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International Journal of Research in Engineering and Science (IJRES) ISSN (Online): 2320-9364, ISSN (Print): 2320-9356 www.ijres.org Volume 9 Issue 1 ǁ 2021 ǁ PP. 10-16

Some Examples and Counterexamples of Advanced Compactness in Topology

Pankaj Goswami Department of , University of Gour Banga, Malda, 732102, West Bengal, India

ABSTRACT: The examples and counter examples are always usefull for better comprehension of underlying concept in a theorem or definition .Compactness has come to be one of the most importent and useful topic in advanced mathematics.This paper is an attempt to fill in some of the information that the standard textbook treatment of compactness leaves out, and giving some constructive significative counterexamples of Advanced Compactness in Topology . KEYWORDS: open cover, compact, connected, , example, counterexamples, continuous , locally compact, countably compact, , sequentially compact, lindelöf , paracompact ------Date of Submission: 15-01-2021 Date of acceptance: 30-01-2021 ------

I. INTRODUCTION We begin by presenting some definitions, notations,examples and some theorems that are essential for concept of compactness in advanced topology .For more detailed, see [1-2], [7-10] and others, in references .If X is a closed bounded subset of the ℝ, then any family of open sets in ℝ whose union contains X has a finite sub family whose union also contains X . If X is a space or topological space in its own right , then the above proposition can be thought as saying that any class of open sets in X whose union is equal to X has a finite subclass whose union is also equal to X. The topological space for which the conclusion of this theorem holds are known as compact topological space. In particular that collection of open sets is called as an open cover.

Defination 1.1. Let X be a topological space and A be a subset of X. An open cover of A is collection = {Gi} of open sets in X whose union contains A; that is,

A subcollection ʹ of whose union is also contains A; that is, is called as subcover of A in .

Example 1.2. Let A = (0,1) ⊂ ℝ then for each n ℕ , Gn = ( is an open cover of A that is

.

Defination 1.3. : A topological space X is said to be compact if every open covering of X contains a finite subcover that also covers X.

Theorem 1.4.A subser A of R is compact iff A is bounded and closed.This is known as Heine- Borel theorem. In particular a subset of ℝ is compact iff it is closed and bounded. The proof is given in [10] and also in others.

Example 1.5. 1. Any closed bounded subset [a, b] ⊂ ℝ is compact by Heine –Borel theorem. 2.The real line ℝ is not compact,for it, each n ℕ ,Gn = (-n, n) is an open cover which has no finite sub cover.

3. A = (0,1) ⊂ ℝ then for each n ℕ , Gn = ( is an open cover of A that is

, clearly it has no finite sub cover, so A is not compact.

In particular for any open A = (a, b) ⊂ ℝ , for each n ℕ ; Gn = (a -ϵn, b − ϵn),ϵn =

,where c be any point in (a,b); is an open cover of (a, b) which has no finite subcover. www.ijres.org 10 | Page Some Examples and Counterexamples in Advanced Compactness of Topology

4.Any finite discrete topological space is compact.As there are only finite number of open sets. 5.Any set X with cofinite topology is compact. For it , if U is open in X ,then X-U is finite and so X-U can be covered by a finite number of open sets say G1, G2, ..., Gn . Then for the collection Cn= {U, G1, G2, ..., Gn } is a open cover for X which is also finite. Hence any set X with cofinite topology is compact. 6. is compact scince it is closed bounded subset of ℝ .

Theorem 1.6. Every closed subset of is compact. Proof is given in [10] and also in others.

Theorem 1.7. Continuous image of compact spaces is compact. Proof is given in [10] and also in others.

Definition1.8. A topological space X is said to be sequentially compact iff every sequence in X has a convergent sub-sequence.

Example 1.9. 1.The closed interval [a,b] in ℝ is sequentially compact ,since any sequence in [a,b] is bounded ,and thus has a convergent subsequence.In particular compact metric spaces is sequentially compact. 2.The real line ℝ is not sequentially compact.The sequence {n} has no convergent sub-sequence.So in particular ℝ is not sequentially compact. 3.Any finite compact so sequentially compact but infinite discrete space is not sequentially compact.

4.The interval (0,1) is not sequentially compact .The sequence { } has no convergence sub-sequence in (0,1).

Definition1.10. A topological space X is said to be locally compact iff every point in X has at least one nbd whose is compact.

Example 1.11. 1.The real line ℝ is locally compact .For any point x in R there is an open interval (a,b) containing x whose closure [a,b] is compact. 2.In particular ℝ is locally compact. 3.The ℚ set of all rational numbers with usual metric is not locally compact.For it closed intervals [a,b]∩ ℚ in ℚ are not compact ,it follows that all compact subsets of ℚ have empty .Hence ℚ is in locally compact. 4.ℝ ,ℝ with lower limit topology constructed from the interval {[a,b): a

Definition1.12. A topological space X is said to be countably compact iff every countable open cover of X has a finite subcover.

Example 1.13. 1. Any compact spaces are the example of .

Definition1.14. A topological space X is said to be limit point compact(Bolzano- Weierstrass Property ) BWP if each infinite subset of X has a limit point in X.

Example 1.15. 1.Every closed bounded interval [a,b] of ℝ , is limit point compact as each infinite subset of it has a limit point in it. 2.ℝ is not limit point compact ,as set of integers has no limit point on it. 3.In particular any compact spaces are limit point compact.

Definition1.16. A collection = {Gi : i Λ } of subsets of a topological space X is said to be locally finite (nbd-finite) if each point x in X has an open neighbourhood having non-empty intersection with at most finitely many members of .

Example 1.17. Let X = ℝ ∪ {0} , where ℝ is the set of all positive real numbers.Let X have the usal www.ijres.org 11 | Page Some Examples and Counterexamples in Advanced Compactness of Topology topology .Let An = { x X : x≥ n} for each non-negative integer n . Consider collection = { An :n=0,1,2,…}.Then is a locally- finite family in X.

Definitin1.18. Let be a covering of a topological space X. A collection is called a refinement of if covers X and each member of is contained in some member of .

Definition1.19. A topological space X is said to be Paracompact if every open covering of X has a locally finite refinement which is also an open covering of X.

Example 1.20. 1.Every compact space is paracompact .For it let be an open cover of a compact space X, then contains a finite open subcover for X, that is has a finite open refinement , which covers X ⇒ has a locally finite refinement wcich covers X. 2.Every metric space is para compact space.Proof is given by A.H.Stone see in [9]. 3.Any infinite set carrying the particular point topology is not paracompact (see [10] for the definition of particular point topology) , in this case there is a open cover which has no nbd finite refinement. 4.The sorgenfrey lines is paracompact ,but not compact neither locally compact. 5.Every discrete space is paracompact as for it ,every open cover of it has a nbd-finite open refinement. 6.The real line ℝ is paracompact,but not compact.For it by A.H Stone [9], every is paracompact.

Definition1.21. A topological space X is said to be Lindel¨of if and only if every open cover of X X has a countable sub-cover.

Example 1.22. Every compact space is Lindel¨of . For it if X is compact ,then every open cover has finite subcover i.e has countable sub-cover.

Theorem1.23. Let (X, d) be a metric space.Then the following statements are equivalent 1.(X, d) compact.

2.(X, d) is countably compact.

3.(X, d) has the Bolzano-Weierstrass property.

4.(X, d) is sequentially compact.

The proof is obvious ,see[10]

Theorem1.24. For a topological spaces , compactness ⇒ lindelof sequentially compact ⇒ countabe compact compactness ⇒ countably compact. Proof is given see in [2].

II. COUNTEREXAMPLES 2.1.A non compact subset ℝ of in which open cover admits finite subcover.

Let (0, 1) as subset of a ℝ and take Cn := (0, - ) ∪ ( , 1) is open cover of (0, 1) such that

(0,1) ∪ .

Now also we can see that (0,1) ⊂ ∪ ∪ ∪ .

So (0,1) ∪ .

So the open cover has finite sub cover, yet (0,1) is not compact.

For compactness every open cover has finite sub-cover ,for it see that

(0,1) ⊂ ℝ , then for each n ℕ , Gn = ( is an open cover of (0,1) that is

and has no finite sub-cover. www.ijres.org 12 | Page Some Examples and Counterexamples in Advanced Compactness of Topology

2.2.An union of infinite number of compact sets, that is not compact.

Let = [-n,n] for all n ℕ ,then each is compact ,but =ℝ ,which is not compact.

2.3.A subset of ℝ i) which is compact but not connected ii) connected but not compact iii) both compact and connected iv) neither compact nor connected

Consider the subset { : n ℕ }∪ {0} , which is closed,bounded so compact and discrete set containing more that one point so this subset is compact but not connected. The subset (0,1) is clearly connected but not compact. Consider closed bounded interval [a,b] ℝ which is clearly compact and connected.

Consider the subset { : n ℕ } which is closed but not bounded so not compact also a discrete set with containing more than one point so not connected.

2.4.A construction of a compact set in ℝ whose set of limits points are denumerable. Let (X,d) be any metric space and , let ( ) be a sequence converging to x X Then the set { ∪ x} is compact ,whose set of limits is denumerable.

In particular the set { : n ℕ }∪ {0} in ℝ is compact having denumerable limit points ,as zero is the only limit point.

2.5.An onto continuous map from a open non-compact subset of ℝ to a compact interval of ℝ.

Let,f:(0,1)→ [0, 1] defined as f(x) = |sin(2πx)| ,then f(1/2)=0 and f(1/4)= 1 and by intermediate value theorem f assumes every values on [0,1].So f is a onto continuous map from a non compact open subset to a compact interval of ℝ . So there exist a continuous onto map f:(a,b) → [a,b] ,for all a,b [-1,1] with a ≠ b.

2.6.A non-compact metric space in which every continuous function is unifromly continuous.

Consider (ℕ, d) with usal metric d; c ℕ is a isolated point.Then for δ > 0, N (c, δ) ∩ ℕ=c. So for x N (c, δ) ∩ ℕ we have x=c.

Let ϵ> 0 and f be a continuous function on it. Then |f (x) − f (c)| < ϵ for all x N (c, δ) ∩ℕ.

Therefore f is unifromly continuous.So in this metric space every continuous function is constant i.e unifromly continuous.

2.7.A non compact metric space in which every continuous function has unique fixed point. Consider (ℕ, d) with usal metric d; c ℕ is a isolated point.Then for δ > 0, N (c, δ) ∩ ℕ=c. So for x N (c, δ) ∩ ℕ we have x=c.

Then every continuous function is a constant function ,say y(x)=f(x)=k ,which is parallel to x-axis, so f has unique fixed point.

2.8.A non compact metric space(X,d) in which, a real valued continuous and bounded function (i) attains its supremum and infimum (ii) attains its infimum but not supremum (iii)attains its supremum but not infimum (iv) attains neither supremum nor infimum. Let X=(0,1) with usual metric d, then X is non compact and take real valued bounded continuous function on X as f:(0,1)→ [0, 1] by f(x) = |sin(2πx)| ,then f(1/2)=0 and f(1/4)= 1 and by intermediate value theorem f assumes every values on [0,1].So f attains both supremum and infimum. Let X=[0,1) with usual metric d, then X is non compact and take real valued bounded continuous on X as f:[0,1) → [0,1] by f(x)=x ,then f attains its infimum but not supremum. Let X=(0,1] with usual metric d, then X is non compact and take real valued bounded continuous on X as f:(0,1] → [0,1] by f(x)=x ,then f attains its supremum but not infimum. Let X=(0,1) with usual metric d, then X is non compact and take real valued bounded continuous on X as f:(0,1) → [0,1] by f(x)=x ,then f attains neither its supremum but not infimum. 2.9.A metric space in which closed bounded subset is not compact. Let X=[0,1] with discrete metric then the subset A= {1,1/2,1/3,… } is then A is bounded by the given metric d, and as the given metric d is discrete so A has empty interior and also d discrete ,so any subset of X is open as www.ijres.org 13 | Page Some Examples and Counterexamples in Advanced Compactness of Topology well as closed ,so clearly A is closed. But A is a infinite subset of X so for given any open cover of A we cannot have finite subcover. So A is closed and bounded but not compact.

Another example is (ℚ ,d) with d as usual metric ; consider a subset A = ∩ ℚ ,then A is bounded by given metric d,also A is closed as A has empty interior .

But A is not compact ,for it the open cover Gn = ( + ) has no finite subcover.

2.10.The intersection of two compact subsets of a topologycal space which is not compact Let ℝ be the set of all with the usal topology.Let S={0,1} have the indiscrete topology. We set X= ℝ S and A= ([a,b] {0})∪ ((a,b) {1}) B= ((a,b) {0})∪ ([a,b] {1})

Clearly A and B are subsets of X. Since [a,b] is compact in ℝ and {0} is compact in S it follows that [a,b] {0} is compact in X.Also for any r (a,b), every containing (r,0) contains (r,1).Thus A compact.Similarly we can show that B is compact. But we see that A∩ B = a b {0,1}.Since the open interval (a,b) is not compact in ℝ ,it follows that A∩ B is not compact in X.

2.11.A compact subset of a topological space that is not closed.

Let X ={a,b,c,d} and τ = {ϕ ,X,{a},{d},{a,d},{b,c}{a,b,c}{b,c,d}} be a topology on X.

Consider the subset A= {a,c,d} therefore there exist τ –open sets {a,d} and {b,c} such A ⊂ {a,d}∪ {b,c}.

So A is compact,but A is not τ –closed because X-A={b} is not τ –open set.

2.12.A topological space in which closer of a compact set not compact. Consider a particular point topology on an infinite set ,say X ={p}∪ ℕ topolozied so that non- empty open sets are the sets containg p. The singleton set {p}is compact (finite sets are compacr ) ,but { }= the smallest containing {p} = X, and X is not compact because the open cover ={p,n} such that X= ℕ ,has no finite sub cover.

2.13. that is not compact. Consider any discrete space X where X is infinite.Then X is not compact ,as the collection of open cover contains infinitely many open sets ,which has no finite subcover. But X is locally compact as if x be any point of X ,then {x} is an nbd of x whose closure is {x},being finite i.e compact.

2.14.A continuous map that not preserved locally compactness. Consider S ⊂ ℝ be the graph of sin(1/x), x (0, 1).

Then Y= {(0, 0}∪S is not locally compact at (0,0) for it, any nbd U of (0,0) contains an infinite subset without limit points ,the intersection of S and a horizontal straight line,so U can not be contained in any compact subset of S.

On the other hand {(0, 0}∪S is the image of a continuous map f defined on the locally compact hausdroff space X= { -1}∪ (0,1) : f:X by f(-1)= (0,0) and f(x)=(x,sin(1/x)) for all x (0,1). .For figure see in [2],p-274. 2.15.A countably compact topological space that is not sequentially compact. For topological spaces , sequentially compact ⇒ countably compact.But the converse is not true. Let S be the set of all strictly increasing sequence of natural number and for each s S let www.ijres.org 14 | Page Some Examples and Counterexamples in Advanced Compactness of Topology

Xs = {0, 1} with the discrete topology putting X = .Then X is compact by Tychonoff’s theorem so countably compact.

Define {xn}n N as follows: let s S with s {nk}n N then define (xn)s = 0 if n = nk, k is even and (xn)s = 1 if n = nk, k is odd. Now we show that it has a subsequence that is not convergent in X.

Let s S with s = {nk}n N then (xnk )s = 0 for k even and (xnk )s = 1 for k is odd. Thus {xnk }k N does not converge in X = as it does not convergent componentwise. So X is countably compact but not sequentially compact.

2.16.A condition for which every countably compact topological space is sequentially compact. If a topological space X satisfy first axiom of cuontability i.e if X is first countable then X is countably compact ⇔ every sequence in X has a cluster point ⇔ every sequence in X has a convergent subsequence(since X is a first axiom space) ⇔ X is seque tially com act So a topological space satisfying first , is sequentially compact iff it is countably compact.

2.17.Example of a countable compact topological space that is not compact. Consider ω1 the space of countable ordinal numbers,see[2],p-38;with the .Then the cover

= with open sets does not have a finite subcover,therefore is not compact.

But ω1 is countably comact ,for it suppose A ω1 is infinite.Then A contains an increasing sequence (αn)n<ω and hence A has the limit point β := supnαn ω1 i.e every countable infinite subset has a limit point on it;thus countably compact.

2.18.Example of a lindel¨of space that is not compact. ℝl , ℝ with lower limit topology having basis of the form [a, b) ,so for it there is an open cover by its basis elements which has no finite sub-cover so ℝ l is not compact. But it can be easyly shown that every open covering of ℝ by basis elements has a countable subcover. For it A = { [ai ,bi) |i J} be such a cover and consider the union of intervals C= ; C is an open cover for ℝl .

Then C is open in the standered topology on ℝ so it has a countable subcover{ (an, bn)}.Then {[an, bn)} a countable subcover of C ,so in ℝl every open cover by its basis elements has countable subcover so ℝl is Lindel¨of but not compact.

2.19.Example of a that is not compact. Let X be infinite discrete space .Then clearly it is not compact. But it is paracompact ,for it let be any open cover of X.Consider = {{x}:x X},then is an open cover of X.It is a refinement of . At each point x X there is a nbd {x} which meets only one member of .Hence is nbd-finite. Thus every open cover of X has a nbd-finite open refinement.Hence paracompact.

2.20.Example of a locally compact space that is not paracompact. Consider open ordinal space [0, ) see in [2],p-38 ; as an ordered space ,it is Hausdroff. As an open subspace of the compact space [0, ] ,it is locally compact ,as for each α < , [0,α ] is a compact neighbourhood of α. But not paracompact ,for it suppose that is an open refinement of open cover {[0,α ): α < }. Setting =0, for each i pick so that , and let be the least ordinal not in . Then is an increasing sequence of ordinals in ,and so has a limit(supremum) α .But for any open nbd of α contains infinitely many ,and therefore meets infinitely many so could not be locally finite. III. CONCLUSIONS Charecterization of compactness by open cover play great roles to advanced concept of topology, it helps to developing new ideas to construct new research work. For example on some certain conditions www.ijres.org 15 | Page Some Examples and Counterexamples in Advanced Compactness of Topology on open cover for compactness we have types compactness say Locally compact,Sequentialy compact,Countably compact, Limit point compact,Paracompactness etc,from which with some certain conditions any one can do further research in advanced topology and counterexamples help to developing the level of understanding of mathematical concepts.

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