1Metric spaces
IB Metric and Topological Spaces
1.1 Definitions
As mentioned in the introduction, given a set
X
, it is often helpful to have a
notion of distance between points. This distance function is known as the metric.
Definition (Metric space). A metric space is a pair (
X, d
X
) where
X
is a set
(the space) and
d
X
is a function
d
X
:
X ×X → R
(the metric) such that for all
x, y, z,
– d
X
(x, y) ≥ 0 (non-negativity)
– d
X
(x, y) = 0 iff x = y (identity of indiscernibles)
– d
X
(x, y) = d
X
(y, x) (symmetry)
– d
X
(x, z) ≤ d
X
(x, y) + d
X
(y, z) (triangle inequality)
We will have two quick examples of metrics before going into other important
definitions. We will come to more examples afterwards.
Example.
– (Euclidean “usual” metric) Let X = R
n
. Let
d(v, w) = |v − w| =
v
u
u
t
n
X
i=1
(v
i
− w
i
)
2
.
This is the usual notion of distance we have in the
R
n
vector space. It is
not difficult to show that this is indeed a metric (the fourth axiom follows
from the Cauchy-Schwarz inequality).
– (Discrete metric) Let X be a set, and
d
X
(x, y) =
(
1 x 6= y
0 x = y
To show this is indeed a metric, we have to show it satisfies all the axioms.
The first three axioms are trivially satisfied. How about the fourth? We
can prove this by exhaustion.
Since the distance function can only return 0 or 1,
d
(
x, z
) can be 0 or 1,
while
d
(
x, y
) +
d
(
y, z
) can be 0, 1 or 2. For the fourth axiom to fail, we
must have RHS
<
LHS. This can only happen if the right hand side is 0.
But for the right hand side to be 0, we must have
x
=
y
=
z
. So the left
hand side is also 0. So the fourth axiom is always satisfied.
Given a metric space (
X, d
), we can generate another metric space by picking
out a subset of X and reusing the same metric.
Definition (Metric subspace). Let (
X, d
X
) be a metric space, and
Y ⊆ X
.
Then (
Y, d
Y
) is a metric space, where
d
Y
(
a, b
) =
d
X
(
a, b
), and said to be a
subspace of X.
Example.
S
n
=
{
v
∈ R
n+1
:
|
v
|
= 1
}
, the
n
-dimensional sphere, is a subspace
of R
n+1
.
Finally, as promised, we come to the definition of convergent sequences and
continuous functions.
Definition (Convergent sequences). Let (
x
n
) be a sequence in a metric space
(
X, d
X
). We say (
x
n
) converges to
x ∈ X
, written
x
n
→ x
, if
d
(
x
n
, x
)
→
0 (as a
real sequence). Equivalently,
(∀ε > 0)(∃N)(∀n > N) d(x
n
, x) < ε.
Example.
–
Let (v
n
) be a sequence in
R
k
with the Euclidean metric. Write v
n
=
(
v
1
n
, ··· , v
k
n
), and v = (
v
1
, ··· , v
k
)
∈ R
k
. Then v
n
→
v iff (
v
i
n
)
→ v
i
for
all i.
–
Let
X
have the discrete metric, and suppose
x
n
→ x
. Pick
ε
=
1
2
.
Then there is some
N
such that
d
(
x
n
, x
)
<
1
2
whenever
n > N
. But if
d
(
x
n
, x
)
<
1
2
, we must have
d
(
x
n
, x
) = 0. So
x
n
=
x
. Hence if
x
n
→ x
,
then eventually all x
n
are equal to x.
Similar to what we did in Analysis, we can show that limits are unique (if
exist).
Proposition. If (
X, d
) is a metric space, (
x
n
) is a sequence in
X
such that
x
n
→ x, x
n
→ x
0
, then x = x
0
.
Proof.
For any
ε >
0, we know that there exists
N
such that
d
(
x
n
, x
)
< ε/
2 if
n > N. Similarly, there exists some N
0
such that d(x
n
, x
0
) < ε/2 if n > N
0
.
Hence if n > max(N, N
0
), then
0 ≤ d(x, x
0
)
≤ d(x, x
n
) + d(x
n
, x
0
)
= d(x
n
, x) + d(x
n
, x
0
)
≤ ε.
So 0 ≤ d(x, x
0
) ≤ ε for all ε > 0. So d(x, x
0
) = 0, and x = x
0
.
Note that to prove the above proposition, we used all of the four axioms. In
the first line, we used non-negativity to say 0
≤ d
(
x, x
0
). In the second line, we
used triangle inequality. In the third line, we used symmetry to swap
d
(
x, x
n
)
with
d
(
x
n
, x
). Finally, we used the identity of indiscernibles to conclude that
x = x
0
.
To define a continuous function, here, we opt to use the sequence definition.
We will later show that this is indeed equivalent to the
ε − δ
definition, as well
as a few other more useful definitions.
Definition (Continuous function). Let (
X, d
X
) and (
Y, d
Y
) be metric spaces,
and
f
:
X → Y
. We say
f
is continuous if
f
(
x
n
)
→ f
(
x
) (in
Y
) whenever
x
n
→ x (in X).
Example. Let
X
=
R
with the Euclidean metric. Let
Y
=
R
with the discrete
metric. Then
f
:
X → Y
that maps
f
(
x
) =
x
is not continuous. This is since
1/n → 0 in the Euclidean metric, but not in the discrete metric.
On the other hand,
g
:
Y → X
by
g
(
x
) =
x
is continuous, since a sequence
in Y that converges is eventually constant.