ECARES Université Libre de Bruxelles MATH CAMP Basic Topology

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1 ECARES Université Libre de Bruxelles MATH CAMP 03 Basic Topology Marjorie Gassner Contents: - Subsets, Cartesian products, de Morgan laws - Ordered sets, bounds, supremum, infimum - Functions, image, preimage, limits, continuity - Finite, countable and uncountable sets - Metric spaces, limit points, open sets and closed sets etc. - Compact sets, the Weierstrass theorem

2 Subset : if A and B are sets, if a A a B, then A is a subset of B, denoted by A B (or B A). If there is an element b Note: A = B iff A B and B A! B such that b A, then A is a proper subset of B. Cartesian product: if A and B are two sets, then the Cartesian product of A and B, denoted AxB, is AxB = {( a,b) : a A andb B}. Example: IR Extension: A xa xa 3 x xa n = n Ai = {( a,a,a 3,...,a n): ai Ai i} i= Example: IR n De Morgan laws: Let A and D be subsets of B and c c c (A D) = A D and Proof: exercise. c A and c c c (A D) = A D. c D their respective complements with respect to B.

3 Ordered sets Let S be a set. An order on S is a relation < on S such that () for all a in S, a< a () for all a,b,c in S, if a< b and b< c, then a < c (3) for all a,b in S, a< b and b< a, then a = b In other words, < is reflexive, transitive and antisymmetric. Examples: in IR or in Q I (divides) in IN 0 An ordered set is a set on which an order is defined: if < is an order on S, then S,< is an ordered set. An order < on S is total iff for all a,b in S: a< b or b < a. Exercise: Let IR, where is defined as follows: For all (x,y), (z,t) in IR, (x,y) (z,t) x z and y t ( in IR!!). Prove that this defines an order on IR. Is it total? Represent graphically E = {(x,y) IR :(x,y) (,)} Supremum, infimum, maximum, minimum Let S, be an ordered set and E S. E is bounded above if there exists α in S such that α x for all x in E. α is then an upper bound of E. E is bounded below if there exists β in S such that β x for all x in E. β is then a lower bound of E. If S, is an ordered set, E S and E is bounded above, then If there exists α in S such that α is an upper bound of E and if γ < α, γ is not an upper bound of E then α is the least upper bound of E, called the supremum of E, denoted by sup E. If sup E exists and sup E E, then sup E is the maximum of E, denoted by max E. 3

4 If S, is an ordered set, E S and E is bounded below, then If there exists β in S such that β is a lower bound of E and if γ > β, γ is not a lower bound of E then β is the greatest lower bound of E, called the infimum of E, denoted by inf E. If inf E exists and inf E E, then inf E is the minimum of E, denoted by min E. Exercises: () In IR,, let E = { n : n IN0 }. Determine the set of upper bounds of E, the set of lower bounds of E as well as, if they exist, sup E and inf E, max E and min E. let E = (7, + ). Same questions. () In IR, where is defined as follows: For all (x,y), (z,t) in IR, (x,y) (z,t) x z and y t. Let E = {(x,y) IR :x and y }. Is E bounded above? Bounded below? If they exist, find inf E and sup E, max E and min E. Important: in IR,, all subsets that are bounded above have a (unique) supremum all subsets that are bounded below have a (unique) infimum. In Q,, this is not the case, e.g. the set E = {p Q + : p < } is bounded above in Q, but has no supremum in Q and the set F = {p Q + : p > } is bounded below in Q, but has no infimum in Q (Prove it (difficult!)! Hint: show that there is no p in Q such that p = and then let q = p - p. p+ Show that for all p in F, q < p and q belongs to F and that for all p in E, q > p and q belongs to E. This means that E has no largest member and F has no smallest member). Note: Q is dense in IR i.e. between any two distinct real numbers, there is a rational one! (if x and y are in IR and x < y then, there exists p in Q such that x < p < y). 4

5 Functions If A and B are two sets and if with each element x of A is associated an element of B, denoted by f(x), then f is said to be a function from A to B (or mapping of A into B): f:a B:x f(x). Note: if (at least) one element of A is associated with more than one element of B (a set of elements of B), then the term used is correspondence or point-to-set mapping rather than function. The set A is called the domain of f, and the elements f(x) are the values of f. The set of all values of f is the range or the image of f. If E is a subset of A, f(e), the set of all elements f(x) for x E, is the image of E under f (the range of f is thus the image of A under f!). In general f(a) B. If f(a) = B, then f maps A onto B. If E B, then pre-image of E under f. If y B, then f (E) denotes the set of all x A such that f(x) E. f (y) is the set of all x A such that f(x) = y. If, for each y B, f (E) is the inverse image or f (y) consists of at most one element of A, then f is said to be a one-to-one mapping of A into B. Equivalently, f is a one-to-one mapping of A into B provided that f(x ) f(x ) whenever x x. Finite, countable and uncountable sets If there exists a one-to-one mapping of A onto B, then one says that A and B can be put in one-toone correspondence or that A and B have the same cardinal number or that A and B are equivalent. This is written A ~ B. For any positive integer n (i.e. n IN 0 ), let J n = {,,,n}. For any set A, (a) A is finite if A ~ J n for some n (the empty set is also considered to be finite). (b) A is infinite if A is not finite. (c) A is countable (or enumerable or denumerable) if A ~ IN 0 (d) A is uncountable if A is neither finite nor countable. (e) A is at most countable if A is finite or countable. For two finite sets, A ~ B is equivalent to A and B have the same number of elements. But for A and B infinite?? 5

6 Example: Z (the set of all integers (positive and negative)) is countable! Proof: consider the following arrangements of Z and IN 0 : Z : IN 0 : 0,,-,,-,3,-3,.,, 3,4, 5,6, 7,. n if n even The function f:in 0 Z:n f(n) = n- - if n odd Z ~ IN 0, proving that Z is countable. is a one-to-one mapping of IN 0 onto A, hence Note: here, though IN 0 is a proper subset of Z, they are equivalent. This cannot happen for finite sets. Since a sequence in a set A (or a sequence of elements of a set A) is a function defined on IN 0, and since a countable set can be seen as the range of a sequence of distinct terms, another (equivalent) definition of the concept of countable set is the following: A is an infinite countable set iff its elements can be arranged in a sequence. Counterexample: the set IR of all real numbers is uncountable! Proof? See property (4). Properties: () Any infinite proper subset of a countable set A is countable Proof: assume E A and E infinite. Since A is countable, arrange the elements of A in a sequence {x n } of distinct elements. Construct {n k } as follows: n is the smallest possible integer such that x E. Having chosen n,n,,n k- (k=,3,4, ), let n k be the smallest integer greater than n k- such that a one-to-one correspondence between E and IN 0. n xn k E. Putting f(k) = x nk (k=,,3,.), we obtain () Let {E n }, n=,,3,. be a sequence of countable sets. Let S = U En. Then S is countable. n= (3) Let A be a countable set. A n is also a countable set. Corollary of (3): Q is a countable set! (4) The set of all real numbers between 0 and written only with the digits 0 and is uncountable. From there [0,] is uncountable, hence IR is uncountable. 6

7 Metric spaces A set E is called a metric space if with any two points p and q of E there is associated a real number d(p,q) called the distance from p to q such that () d(p,q) > 0 if p q and d(p,q) = 0 if p = q () d(p,q) = d(q,p) (3) d(p,q) d(p,r) + d(r,q) for any r E. Any function d:exe IR with these three properties is a distance function or a metric on E. Most important metric spaces: IR n with the Euclidian distance i.e. If x and y IR n, with n IN 0, then d(x,y) = x y = n (xi y i). i= In the particular case n =, i.e. in IR, if x and y IR, d(x,y) = (x y) = x y. Open, closed, convex, bounded and compact subsets of IR n Note: all of the concepts defined below can be defined in any metric space. Let c IR n, with n IN 0 and r > 0. The open ball with center at c and radius r is defined as B(c,r) = {x IR n : d(x,c) < r} The closed ball with center at c and radius r is defined as B (c,r) = {x IR n : d(x,c) r}. For n =, B(c,r) = (c-r,c+r) and B (c,r) = [c-r,c+r]. For n = and n = 3, what is an open ball? A closed ball? 7

8 Let E IR n, with n IN 0. A neighborhood of p IR n is a set containing an open ball centered at p. In particular, an open ball centered at p is a neighborhood of p. A point p IR n is a limit point of E iff every open ball centered at p contains a point q p such that q E or, equivalently, A point p IR n is a limit point of E iff p is the limit of a sequence of points of E none of which are p. The set of all limit points in E is denoted by E. If p E and p is not a limit point of E, then p is an isolated point of E. E is closed iff every limit point of E is a point of E. The closure of E is the set E = E E'. A point p is an interior point of E iff there is an open ball B centered at p such that B E. E is open iff every point in E is an interior point of E. The interior of E is the set of all interior points of E. Notation: E o. The complement of E, denoted by E c, is the set of all points p IR n such that p E. E is perfect iff E is closed and every point of E is a limit point of E. E is bounded iff E is entirely included in an open ball (equivalent to: E is bounded iff there exists a real number M and a point q IR n such that d(p,q) < M for all p E). E is dense in IR n iff every point of IR n is a limit point of E or is a point of E (or both!) A subset of IR n is compact iff it is closed and bounded. Note that the basic definition of a compact subset of a metric space is more complicated, but boils down to this one for IR n! Given any finite number of points x, x, x 3,., x n with x i IR n, a point z IR n is a convex combination of the points x, x, x 3,., x n if there exists a λ (IR + ) n such that n z = λi.xi. i= E is convex iff the convex combination of any two points of E is also contained in E i.e. n λi = such that i= x,x E, [0,] :.x ( ).x E λ λ + λ. Geometrically, E is convex iff the segment joining any two points of E is entirely included in E. 8

9 Properties: () A set E is open iff E c is closed. A set E is closed iff E c is open. () Any open ball is an open set, any closed ball is a closed set. (3) Any union of open sets is open. Any intersection of closed sets is closed. The intersection of any finite number of open sets is open. The union of any finite number of closed sets is closed. 9

10 Limits and continuity of functions () Limits Let f:d n m IR IR : x f(x) = (f (x),f (x),,f m (x)) and a, a limit point of D. We have x a m lim f(x) = b ( IR ) iff ε > 0, δ > 0 : 0 < d (x,a) < δ and x D d (f(x),b) < ε. n m Note: If m =, the definition becomes: lim f(x) = b ( IR) iff ε > 0, δ > 0 : 0 < d n(x,a) < δ and x D f(x) b < ε x a Furthermore, if m = : lim f(x) =+ (resp. ) iff K > 0, δ > 0:0< d n(x,a) < δ and x D f(x) > K (resp.f(x) < K). x a This definition can be rewritten in terms of sequences as follows: x a m lim f(x) = b ( IR ) iff for any sequence (x n) n IN such that 0 lim xn n + = awith x n a for all n, lim f(x n) = b. n + () Continuity Let f:d n m IR IR : x f(x) = (f (x),f (x),,f m (x)) and a D, f is continuous at a iff lim f(x) = f(a) x a which is equivalent to ε > 0, δ > 0 : 0 < d n(x,a) < δ and x D d m(f(x),f(a)) < ε. If f is continuous at every point of D, then f is said to be continuous on D. An equivalent definition (not easy to prove!!) is that Let f:d n m IR IR : x f(x) and a D, f is continuous on D iff set V in IR m. f (V) is open in IR n for every open 0

11 Weierstrass s theorem n If f: D IR IR: x f(x) is continuous on D and D is a compact set, then, if M = sup f(x) x D there exist x, x and m = and its minimum (at x ). inf f(x) x D D such that f(x ) = M and f(x ) = m. In other words, f reaches its maximum (at x ) Reference: W. Rudin: Principles of Mathematical Analysis, McGraw-Hill

12 Complement: point-to-set mappings A point-to-set mapping of A into B is a mapping f:a!(b): x f(x) B. The graph of f is the set of all pairs (x,y) such that y f(x). Example: f: [0,]!([0,]): x [ x,x]. Let f be a point-to-set mapping of A into B. Let S be the set of limit points of all sequence paths in the graph of f as x tends to x 0. f is upper semicontinuous at x 0 iff S f(x 0 ). f is lower semicontinuous at x 0 iff f(x 0 ) S. A point-to-set mapping is continuous at x 0 iff it is upper semicontinuous and lower semicontinuous at x 0.