nothing to prove.
So, we can assume that x 6= y. Hence, δ(x, y) = 1. The point z can
be equal to x or different from x. In the first case, z is different from
y and we have δ(z, y) = 1. Hence the inequality holds. In the second
case, δ(x, z) = 1 and the inequality holds again.
Solution of 7.2 Ex. 13:
If ≤ 1, the ball B (x) is equal to {x}. Therefore, if (xn ) converges
to x, there exists N such that n ≥ N implies xn = x.
Therefore, except first N terms in the sequence (xn ), all xn are equal
to x.
On the other hand, all such sequences clearly converge to x.
Solution of 7.3 Ex. 2:
Let x in Rd . Consider the set U = Rd − {x}. If y ∈ U , y 6= x and
δ(x, y) > 0. Let = δ(x, y). Then x is not in B (y), so B (y) is in U .
Since y ∈ U is arbitrary, it follows that U is an open set. Therefore
{x} is a closed set.
Hence, we proved that points are closed in Rd . Since finite unions of
closed sets are closed, finite sets of points are also closed.
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