It follows that
¯
A
∩
Y
is a subset of all closed subsets of
Y
containing
A
, thus is contained in
¯
A
Y
. The result follows.
3. Let
X
1
, . . . , X
n
be metric spaces with distance functions
d
1
, . . . , d
n
respec-
tively. Let
X
=
X
1
× · · · ×
X
n
=
{
p
= (
p
1
, . . . , p
n
) :
p
i
∈
X
i
, i
= 1
, . . . , n
}
.
Consider the following three functions from
X
×
X
to
R
. In the definition,
p
= (
p
1
, . . . , p
n
),
q
= (
q
1
, . . . , q
n
)
∈
X
.
δ
1
(
p, q
)
=
n
X
i
=1
d
i
(
p
i
, q
i
)
.
δ
2
(
p, q
)
=
ˆ
n
X
i
=1
d
i
(
p
i
, q
i
)
2
!
1
/
2
.
δ
∞
(
p, q
)
=
max
{
d
i
(
p
i
, q
i
) :
i
= 1
, . . . , n
}
.
(a) Prove all three metrics for
X
are equivalent and a sequence
{
p
m
}
∞
m
=1
,
where
p
m
= (
p
m
1
, . . . , p
mn
) converges to
p
= (
p
1
, . . . , p
n
) if and only
if each sequence
{
p
mi
}
∞
m
=1
converges in
X
i
for
i
= 1
, . . . , n
.
(b) Prove that all three metrics for
X
have the same Cauchy sequences.
(
Equivalent metrics do NOT necessarily have the same Cauchy
sequences
so this is not an immediate consequence of the previous
point.
It is however essentially the same proof as one uses for the
previous point.)
(c) Prove that
X
is complete in any one of the three metrics if and only
if it is complete in the others, and this happens if and only if each
one of the spaces
X
1
, . . . , X
n
is complete.
(d) Explain how all this applies to
R
n
.
In particular, show that
R
n
is
complete.
Solution.
It will be convenient to prove first the following lemma:
3