2.Some set theory.
Answer:
Definition: Let S and T be set:
S contains T, and we write S c T or T c S, if every number of T is also in S. in this case, T is a subset of S.
S\T is the set of elements that are in S, but not in T.
S equals T, and we write: S=T, is S contains T, and T contains S, thus, S=T if and only if S and T have the same members.
The compecment of S denoted by Sc , is the set of elements in the universal set that are not in S.
The union of S and T, denoted by S
T, is the set of elements in at least one of S and T.The intersection of Sand T , denoted by S
T is the set of element in both S and T. If S
T
0(the empty set), then S and T are disjoint sets.
Definition If x0 is real number and, ε>0, then the open interval (x0-ε,x0+ε) in an neighborhood of x0.
If the set S contains an ε-neighborhood of x0 , then S is a neighborhood of x0, and x0 is an interior point of S.
Definition. Let S be subset of R. Then (a) x0 is a limit-point of S, if every deleted neighborhood of x0 contains a point of S.
(b) x0
is boundary point of S, if every neighborhood of x0
contains a least one point in S and one not in S. The set of
boundary points of S denoted by
S
The closure
of S, denoted by
=S
S
(c)x0 is exterior to S, of x0 is in the interior of Sc
Theorem. A set Sis closed, if and only if no point of Scis a limit point of S.
Corollary. A set is closed, if and only if it contains all its limit points.
Definition . A collection H of open sets, is an open covering of a set S, if every point in S is contained
in a set H belonging to H, that is, if Sc {H|HcH}
3.1Heine-borel Theorem. Bolzano-Weierstrass Theorem.
Answer:
If H is an open covering of a closed and bounded subset S of the real line, then S has an open covering H consisting of finitely many open sets belonging to H.
Proof:
Since S bounded, it has an infimumα
and supremum β , and, since S is closed,α and β belong to S.
Define St=S
[α;t]
for
≥α.
and let F={t|α<t<β and finitely many sets form H cover St}
Since Sβ =S1 the theorem will be proved, if we can show that β F. To do this, we use the completeness of the reals.
Since α S, Sαis the singleton set {α}, which is contained in some open set Hαform H, because H covers S; therefore, αcF.
Since, F is nonempty and bounded above by β, it has supremumλ. First, we wish to show that λ=β by definition of F , it suffices to rull out the possibility that α<β.
We consider two cases:
Case 1.
Suppose that α<β and λ
S.
Then, since S is closed
λ is not a limitpointof S. Consequently, there is an ε>0 such
that [λ-ε, λ+ε]
S=
, so Sλ-ε=Sλ+ε
However, the definition of λ implies that Sλ-ε has a finite subcovering from H, while Sλ+εdoes not. This is a contradiction.
3.2 Case 2.Suppose that λ<β and λ S. Then there is an open set Hλ in H that contains λ and along with λ, an interval [λ-ε, λ+ε] for some positive ε. Since Sλ-ε has a finite covering.{H1,…,Hn} of sets from H, it follows, that Sλ+ε has the finite covering{H1,….,Hn,Hλ}. This contradicts the definition of λ.
Now we know that α=β, which is in S. Therefore, there is an open set Hβ in H that contains β and along with, an interval of the from [β-ε,β+ε], for some positive ε.
Since Sβ-ε is covered by a finite collection of set {H1,…,Hk}, Sp is covered by the finite collection {H1,…,Hk,Hβ}. Since Sβ=S1 .
Henceforth , we will say a closed and bounded set is compact.
As an application of the Heine-Borel theorem, we prove the following theorem of Bolzano andWeierstrass.
Bolzano-Weierstrass theorem. Every boumded infinite set of real numbers has at least one limit point.
Proof: We will show that a bounded nonempty set without a limit point can contain only a finite number of points.
If S has no limitpoints,then S is closed and every point x of S has an open neighborhood Nx that contains no point of other than x. The collection H={Nx| x S} is an open covering for S. Since S is also bounded Heine-Borel theorem implies that S can be covered by a finite collection of sets from H1,…. say Nx1,…,Nxn. Since these set contain only x1,x2,…xn from S, it follows that S={x1,…,xn}