4ℝn as a normed space
IB Analysis II
4.3 Sequential compactness
In general, there are two different notions of compactness — “sequential com-
pactness” and just “compactness”. However, in normed spaces (and metric
spaces, as we will later encounter), these two notions are equivalent. So we will
be lazy and just say “compactness” as opposed to “sequential compactness”.
Definition ((Sequentially) compact set). Let
V
be a normed vector space. A
subset
K ⊆ V
is said to be compact (or sequentially compact) if every sequence
in K has a subsequence that converges to a point in K.
There are things we can immediately know about the spaces:
Theorem. Let (V, ∥ · ∥) be a normed vector space, K ⊆ V a subset. Then
(i) If K is compact, then K is closed and bounded.
(ii)
If
V
is
R
n
(with, say, the Euclidean norm), then if
K
is closed and bounded,
then K is compact.
Proof.
(i)
Let
K
be compact. Boundedness is easy: if
K
is unbounded, then we can
generate a sequence x
k
such that
∥
x
k
∥ → ∞
. Then this cannot have a
convergent subsequence, since any subsequence will also be unbounded,
and convergent sequences are bounded. So K must be bounded.
To show
K
is closed, let y be a limit point of
K
. Then there is some
y
k
∈ K
such that y
k
→
y. Then by compactness, there is a subsequence
of y
k
converging to some point in
K
. But any subsequence must converge
to y. So y ∈ K.
(ii)
Let
K
be closed and bounded. Let x
k
be a sequence in
K
. Since
V
=
R
n
and
K
is bounded, (x
k
) is a bounded sequence in
R
n
. So by Bolzano-
Weierstrass, this has a convergent subsequence x
k
j
. By closedness of
K
,
we know that the limit is in K. So K is compact.