2. Metric Spaces. 2.1 Definitions etc.

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1 2. Metric Spaces 2.1 Definitions etc. The procedure in Section for regarding R as a topological space may be generalized to many other sets in which there is some kind of distance (formally, sets with a metric). For example inr 2 the distance between (x 1, y 1 ) and (x 2, y 2 ) is (x 1 x 2 ) 2 + (y 1 y 2 ) 2. Notation For sets X and Y the product X Y is {(x, y) x X, y Y }, i.e. the set of pairs (x, y) with first entry in X and the second entry in Y. Definition Let X be a set. A metric on X is a map d : X X R satisfying the properties: (i) d(x, y) for all (x, y) X X and d(x, x) = if and only if x = y; (ii) d(x, y) = d(y, x) for all x, y X; (iii) d(x, z) d(x, y) + d(y, z) for all x, y, z X. Definition A metric space if a pair (X, d), where X is a set and d is a metric on X. Condition (iii) is called the triangle inequality. Examples 1. X =R, d(x, y) = x y. 16

2 (Properties (i),(ii) are obvious and x z = (x y) + (y z) x y + y z proves (iii).) 2. X =C, d(z, ω) = z ω (where z = x 2 + y 2 for z = x + iy). Exercise B 2 AC. Suppose A, B, C R and Ax 2 2Bx + C for all x R. Show A and 3. Let X =R n, where n is a positive integer. For x = (x 1, x 2,..., x n ) and y = (y 1, y 2,..., y n ) let d(x, y) = n (x i y i ) 2. Lets prove (iii) in this case ((i) and (ii)) are obvious. We have to prove that n (x i z i ) 2 n (x i y i ) n + n (y i z i ) 2 for all real numbers x i, y i, z i. Let s i = x i y i, t i = y i z i so the above is true provided that, for all real numbers s i, t i, we have n (s i + t i ) 2 n s 2 i + n t 2 i. This true provided that the inequality is still satisfied when we square both sides, i.e. we need only show n (s i + t i ) 2 n n s 2 i + t 2 i + 2 n i.e. n s it i n s2 i n t2 i, or s 2 i n n n n ( s i t i ) 2 ( s 2 i )( t 2 i ) ( ). This is a well known inequality (called the Cauchy-Schwartz inequality) and we now demonstrate its validity. We have n (t i xs i ) 2 17 t 2 i

3 i.e. n n x 2 s 2 i 2x s i t i + n t 2 i for all x R. Setting A = n s2 i, B = n s it i, C = n t2 i we get B2 AC, by the exercise above. Hence ( ) holds. 4. Take X =R n again with d(x, y) = max{ x i y i : 1 i n} for x = (x 1, x 2,..., x n ), y = (y 1, y 2,..., y n ), where max{a 1, a 2,..., a r } denotes the maximum value of real numbers a 1, a 2,..., a r. Again (i),(ii) are obvious. Now show (iii). Let z = (z 1, z 2,..., z n ) then we have for some j and we have d(x, z) = max{ x i z i : 1 i n} = x j z j x j z j x j y j + y j z j max{ x i y i : 1 i n} + max{ y i z i : 1 i n} = d(x, y) + d(y, z) so that d(x, z) d(x, y) + d(y, z), as required. 5. Take X = R n again with d(x, y) = n x i y i. It is easy to check that this is metric. So we now have three ways to regard R n as a metric space. Note that if n = 1 the metrics are all the same, otherwise they are different. 6. Let X = C[, 1], the set of all continuous real-valued functions on the closed interval [,1]. For f, g X we define d(f, g) = 1 f(t) g(t) dt. To check property (i) we shall need the following: 18

4 (*) If q : [,1] R is continuous, if q(t) for all t [,1] and if 1 f(t)dt = then q(t) = for all t [,1]. So lets prove (*). Assume q(t) for some t and derive a contradiction. Thus q(t ) for some t [, 1]. Let α = q(t ). Since q is continuous there exists a δ > such that q(t) q(t ) < α/2 for all t [,1] such that t t < α/2 (1). Since t [,1] we can choose δ so small that either [t δ/2, t ] [,1] or [t, t + δ/2] [,1] (or both). By (1) we have α/2 < q(t) q(t ) < α/2, i.e. α/2 < q(t) α < α/2 for all t [, 1] such that t t < δ/2. Thus we have q(t) α/2 for all such t. Hence if [t δ/2, t ] [,1] then - a contradiction. 1 q(t)dt t t δ/2 Similarly, if [t, t + δ/2] [,1] we get 1 q(t)dt t +δ/2 t q(t)dt q(t)dt - a contradiction. So the claim (*) is proven. t t δ/2 t +δ/2 t α δ/2.α dt = 2 2 α α/2.δ dt = 2 2 Now check that d(f, g) = 1 f(t) g(t) dt is a metric. (i) Clearly d(f, g). If d(f, g) = then 1 f(t) g(t) dt = so that 1 q(t) =, where q(t) = f(t) g(t) and by (*) we get q(t) = for all t, i.e. f(t) = g(t) for all t, i.e. f = g. (ii) d(f, g) = d(g, f) is clear. (iii) For f, g, h X we have d(f, h) = = f(t) h(t) dt f(t) g(t) + g(t) h(t) dt ( f(t) g(t) + g(t) h(t) )dt = d(f, g) + d(g, h). 19 > >

5 7. Let X be any set. Define d : X X R by { 1, if x y; d(x, y) =, if x = y. It is easy to check that d is a metric. It is called the discrete metric on X. 8. Let X be a set and let x be some fixed element of X. We define d : X X R be d(x, y) = { 2, if x y, x x and y x ; 1, if x x and y = x or if x = x and y x ;, if x = y. Check d is a metric. It is called the star metric. 9. More generally let G = (V, E) be a connected graph (with vertex set V and edge set E). For x, y V we define d(x, y) to be the smallest length n of a sequence of vertices v, v 1,..., v n such that v = x, v n = y and {v i, v i+1 } is an edge for i < n. (So v, v 1,..., v n is an edge sequence joining x to y.) Then d : V V R is a metric on V. Suppose that (X, d) is a metric space. We want to regard X as a topological space. For x X and r > we define B d (x; r) = {y X d(x, y) < r} the open ball around x of radius r. We define U X to be open if for each x U there exists some r > such that B d (x; r) U. (Since we haven t yet proved that the open sets form a topology we call them open rather than open.) Write simply B(x; r) for B d (x; r) when no confusion is likely to result. 2

6 (2.1a) Proposition The open sets form a topology. Proof (i) is open (there is nothing to check) and X is open since for each x X we have B(x;1) X. (ii) Suppose that U and V are open. We must show that U V is too. Let x U V. Since U is open we have B(x; r 1 ) U for some r 1 >. Since V is open we have B(x; r 2 ) V for some r 2 >. Thus B(x; r) = B(x; r 1 ) B(x 2 ; r 2 ) U V, where r = min{r 1, r 2 } and so U V is open. (iii) Suppose that {U i i I} is a collection of open sets. Let U = i I U i and let x U. Then we have x U j for some j I and hence B(x; r) U j for some r > and we get B(x; r) U j U and U is open. Note Since open sets form a topology we call them open sets now. Remark For x X and r >, B(x; r) is open. Proof Let U = B(x; r). Let y U. Then d(x, y) = α, say, with α < r. Then B(y; r α) U (for t B(y; r α) implies d(x, t) d(x, y) + d(y, t) < α + (r α) = r so B(y; r α) U). Definition Let (X, d) be a metric space. Let T d be the collection of subsets U of X such that for each x U there exists r > with B(x; r) U. Then (X, T d ) is called the topological space defined by the metric d and call T d the topology on X defined by d. Sometimes different metrics on a set give rise to the same topology. Definition Let d 1, d 2 be metrics on a set X. We say that d 1 and d 2 are equivalent if they define the same topology, i.e. if T d1 = T d2. (2.1b) Proposition Suppose that metrics d 1, d 2 on X are such that for some λ > we have 1 λ d 1(x, y) d 2 (x, y) λd 1 (x, y) for all x, y X. Then d 1 and d 2 are equivalent. Proof Let T 1 be the topology defined by d 1 (i.e. T 1 = T d1 ) and let T 2 be the topology defined by d 2 (i.e. T 2 = T d2 ). We must show that a subset U of X belongs to T 1 if and only if it belongs to T 2. 21

7 Suppose U belongs to T 1. Let x U. There there exists some r > such that B d1 (x; r) U, i.e. {y X d 1 (x, y) < r} U. Consider B 2 (x; r/λ). If y B d2 (x; r/λ) then d 2 (x, y) < r/λ. But 1 λ d 1(x, y) d 2 (x, y) and so, for y B d2 (x; r/λ) we have d 1 (x, y) λd 2 (x, y) < λ.r/λ = r. Hence y B d1 (x; r) whenever y B d2 (x; r/λ). But B d1 (x; r) U and so B d2 (x; r/λ) B d1 (x; r) U. Thus, for x U, there exists some r > (namely r = r/λ) such that B d2 U. Thus U is open in the topology determined by d 2, i.e. U T 1 implies U T 2. Now suppose that U T 2. For x U there exists some r > with B d2 (x; r) U. Now if d 1 (y, x) < r/λ we have d 2 (x, y) λd 1 (x, y) < λ. r λ = r so B d1 (x; r/λ) B d2 (x; r) U, and so U T 1. Thus U T 1 if and only if U T 2 and hence T 1 = T 2. (2.1c) Proposition The three metrics onr n, of examples 3,4,5 above, are equivalent. Proof We have three metrics d 1, d 2, d 3 given by d 1 (x, y) = n (x i y i ) 2 d 2 (x, y) = max{ x i y i : i = 1,2,..., n} n d 3 (x, y) = x i y i for x = (x 1, x 2,..., x n ), y = (y 1, y 2,..., y n ). Note that d 1 (x, y) = n (x i y i ) 2 n d 2 (x, y) (since x i y i d 2 (x, y) for all i) = nd 2 (x, y) nd 2 (x, y) 22

8 so that Also we have d 1 (x, y) nd 2 (x, y) (1). d 2 (x, y) = max{ x i y i : i = 1,2,..., n} = x j y j for some j = (x j y j ) 2 n (x i y i ) 2 = d 1 (x, y) and so d 2 (x, y) d 1 (x, y) and certainly Combining (1) and (2) we get and so d 1 and d 2 are equivalent, by (2.1b). Clearly d 2 (x, y) d 3 (x, y) and so d 2 (x, y) nd 1 (x, y) (2). 1 n d 2(x, y) d 1 (x, y) nd 2 (x, y) 1 n d 2(x, y) d 3 (x, y) (3). Also, d 3 (x, y) = n x i y i and each x i y i d 2 (x, y) so that Combining (3) and (4) we have so that d 2 and d 3 are equivalent. d 3 (x, y) nd 2 (x, y) (4). 1 n d 2(x, y) d 3 (x, y) nd 2 (x, y) We have now shown that d 1 and d 2 define the same topology and that d 2 and d 3 define the same topology and hence d 1, d 2, d 3 all define the same topology, i.e. d 1, d 2, d 3 are equivalent. Remark There are three obvious ways to define distance inr n (i.e. d 1, d 2, d 3 ) but only one obvious topology (involving distance). So the topology seem to be more natural than distance. 23

9 Definition We call the topology onr n defined by d 1, d 2 and d 3 the natural topology. When is a map between metric space continuous? Continuity is characterized by the ɛ, δ condition just as it is for maps fromr tor. (2.1d) Proposition Let (X 1, d 1 ), (X 2, d 2 ) be metric spaces and let f : X 1 X 2 be a map. Then f is continuous if and only if for each x X 1 and ɛ > there exists a δ > such that d 2 (f(x), f(y)) < ɛ whenever d 1 (x, y) < δ. Proof ( ) Suppose f is continuous and let x X 1, ɛ >. Then V = B d2 (f(x); ɛ) is open in X 2. Let U = f 1 V. Then U is open in X 1, by the continuity of f. Now x U (since f(x) V ) so we have B d1 (x; δ) U for some δ >. Thus, for any y B d1 (x; δ), we have f(x) B d2 (f(x); ɛ), in other words we have d 2 (f(x), f(y)) < ɛ whenever d 1 (x, y) < δ. ( ) Suppose f satisfies the ɛ, δ condition. Let V be open in X 2 and let U = f 1 V. We must show that U is open. Let x U. Then f(x) V and since V is open there is an ɛ > such that B d2 (f(x); ɛ) V. By hypothesis there exists δ > such that d 2 (f(x), f(y)) < ɛ whenever d 1 (x, y) < δ, i.e. f(y) B d2 (f(x); ɛ) whenever y B d1 (x; δ), in other words B d1 (x; δ) f 1 B d2 (f(x); ɛ). Thus we have B d1 (x; δ) f 1 B d2 (f(x); ɛ) U. Hence U is open, i.e. f 1 V is open in X 1 whenever V is open in X 2. Hence f is continuous. Before going further with metric spaces we make a useful general observation on topological spaces. (2.1e) Lemma A subset V of a topological space X is open if and only if for each x V there exists an open set U x of X with x U x V. Proof ( ) Assume V is open. Then we can take U x = V for each x V. ( ) Assume that for each x V there exists an open set U x with U x V. Then we have V = x V U x, so V is a union of open sets and hence open. We now return to metric space theory. Let (X, d) be a metric space and let Z be a subset of X. Defining d Z : Z Z R to be the restriction of d (i.e. d Z (p, q) = d(p, q) for p, q Z) we have a metric space (Z, d Z ). We can regard Z as a topological space via the subspace topology induced from X, or we can regard Z as a topological space with topology induced by the metric d Z. In fact these topologies are the same. 24

10 We check this now. Let T be the topology on X determined by the metric d. Let S 1 be the subspace topology on Z and let S 2 be the topology on Z determined by the metric d Z. We have to show that every set is S 1 = S 2, in other words that every set in S 1 is also in S 2, and vice versa. So let U S 1. Then U = V Z for some V T. Let x U. Then x V so we have B d (x; r) V for some r >. Hence B dz (x; r) = Z B d (x; r) Z V = U. This proves that U S 2. Suppose conversely that U S 2 and let x U. Then we have B dz (x; r) U for some r >, in other words Z B d (x; r) U. Putting U x = Z B d (x; r) we have that U x is open in the subspace topology S 1. By (2.1e) we have that U is open in the subspace topology S 1, i.e. U S 1, as required. In particular any subspace Z of R has topology given by the usual distance d(x, y) = x y, for x, y Z. Consider f :R (, ) given by f(x) = 1 Then f is continuous at x <, since 1 x at x > as 1 + x is continuous. At x =, f() = 1 and f(y) f() = { 1/(1 x), x ; 1 + x, x. is continuous on (, ) and f is continuous y 1 y < y if y < and f(y) f() = y if y >. Hence, given ɛ >, we have f(y) f() < ɛ for all y such that y < ɛ. So the ɛ, δ condition is satisfied at. Hence f :R (, ) is continuous. Leave it to you to check that the function g : (, ) R given by g(y) = { y 1 y, if < y 1; y 1, if y 1 is the inverse of f and is continuous. Hence f is a homeomorphism. (You could also use the maps log and exp to get inverse homeomorphisms (, ) R andr (, )). (2.1f) Proposition All open intervals (a, b) (with a < b), (a, ), and (, a) (for a R ) are homeomorphic tor. 25

11 Proof The map f : (,1) (1, ), f(x) = 1/x is a homeomorphism, with inverse g(x) = 1/x. Hence (,1) is homeomorphic to (1, ). Also (1, ) is homeomorphic to (, ) via the maps f : (1, ) (, ), f(x) = x 1 and g : (, ) (1, ), g(x) = x+1. HenceR is homeomorphic to (,1). We have already seen that all (a, b) are homeomorphic to (,1) so that all (a, b) are homeomorphic tor. Now (a, ) is homeomorphic to (, ) via the map f(x) = x a (with inverse g(x) = x+a). So (a, ) is homeomorphic to (, ) which is homeomorphic tor and hence (a, ) is homeomorphic tor. Finally, (, a) is homeomorphic to ( a, ) via f(x) = x (with inverse g(x) = x) and ( a, ) is homeomorphic tor. Hence (, a) is homeomorphic tor. Hence all open intervals (bounded or not) are homeomorphic tor. 26

12 2.2 Closed Sets Definition Let X be a topological space. A subset Z of X is closed if its complement C X (Z) = {x X x Z} is an open set. Example For a < b the set [a, b] is closed inr since C R ([a, b]) = (, a) (b, ) is the union of two open sets hence open. (2.2a) Example In a metric space X the closed ball E(x; r) = {y X d(x, y) r} is closed, for all x X, r >. Proof Let U = C X (E(x; r)) = {y X d(x, y) > r}. We shall show that U is open (and hence E(x; r) is closed). Let y U and put s = d(x, y) r. We claim that B(y; s) U. For z B(y; s) we have d(x, y) d(x, z) + d(z, y) and d(z, y) < s = d(x, y) r so d(x, y) < d(x, z) + d(x, y) r and hence d(x, z) > r, i.e. z U. Hence we have B(y; s) U and U is open so E(x; r) is closed. 27

13 (2.2b) Let X be a set and let {A i i I} be a collection of subsets of X. Then we have: (i) C X ( i I A i) = i I C X(A i ); and (ii) C X ( i I A i) = i I C X(A i ). Proof (i) x LHS x i I A i for each i I we have x A i for each i I we have x C X (A i ) x i I C X(A i ) = RHS. (ii) x LHS x i I A i x A i for some i I x C X (A i ) for some i I x i I C X(A i ) = RHS. (2.2c) Proposition Let X be a topological space. Then: (i) and X are closed; (ii) if Y and Z are closed sets then Y Z is closed; (iii) if {Z i i I} is any collection of closed sets then i I Z i is closed. Proof (i) C X ( ) = X which is open, so is closed. C X (X) = which is open so X is closed. (ii) C X (Y Z) = C X (Y ) C X (Z), the intersection of two open sets hence open and so Y Z is closed. (iii) C X ( i I Z i) = i I C X(Z i ), a union of open sets and hence open. Thus i I Z i is closed. Note that in general there are subsets which are neither open nor closed, e.g. (,1] is not open since there is no r > such that B(1; r) (, 1] and it is not closed since C X ((,1]) = (,] (1, ) and there is no r such that B(; r) C X ((,1]). Definition Let X be a topological space and let V be a subset of X. Let V be the intersection of all closed subsets which contain V. The set V is called the closure of V. (2.2d) Note V is closed and if Z is any closed set containing V then V Z. So V is the smallest closed set containing V. This is because, by definition, V is an intersection of closed sets hence closed by (2.2c)(ii). If V Z with Z closed then V is an intersection of closed sets one of which is Z so V Z. Example Let X = {a, b, c, d, e} and let T consist of the empty set together with all subsets of X which contain a. Find the closure of {a, b} and of {c, d, e}. 28

14 Definition Let X be a topological space and let V be a subset of X. We say that V is a dense subset of X (or V is dense in X) if V = X. (2.2e) Proposition The set of rational numbers is dense inr. We will give the proof of this after the following exercise. (2.2f) Exercise Between any two real numbers there is a rational number. Proof Let α, β R with α < β. Case (i) α is rational. Then α+1/n < β for some positive integer n and α+1/n is rational. Case (ii) α irrational. Choose a positive integer q such that q(β α) > 1. Thus we have 1 + qα < qβ. Let p be the smallest integer greater than or equal to qα. So we have p 1 < qα and so p < 1 + qα < qβ. Hence we have qα p < qβ. But qα p (because α is irrational) so that qα < p < qβ and α < p/q < β. We now prove (2.2e). Let Q be the set of rational numbers. Let Z = Q and assume for a contradiction that Z R. Let U be the complement of Z inr. Then U and so there is some x U and we have B(x; r) U. Thus there are no rational numbers between x r and x + r. This is not true, by the Exercise, so we must have Z =R, i.e. Q =R, i.e. Q is dense. 2.3 Closed sets in a metric space 29

15 We shall describe the closure of a subset of a metric space in terms of limits of elements. Definition Let (X, d) be a metric space and let (x n ) be a sequence of points in X. We say that (x n ) converges to x X, and write x = lim x n if for every ɛ > there exists an integer N such that d(x, x n ) < ɛ for all n N. We say that x is a limit of (x n ) and write x n x. (2.3a) A sequence has at most one limit. Proof Let (x n ) be a sequence. Suppose x, y X are both limits of the sequence. Suppose for a contradiction that x y. Let r = d(x, y). Take ɛ = r/2. Then there exists some N 1 such that d(x n, x) < r/2 for all n N 1 and there exists some N 2 such that d(x n, y) < r/2 for all n N 2. Let n be an integer such that n N 1 and n N 2. Then we have d(x n, x) < r/2 and d(x n, y) < r/2. Thus we get r = d(x, y) d(x, x n ) + d(x n, y) < r/2 + r/2 = r and so r < r, a contradiction. Thus we have x = y. Remark (x n ). In view of (2.3a) if (x n ) converges to x we now call x the limit of the sequence Examples x n = 1/n converges to (i) Taking X =R with the usual metric we have all the usual examples, e.g. Not every sequence converges, e.g. taking x n = ( 1) n, the sequence is 1,+1, 1,+1,... (which does not converge). (ii) Take X = C[, 1] (our Example 6 of a metric space, in Section 2.1). Let f n = 1 n sin(nx). We claim f n converges to (the zero function). We have d(f n,) = 1 1 = f n (x) dx = 1 n. 1 1 n sin(nx) dx 1 n dx Hence we have d(f n,) 1 n. For ɛ >, choose N > 1 ɛ then, for all n N, we get n > 1 ɛ so 1 n < ɛ and d(f n,) 1 n < ɛ. Hence f n. 3

16 (2.3b) Lemma Let Z be a closed subset of a metric space X. If x X is the limit of a sequence of points in Z then x Z. Proof We assume x is the limit of the sequence (x n ), with x n Z for all n 1. If x Z then x U, where U is the complement of Z in X. Since U is open we would then have B(x; r) U for some r >. But since x = lim x n, for some N we have x n B(x; r) for all n N. In particular x N B(x; r) U. But x N Z and U is the complement of Z so this is impossible. We can now describe the closure in terms of limits. (2.3c) Proposition Let S be a subset of a metrix space X. Then S is the set of all x X such that x is the limit of some sequence of elements of S. Proof If x is the limit of a sequence of points in S then x is the limit of a sequence of points in S so x S, by (2.3b). We now suppose that x S and show that x can be written as the limit of a sequence of points in S. If x S we have x = lim x n, where all x n = x. So now suppose x S\S. Let n 1 and consider U = B(x; 1/2 n ). Then C X (U) is closed. If S C X (U) then S C X (U) (since S is the smallest closed set containing S). But this is not true because x S would then give x C X (U) (whereas in fact x U). So S is not contained in C X (U), i.e. the there is an element in both S and U. Choose one and call it x n. Then x n S and d(x, x n ) < 1/2 n. This gives x = lim x n, with all x n S, as required. Definition Let S be a subset of a metric space X. A limit point of S is the limit of a convergent sequence of points of S. Example Let X =R (with the natural topology) and S = {1,1/2,1/3,...}. Then is a limit point of S (because = lim1/n). We can make (2.3c) a bit more explicit as follows. (2.3d) Proposition Let S be a subset of a metrix space X. Then S is the union of S and the set of all limit points of S. 31

17 Proof Let S be the union of S and the set of all limit points of S. Then S S, by (2.3c). We need to show S S. We know that x = lim x n for some convergent sequence (x n ) of elements of S (by (2.3c)). We have to show x S. There are two cases to consider. Case (i) : Suppose the set {x 1, x 2, x 3,...} is finite. (For example inr this could happen if x 1 = 2, x 2 = 33, x 3 = 7 and x n = 13 for all n 4.) Let α 1, α 2,..., α r be elements of S such that, for each n, we have x n is one of α 1, α 2,..., α r. Let ɛ = min{d(x, α 1 ), d(x, α 2 ),..., d(x, α r )}. If ɛ = then x = α i for some i, in particular x = α i X, as required. If ɛ > then, by definition of convergence, there exists an N such that d(x n, x) < ɛ for all n N. Thus d(x, x N ) < ɛ. But x N = α j for some j so we get d(x, α j ) < ɛ. But ɛ is the minimum among d(x, α 1 ), d(x, α 2 ),..., d(x, α r ) so this is impossible. So this never happens, i.e. we must have ɛ = and x = α i for some i, so x S. Case (ii) : Suppose the set {x 1, x 2, x 3,...} is infinite. Let y 1 = x 1. Let y 2 be the first x n different from y 1, and generally let y r be the first x n different from y 1, y 2,..., y r 1. [For example if our sequence is (1,1,2, 1,3,3,4,2,...) then y 1 = 1, y 2 = 2, y 3 = 1, y 4 = 3, y 5 = 4,....] By construction all the y r are distinct and, for each r we have y r = y n for some n r. Let ɛ >. Then there exists N such that d(x n, x) < ɛ for all n N. Hence, for r N, we have d(x, y r ) = d(x, x n ) for some n r and hence n N < ɛ. Hence y r x. Thus (y r ) is a sequence of distinct terms in S converging to x. Hence x is a limit point of S and so x S. Example 1 Lets work out the closure of S = {1,1/2,1/3,...} inr. First work out its limit points. There is an obvious one, namely = lim1/n. Suppose that x is a limit point of S, so that x = lim x n for s sequence (x n ) of distinct points of S. Let ɛ >. Then there exists a positive integer N 1 such that x x n < ɛ for n > N 1. Let K be a positive integer such that 1/K < ɛ. Since the elements x n are distinct we have x n {1,1/2,...,1/K} for only finitely many values of n. So we may choose N 2 such that for all n N 2 we have x n {1, 1/2,..., 1/K}. So if n > N 2 then x n = 1/r for some r > K and so x n = 1/r < 1/K < /ep. Now let N be the maximum of N 1 and N 2. For 32

18 n > N we have x n x < ɛ and x n < ɛ so that x = (x x n ) + x n < ɛ + ɛ = 2ɛ. But this is true for all ɛ > so x = and hence x =. Thus is the only limit point of S. Suppose now that α is a limit point and α = lim x n, where (x n ) is a sequence of distinct points in S. Let ɛ > and let K be a positive integer such that 1/K ɛ. Since (x n ) is a sequence of distinct terms, x n can only belong to {1,1/2,1/3,...,1/K} for only finitely many values of n, say x n1, x n2,..., x nr {1,1/2, 1/3,...} and x n {1,1/2,1/3,...} if n is different from n 1, n 2,.... Thus if n is different form n 1, n 2,..., n r then x n = 1/k, for some k > K and x n = 1/k < 1/K ɛ. Thus x n < ɛ for all n N = max{n 1, n 2,..., n r } and so x n, i.e. = lim x n. Hence is the only possible limit point of S. So {x R x is a limit point of S} = {} and, by (2.3d), we have S = {,1, 1/2, 1/3,...}. Example 2 Let S = (,1). Find the set of limit points of S, and find S. First note that [,1] is closed and so S [,1], (2.2d). Hence, by (2.3d) Theorem, every limit point of S belongs to [.1]. Note also = lim 1/n, 1 = lim(1 1/2n) so that and 1 are limit points of S. Let α (,1). Then α = lim x n, where x n = α α/2n, so that every α (,1) is a limit point. Hence [,1] is the set of limit points of (,1) and so S = [, 1], by (2.3d)Theorem. 33

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