2.1 KiB
Compactness
Definition 1: a metric space
X
is compact if every sequence inX
has a convergent subsequence. A subsetM
ofX
is compact if every sequence inM
has a convergent subsequence whose limit is an element ofM
.
A general property of compact sets is expressed in the following proposition.
Proposition 1: a compact subset
M
of a metric space(X,d)
is closed and bounded.
??? note "Proof:"
Will be added later.
The converse of this proposition is generally false.
??? note "Proof:"
Will be added later.
However, for a finite dimensional normed space we have the following proposition.
Proposition 2: in a finite dimensional normed space
(X, \|\cdot\|)
a subsetM \subset X
is compact if and only ifM
is closed and bounded.
??? note "Proof:"
Will be added later.
A source of interesting results is the following lemma.
Lemma 1: let
Y
andZ
be subspaces of a normed space(X, \|\cdot\|)
, suppose thatY
is closed and thatY
is a strict subset ofZ
. Then for every\alpha \in (0,1)
there exists az \in Z
, such that
\|z\| = 1
,\forall y \in Y: \|z - y\| \geq \alpha
.
??? note "Proof:"
Will be added later.
Lemma 1 gives the following remarkable proposition.
Proposition 3: if a normed space
(X, \|\cdot\|)
has the property that the closed unit ballM = \{x \in X | \|x\| \leq 1\}
is compact, thenX
is finite dimensional.
??? note "Proof:"
Will be added later.
Compact sets have several basic properties similar to those of finite sets and not shared by non-compact sets. Such as the following.
Proposition 4: let
(X,d_X)
and(Y,d_Y)
be metric spaces and letT: X \to Y
be a continuous mapping. LetM
be a compact subset of(X,d_X)
, thenT(M)
is a compact subset of(Y,d_Y)
.
??? note "Proof:"
Will be added later.
From this proposition we conclude that the following property carries over to metric spaces.
Corollary 1: let
M \subset X
be a compact subset of a metric space(X,d)
over a fieldF
, a continuous mappingT: M \to F
attains a maximum and minimum value.
??? note "Proof:"
Will be added later.