5Metric spaces
IB Analysis II
5 Metric spaces
We would like to extend our notions such as convergence, open and closed
subsets, compact subsets and continuity from normed spaces to more general
sets. Recall that when we defined these notions, we didn’t really use the vector
space structure of a normed vector space much. Moreover, we mostly defined
these things in terms of convergence of sequences. For example, a space is closed
if it contains all its limits, and a space is open if its complement is closed.
So what do we actually need in order to define convergence, and hence all
the notions we’ve been using? Recall we define x
k
→
x to mean
∥
x
k
−
x
∥ →
0
as a sequence in
R
. What is
∥
x
k
−
x
∥
really about? It is measuring the distance
between x
k
and x. So what we really need is a measure of distance.
To do so, we can define a distance function
d
:
V ×V → R
by
d
(
x, y
) =
∥x−y∥
.
Then we can define x
k
→ x to mean d(x
k
, x) → 0.
Hence, given any function
d
:
V × V → R
, we can define a notion of
“convergence” as above. However, we want this to be well-behaved. In particular,
we would want the limits of sequences to be unique, and any constant sequence
x
k
= x should converge to x.
We will come up with some restrictions on what
d
can be based on these
requirements.
We can look at our proof of uniqueness of limits (for normed spaces), and
see what properties of
d
we used. Recall that to prove the uniqueness of limits,
we first assume that x
k
→ x and x
k
→ y. Then we noticed
∥x − y∥ ≤ ∥x − x
k
∥ + ∥x
k
− y∥ → 0,
and hence
∥x −y∥
= 0. So
x
=
y
. We can reformulate this argument in terms of
d. We first start with
d(x, y) ≤ d(x, x
k
) + d(x
k
, y).
To obtain this equation, we are relying on the triangle inequality. So we would
want d to satisfy the triangle inequality.
After obtaining this, we know that
d
(
x
k
, y
)
→
0, since this is just the
definition of convergence. However, we do not immediately know
d
(
x, x
k
)
→
0,
since we are given a fact about
d
(
x
k
, x
), not
d
(
x, x
k
). Hence we need the property
that d(x
k
, x) = d(x, x
k
). This is symmetry.
Combining this, we know that
d(x, y) ≤ 0.
From this, we want to say that in fact,
d
(
x, y
) = 0, and thus
x
=
y
. Hence
we need the property that
d
(
x, y
)
≥
0 for all
x, y
, and that
d
(
x, y
) = 0 implies
x = y.
Finally, to show that a constant sequence has a limit, suppose
x
k
=
x
for all
k ∈ N
. Then we know that
d
(
x, x
k
) =
d
(
x, x
) should tend to 0. So we must have
d(x, x) = 0 for all x.
We will use these properties to define metric spaces.