METRIC SPACE- BOUNDEDNESS, COMPACTNESS AND CONNECTEDNESS
Objective
To study the bounded sets, compact sets, connected sets and the properties of such sets.
Modules
Module 1- Boundedness
Module II- Compactness
Module III- Compactness and continuity
Module IV- Connectedness
MODULE I- BOUNDEDNESS
Definition 1
A set is bounded when the distances between any two points in the set have an upper bound. The least such upper bound is called the diameter of the set.
.
Example: The set is bounded in R, because for any two elements in ; infact diameter . The set R is unbounded, because can be made as large as needed.
Note that boundedness is a property of how large a set is. However it is not a good metric property: if is bounded and f is a continuous function then need not be bounded. Thus a set may be bounded in one metric space, but not remain so when is deformed continuously to another.
Proposition 1
A set is bounded when it can be covered by a ball.
i.e., .
For example, the set is bounded, because it can be covered by the ball .
Definition 2
A set is totally bounded when it can be covered by a finite number of - balls , however small .
i.e., .
Example: The set is totally bounded, because it can be covered by the balls, for , where .
Proposition 2
A totally bounded set is bounded
Proof.
Let be a totally bounded set. Therefore we can cover it by finite number of - balls. In particular by 1- balls.
.
Since there are only a finite number of these balls, we can find the maximum distance between their centers
. Now given any two points, they must be covered by two of these balls and , say. Therefore, using the triangle inequality twice
.
That is, is an upper bound for the distance between two points in.
MODULE II- COMPACTNESS
We define compact set to be the closed and bounded set in. However, the definition of compact is not exactly the same in metric space.
Definition 3
The set is compact if any infinite sequence in has convergent subsequence in.
From the definition it is very easy to see that a compact set is closed in any metric space. However, in general, close and bounded does not imply compact.
We can see that we need the concept of convergence (hence metric) in the definition 3.
Definition 4
Let be a collection of some open sets. If, then we say that is an open cover of .
is called a finite cover of ifis an open cover of with finitely many elements.
is called a sub cover of with respect to if and.
For example, is an open cover of but not of.
If we add the intervals and to, then is an open cover of .
Theorem 1
is compact if and only if every open cover of has a finite subcover.
The second part of the theorem means that for an open cover of, there exists with finitely many elements, where D is a cover of . Here may have uncountably many elements.
Proof.
Suppose is compact and there is an open cover which cannot have a finite subcover.
Exclude those elements in, which is a subset of the union of other elements. Therefore, for any, we can choose such that if, . Now, for each, pick in such a manner. As no finite cover with respect to exists, forms an infinitely sequence. Hence there is a subsequence converging to in. Now, for some. This contains infinitely many, contradicting the choice of .
Conversely, suppose there is an infinite sequence which does not have convergent subsequence. Then for each and for any, there is such that contains no, or if, there is such thatcontain one only. Now is an open cover of, which does not have finite sub cover, a contradiction. This completes the proof.
Example:is not compact. Because, the cover of balls for has no finite subcover. Similarly the sets and are not compact. But the sets are compact.
Proposition 3
Any closed subset of a compact set is compact.
First proof.
For an infinite sequence in, it has a subsequence converging to in. Then is a
limit point of , hence in . This proves the compactness of.
Second proof.
Suppose, where is open. Let, which is open. Then and together form an open cover of. Since is compact, from this open cover there exists a finite subcover, which means. Hence. We have found a finite subcover of.
Remark: The second proof does not involve metric. It will be the proof of the theorem in topological spaces.
Corollary
Any intersection (including uncountable) of compact sets is compact.
Proposition 4
Finite union of compact sets is compact.
Proof.
Let, where is compact and be an open cover of. Now for each, we pick
finite elements of to form a subcover . Nowis a finite subcover of .
MODULE III- COMPACTNESS AND CONTINUITY
Theorem 2
If is continuous on, then is compact in for any compact subset of.
Recall that
Proof.
Take any infinite sequencein. Then there is a subsequence. This impliesin.
Now we turn to see the generalized extreme value theorem.
Theorem 3
Suppose is continuous on compact in, and. Then there exists such that for all.
Proof.
First we show that there exists such that for all. Otherwise, for each positive integerwe have in so that. Since is compact, has a subsequence converging to. Then by the continuity of, for all, a contradiction.
Next, we prove the existence of. As is bounded on, is a real number. By supremum property, there exists such that.has a subsequence converging to , and by the continuity of , .
Remark: Actually we do not need to assume to be a metric space in theorem 3. We can proceed the proof by only considering open sets in.
The extreme value theorem we learnt in lower classes is just the special case of this theorem, when we take (the real set), to be a closed and bounded interval, to be the usual metric and.
Before we end this section, let us go a little bit deeper to the concept of uniform continuity.
Some of the beginning analysis course includes this concept, and some does not. Anyway, let us see the definition first.
Definition 5
A function is said to be uniformly continuous on a set if for every, there exists such that whenever for and in.
Obviously, uniformly continuous are continuous. In general, continuity does not imply uniform
continuity. However, in some conditions, the weaker one implies the stronger.
Theorem 4
If is continuous on a compact, then is uniform continuous on.
Proof.
Suppose it is not the case. Then there exists, such that for each we can pick and
in, where, but . Now has a subsequence in. Thusand.
To reach a contradiction, choose so that implies. Then
for all .
This shows, which is wrong.
MODULE IV- CONNECTEDNESS
Definition 6
A set is disconnected when it can be divided into at least two disjoint non empty subsets, say and, and each subsets can be covered exclusively by open sets, say and.
i.e.,
.
Otherwise, the set is called connected.
Example: Single points are always connected, because they cannot be separated into two non empty sets. Similarly the empty set is connected.
Any subset of the natural numbers is disconnected except the single points and the empty set.
The set of rational numbers is disconnected.
Theorem 5
The connected subsets of are precisely its intervals
Proof.
Let be a connected subset of. Let and let. Then if, it follows that is a non trivial partition of, which is a contradiction for a connected set. Hence, which means that is an interval.
For the converse, first consider an interval. Suppose it to be connected. Then it can be split up into disjoint subsets and, each covered exclusively by andrespectively. Suppose
say. Consider the open set (open in). Then this set is bounded above by and is non empty (because it contains the non empty set). Hence by the least upper bound property of the real numbers has a least upper bound. Now ought to belong to either or. If it belongs to then we should find an -ball around it contained completely in; but this would make a smaller upper bound. If instead it belongs to, then we can find an -ball around it contained completely in; but then there would be points of larger than. As both of these are contradictory, we conclude that the setis connected.
Now consider any other interval. The intervals and are always connected. So suppose the interval contains at least two points and is disconnected. Then we can find two points contained in separating open sets and. But this means that the set is disconnected using the same open sets and, which is false. Hence every interval is disconnected.
Theorem 6
If is continuous and is a connected set, then is connected subset of.
Proof.
Suppose is disconnected into non empty disjoint sets A and B covered exclusively by the open sets and. That is,
; .
Then,
;.
Moreover, and are disjoint andand are open sets.
Hence, disconnectedis connected.
This means that the property of connectedness is preserved by continuous maps. In particular a path which is a continuous function has a connected image. For example, straight line segments, circles and parabolas in are connected.
Corollary (Intermediate value theorem)
If is continuous and then there exists an such that and .
Proof.
We have proved that is a connected set and hence is also a connected set in R. But connected subsets of R are just intervals. Now. i.e., for some.
Similarly a continuous function has the same property: If, then there exists an such that. By the above theorem, this remains true if we substitute by any connected subsets of.
Assignment Questions
- Show that any subset of a bounded set is bounded; the union of two (or finite number of) bounded sets is bounded; repeat the exercise with totally bounded sets.
- Show that in, a bounded set is totally bounded.
- Show that compact sets are closed and bounded. What about the converse?
- Show thatis not compact in by finding an infinite open cover that does not have a finite subcover.
- Show that any intersection and any finite union of compact sets are compact.
- Show that the only connected subsets of are the single points or the empty set.
- Let be a positive real number. Use the Intermediate value theorem to show that exists.
Quiz
- Which one the following is true?
(a)Any subset of a compact set is compact.
(b)Any closed subset of a bounded set is compact.
(c)Any closed subset of a compact set is compact.
- An example for compact set is
(a), the set of real numbers.
(b)Interval.
(c)Interval.
- The connected subsets of R are precisely
(a)Its intervals.
(b)Single points.
(c)and.
Answers
1-(c)2- (b)3- (a)
Glossary
Set: A collection of well defined objects is called a set
Function: A relation from a set X to another set Y which associates every element of X with a unique element of Y is called a function.
Distance: For a set X (more than 1 element), suppose there is a function d, which maps XXto , and satisfy the following three properties:
For x,y and zX,
(i)d(x,y) 0 and d(x,y) =0 if and only if x=y;
(ii)d(x,y)=d(y,x);
(iii)d(x,y)d(x,z)+d(y,z).(triangle inequality)
Then the pair (X,d) is called a metric space, and d is called distance function, or metric.
Ball: For xX, define Bd(x,r)={yX:d(x,y)<r}. Bd(x,r) is called an open ball centered at x with radius r. Any open ball centered at x is called a neighborhood of x.
Interior: For a subset S of X, define the interior of S, denoted by int(S), to be the set of points which have a neighborhood lying inside S.
Open: A set S is said to be open if int(S), or equivalently for every x in S, there exists r>0 such that S.
Convergence: In the metric space (X,d), we said a sequence xnconverges to x, if for every , there exists an integer K such that nkimplies d(xn,x).
Limit point: A point x is said to be the limit point of a set S if there exists a sequence xn which converges to x, where xnSand xnx.
Closed set: The set S is said to be closed if S contains all its limit points.
Continuity: Suppose f is a function from X to Y with metric dxand dyrespectively. f is said to be continuous at x in X if for any , there exists such that dx(x,y)<implies dy(f(x),f(y))< .
Summary
A set is bounded when the distances between any two points in the set have an upper bound. The least such upper bound is called the diameter of the set.
A set is bounded when it can be covered by a ball.
A set is totally bounded when it can be covered by a finite number of - balls , however small . A totally bounded set is bounded.
The set is compact if any infinite sequence in has convergent subsequence in.
Let be a collection of some open sets. If, then we say that is an open cover of. is called a finite cover of if is an open cover of with finitely many elements.
is called a sub cover of with respect to if and.
is compact if and only if every open cover of has a finite subcover.
Any closed subset of a compact set is compact.
Any intersection (including uncountable) of compact sets is compact.
Finite union of compact sets is compact.
If is continuous on, then is compact in for any compact subset of.
Suppose is continuous on compact in, and. Then there exists such that for all.
A function is said to be uniformly continuous on a set if for every, there exists such that whenever for and in.
If is continuous on a compact, then is uniform continuous on .
A set is disconnected when it can be divided into at least two disjoint non empty subsets, say and, and each subsets can be covered exclusively by open sets, say and . Otherwise, the set is called connected. The connected subsets of are precisely its intervals.
If is continuous and is a connected set, then is connected subset of .
Intermediate value theorem: If is continuous and then there exists an such that and.
FAQs
- Find an example of a bounded set and a continuous function such that is not bounded.
Answer:
maps.
- Let be the unit circle in with the usual metric. Does there exists a continuous functionwhich maps onto the real line?
Answer:
No, because the unit circle is compact but the real line is not, while continuous functions
preserve compactness.
- Construct a close and bounded set which is not compact in a metric space .
Answer:
Consider an infinite set X with the discrete metric
.Then X is closed and bounded, but it is not compact, because a sequence in whose all terms are distinct has no convergent subsequence. Alternatively, a less artificial example is the sequence space, which consists of all sequences for which. If, for each positive integer , we define to be the sequence in the space for which the -th entry is 1 and all other entries are zero, then the set of all ’s is a closed and bounded set under themetric and yet it isnot compact (because the sequence in has no convergent subsequence).
- If A and B are connected sets and, then prove thatis connected.
Proof:
Suppose splits up into two parts covered exclusively by open sets U and V. Then A itself would split up into two parts andwere these non empty. But A is known to be connected; hence one of these must be empty, say. Similarly for B, . In other words, and. Hence.
REFERENCE
- W. Rudin, Principles of Mathematical analysis, McGraw Hill book company, Auckland (1985).
- P. K. Jain and K. Ahmad, Metric Spaces, Narosa, New Delhi (1993).
- J. R. Munkers, Topology, Prentice hall of India (1991).
- G. F. Simmons, Introduction to topology and Modern Analysis, McGraw Hill (1963).
- B. K. Tyagi, Metric Spaces, Cambridge University Press, new Delhi (2010).
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