Definition . A collection A = {Uα}α∈J of subsets of X is said to cover X,
or to be a covering of X, if the union of its elements is equal to X, so X
=U{Uα} α∈J. If each element in the collection is an open subset of X, then
we say that A is an open cover.
A sub collection B ⊂ A that also covers X is called a sub cover of A . If,
furthermore, B has finitely many elements, then we call B a finite sub
cover of A .
Definition :. A space X is said to be compact if for each open cover A =
{Uα}α∈J of X there exists a finite sub collection {Uα1 , . . . ,Uαn} that also
covers X. In other words, X is compact if each open cover of X contains a
finite sub cover.
Example: . A finite topological space X is compact, since there are only
finitely many different open subsets
U ⊂ X, so any collection covering X is finite.
Example .: The real line R is not compact, since the open cover A = {(n −
1, n + 1) |n ∈ N} does not have a finite subcover.
Example . The real line R in the indiscrete topology is compact, since the
only open covers
are the collections {R} and {∅,R}, which are finite.
Example: The subspace X = {0} ∪ {1/n | n ∈ N}
of R is compact. Given an open covering A of X, choose U ∈ A with 0 ∈
U. Since U is open in the subspace topology, there is an N ∈ N such that
1/n ∈ U for all n > N. For each 1 ≤ n ≤ N
choose Un ∈ A with 1/n ∈ Un. Then B = {U,U1, . . . ,UN} is a finite
subcover of A .
Lemma . Let Y be a subspace of X. Then Y is compact if and only if each
covering of Y by open subsets of X contains a finite sub collection
covering Y .
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Proof. Suppose that Y is compact, and that A = {Uα}α∈J is a covering of Y
by open subsets of X. Then
{Y ∩Uα}α∈J is an open cover of Y , hence contains a finite subcover {Y
∩Uα1 , . . . , Y ∩Uαn}. Then
{Uα1 , . . . ,Uαn} is a finite sub collection of A that covers Y .
Conversely, suppose that each covering of Y by open subsets of X
contains a finite sub collection covering Y , and let A = {Vα}α∈J be an
open covering of Y . For each α ∈ J we can write Vα = Y ∩ Uα for some
open subset Uα ⊂ X. Then the collection {Uα}α∈J is a covering of Y by
open subsets of X, which we have assumed contains a finite sub
collection
{Uα1 , . . . ,Uαn} covering Y . Then {Vα1 , . . . , Vαn} is a finite sub
collection of A that covers Y . Since A was an arbitrary open cover, it
follows that Y is compact.
Theorem :. The continuous image of a compact space is compact.
Proof. Let f : X → Y be continuous, and assume that X is compact. We
prove that f(X) is a compact subspace of Y . Let A = {Uα}α∈J be a
covering of f(X) by open subsets of Y . Then {f−1(Uα)}α∈J is an open
cover of X. By compactness there exists a finite subcover
{f−1(Uα1 ), . . . , f−1(Uαn)}. Then the sets
{Uα1 , . . . ,Uαn} cover f(X).
Theorem: every compact subset A of a housdorff space X is closed.
Proof: H.W
Example: let T={X,Ø,{a},{a,b}} where X={a,b,c}
Since X is finite then it is compact.
But X is not hausdorff since a≠b ,there isnot disjoint nbds.
Theorem: a topological space X is compact iff every basic open cover of
X has a finite sub cover.
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Proof: let X be compact then every open cover of X has a finite sub
cover , in particular every basic open cover of X have a finite sub cover.
Conversely , suppose that every basic open cover have a finite sub cover
Let C={zλ: λ▲} be any open cover of X
If B={Bα :α} be any open base for X , then each zλ is a union of
members of B and the totality of all members of B is evidently a basic
open cover of X.
By hypothesis this collection of members of B has afinite subcover say
{Bαi :i=1,2,….,n}
For each Bαiin this finite subcover we can select zλi from C such that
Bαizλi
Then {zλi :i=1,2,…,n} is a finite subcoverof C.
Theorem: a topological space X is compact iff every collection of closed
subset of X with the finite intersection property is fixed that is has a
nonempty intersection.
Proof :H.W
Definition: a space X is said to have Bolzano weirstress property
(BWP)iff every infinite set in X has a limite point .aspace with BWP is
also said to be frechet compact.
Theorem: if A is an infinite subset of a compact space X then A has a
limit point in X in other words compact spaces have BWP.
Proof: suppose A has no limit point in X. xXopen nbd Nx/x∩A=Ø.
Now the collection {Nx:xX}forms an open cover of X.
Since X is compact x1,x2,…,xn in X such that
XNx1∪Nx2∪…∪Nxn
ANx1∪Nx2∪…∪Nxn
Since each Nxi contain at most one point of A ,A has at most n point and
so finite which is contradiction
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Definition: a topological space X is said to be locally compact iff every
point in X has at least one neighborhood whose closure is compact.
Theorem: every compact topological space X is locally compact.
Proof: let X be a compact space since X is both open and closed it is
neighborhood of each of its point such that cl(X)=X is compact hence X
is locally compact.
Example: let X is infinite set and T is the discrete topology then X is not
compact since the set of all singleton sets is infinite open cover of X
which have no finite sub cover.
Now let xX , then{x} is nbd of x and cl({x})={x}also {x} is compact
hence every point of X has anbd whose closure is compact
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