MATH 51H Section 4. October 16, Recall what it means for a function between metric spaces to be continuous:

MATH 51H Section 4 October 16, 2015 1 Continuity Recall what it means for a function between metric spaces to be continuous: Definition. Let (X, d X ), (Y, d Y ) be metric spaces. A function f : X Y is continuous at a point a X if for all ε > 0 there exists a δ > 0 such that whenever x X and d X (x, a) < δ, d Y (f(x), f(a)) < ε. We say that f : X Y is continuous if f is continuous at every point a X. So if f : R R, then the statement d X (x, a) < δ translates to x a < δ, and the statement d Y (f(x), f(a)) < ε translates to f(x) f(a) < ε. Here are two examples that demonstrate how to prove that a function is continuous. Proposition (Projection maps to coordinates are continuous). Define f i : R n R by Then f i is continuous. f i (x 1,..., x n ) = x i. Proof. Let a = (a 1,..., a n ) R n. Let ε > 0. We want to find a δ > 0 such that whenever x a < δ, f(x) f(a) < ε. Note that for any x = (x 1,..., x n ) R n, and that f i (a 1,..., a n ) f i (x 1,..., x n ) = a i x i x a = (x 1 a 1 ) 2 + + (x i a i ) 2 + + (x n a n ) 2. Since b 2 0 for all b R, we can drop all of the terms under the square root sign above except for the (x i a i ) 2 term to get an expression that is smaller (or equal): (x1 a 1 ) 2 + + (x i a i ) 2 + + (x n a n ) 2 (x i a i ) 2 = x i a i. But, as we showed above, x i a i = f i (a) f i (x). So, f i (a) f i (x) x a. Take δ = ε. By our manipulations, we see that whenever x a < ε, f i (a) f i (x) x a < ε as well. Since a R n and ε > 0 were arbitrary, f i is continuous. 1

In the example above, we saw that having the condition that f i (a) f i (x) x a was very useful for showing that f i is continuous, for if x a < ε, then certainly f i (x) f i (a) < ε as well. The same argument given at the end of the proof should show that for any g : X Y satisfying f(x) f(y) x y for all x, y X, g should be continuous on X. This is an example of a more general phenomenon: Proposition (Lipschitz implies continuous). Let (X, d X ) and (Y, d Y ) be metric spaces and K > 0. Suppose that f : X Y satisfies d Y (f(x), f(y)) K d X (x, y) for all x, y X. Then f is continuous. Proof. Suppose that a X. Let ε > 0. Then for all x X such that d X (x, a) < ε (so here K we take δ = ε ), we have that d K Y (f(x), f(a)) K d X (x, a) < K ε = ε. So since a X K and ε > 0 were arbitrary, f is continuous. One very important property of real-valued continuous functions from a metric space is that they can be added, multiplied, and multiplied by a scalar to get more continuous functions. Proposition. Let (X, d) be a metric space, f, g : X R be continuous, and c R. Then f + g, cf, and f g are all also continuous functions from X to R. The proof of this proposition is very similar to the proofs about limits of sums, products, and scalar multiples of convergent sequences. For example, if we want to show from scratch that sums of continuous functions are continuous, then we ll end up doing the manipulation (f + g)(x) (f + g)(a) = f(x) + g(x) f(a) g(a) = f(x) f(a) + g(x) g(a) f(x) f(a) + g(x) g(a), and then use that f and g are continuous at a to conclude, just as we had to use that the individual sequences converged for the corresponding proof for sums of convergent sequences. In fact, this similarity is not just cosmetic. You saw in class the statement of the following proposition: Proposition. Let (X, d X ) and (Y, d Y ) be metric spaces an a X. Then f is continuous at a if and only if for every sequence (x n ) n=1 in X which converges to a in X, we have lim n f(x n ) = f(a) in Y. Half of this was done in class and the other half was given to you as an exercise, so I m not going to prove it. But I will show how it gives you for free these properties of continuous functions from the corresponding properties for limits. For example, to show that fg is continuous at a when f, g : X R are continuous, let a n a in X. Then f(a n ) f(a) and g(a n ) g(a) since f and g are continuous at a. Then by the result on the product of convergent sequences, (fg)(a n ) = f(a n )g(a n ) f(a)g(a) = (fg)(a). Recall that any subset of a metric space inherits the metric space structure of the space containing it, and thus can be viewed as a metric space itself. This is one way of thinking about what it means for a function from a subset of a metric space to be continuous. We ll introduce another way to think about continuous functions that the book uses, which you probably saw a version of in high school 2

Definition. Let (X, d) be a metric space. Suppose that A X. A point a A is an isolated point of A if there exists a ρ > 0 such that B ρ (a) A = {a}. For example, if A R is the set A = [0, 1] {2}, then 2 is an isolated point of A because B 1/2 (2) A = (3/2, 5/2) A = {2}. No other points of A are isolated. Definition. Let (X, d X ) and (Y, d Y ) be metric spaces. Suppose that A X, a A is not an isolated point, y Y, and f : A Y. We say that lim x a f(x) = y if for all ε > 0 there exists a δ > 0 such that if x A B δ (a) \ {a}, then d Y (f(x), f(y)) < ε. The first thing to note is that any function f : X Y is automatically continuous at an isolated point of X. Suppose that a X is isolated. Returning to the definition of continuity, for all ε > 0 we have to produce a δ > 0 such that whenever d X (x, a) < δ, it is also true that d Y (f(x), f(a)) < ε. But since a is an isolated point of X, we can find a δ > 0 such that B δ (a) = {a}. Then this δ works for all ε, since d Y (f(a), f(a)) = 0 < ε. Unraveling the definition of the limit of a function at a non-isolated point a, we see that f is continuous at a if and only if lim x a f(x) = f(a). 2 Open and closed sets Recall from class the following definitions: Definition. Let (X, d X ) be a metric space and A X. We say that A is open if for all a A there exists a ρ > 0 such that B ρ (a) = {x X : d(x, a) < ρ} A. We say that a point x X is a limit point of A if there exists a sequence (a n ) where a n A for all n N such that lim n a n = x. We say that A is closed if it contains all of its limit points. That is, if whenever (a n ) is a sequence in A that converges to some x X, then x A. First we ll give some examples of how to go about proving that a set is open. Example. 1. Any subset of a discrete metric space X is open. Let A X, where X has the discrete metric. Recall that this metric is defined by { 0 x = y d(x, y) = 1 x y. To show that A is open, we need to show that around any point in A we can find an open ball centered at that point contained in A. So let a A. Consider the ball B 1 (a). We have B 1 (a) = {x X : d(x, a) < 1} = {a}. Then certainly B 1 (a) = {a} A. Thus, we have found an open ball centered at a which is contained in A. Since a A was arbitrary, this shows that A X is open. 3

2. The upper half plane H = {(x, y) R 2 : y > 0} in R 2 with the usual Euclidean metric is open. To prove this, let (a, b) H. Then by definition, b > 0. Now take ρ = b/2 > 0. Then if (x, y) B ρ ((a, b)), we have d((x, y), (a, b)) = (x a) 2 + (y b) 2 < b/2. As before, note that y b (x a)2 + (y b) 2 < b/2. So b/2 < y b, which implies that y > b/2 > 0. Thus, (x, y) H. Since (x, y) was an arbitrary element of B ρ ((a, b)), we see that B ρ ((a, b)) H. Thus, H is open. Now we ll give some examples of how to go about proving that a set is closed. Example. 1. Let (X, d) be a metric space and a X be a point. Then the singleton set {a} is closed. To show this, we need to show that any limit point of {a} is contained in {a}. So suppose that x X is a limit point of {a}. Then there exists a sequence of a n in {a} such that a n x. But since {a} has one element, a n = a for all n N. Then (a n ) is the constant sequence (a), which converges to a. Thus, x = a, and so certainly x {a}. Since x was an arbitrary limit point of {a}, we have shown that {a} contains all of its limit points. That is, {a} is closed. 2. The x-axis R = {(x, y) : x = 0} R 2 is a closed set when R 2 has the usual Euclidean metric. To see this, let (a, b) R 2 be a limit point of R in R 2. So there exists a sequence ((0, x n )) in R such that (0, x n ) (a, b). Note that (0, x n ) (a, b) = a 2 + (x n b) 2 a 2 = a But as n, by the definition of limit, (0, x n ) (a, b) 0. This shows that a = 0, which is equivalent to a = 0. Thus, (a, b) = (0, b) R. Since (a, b) was an arbitrary limit point of R, we see that R contains all of its limit points, and hence is closed. Recall from class that finite unions of closed sets are closed. So from our first example we can conclude that any finite subset of a metric space is closed. You could have approached the seventh problem on your homework using this fact, since at some point you needed to find an open interval in (0, 1) which contained your irrational number and did not contain any rational number with a small denominator. Finally, we ll do an example of how to show that a set is not open, and also an example of how to show that a set is not closed. Example. The set Q R is neither open nor closed with the standard metric on R. To see that Q R is not open, we just have to show that there is some point in Q such that any open ball centered at that point is not contained in Q. So take 0 Q and consider the open intervals ( ε, ε) for all ε > 0. Recall from the second homework that any open interval contains an irrational number. This means that ( ε, ε) Q, since it contains an irrational number. So Q is not open. To see that Q is not closed, we just have to find a limit point of Q which is not in Q. You also essentially proved this on the second homework when you showed that any open interval contains a rational number. In particular, any interval around 2 contains a rational number. So for all n N, there exists an a n Q ( 2 1/n, 2 + 1/n), which is equivalent 4

to a n satisfying a n 2 < 1/n. This shows that a n 2, so that 2 is a limit point of Q, yet 2 / Q by the first homework. So Q is not closed. Here is another useful result you can use to prove a set is open or closed or not open or not closed. Proposition. Let (X, d) be a metric space. Then A X is open if and only if X \ A is closed: Proof. First suppose that A X is open. We ll show that X \ A is closed. Let b X be a limit point of X \ A. So there exists a sequence (b n ) in X \ A such that b n b. Suppose by way of contradiction that b / X \ A. Then this means that b A. Since A is open, there exists a ε > 0 such that B ε (b) A. By the definition of limit, there exists an N N such that for all n N, d(b n, b) < ε, i.e. b n B ε (b) A. In particular, this says that b N A. But b N X \ A, so we have a contradiction. Thus, we must have b X \ A. Since b was an arbitrary limit point of X \ A, this shows that X \ A contains all of its limit points. That is, X \ A is closed. Now suppose that X \ A is closed. Suppose by way of contradiction that A is not open. This means that there exists some a A such that for all ρ > 0, B ρ (a) A, i.e. B ρ (a) (X \ A). For each n N, pick an a n B 1/n (a) (X \ A), which we can do because this set is nonempty. Then (a n ) is a sequence in X \ A such that for all n N, d(a n, a) < 1/n. Hence, we see that a n a. However, X \ A is closed, and this shows that a is a limit point of X \ A, so a X \ A. But a A, so this gives a contradiction. Thus, A must be open. Recall that given a function f : A B and a subset of the codomain S B, the inverse image of S under f is defined to be f 1 (S) = {x A : f(x) S}. The next proposition, which was mentioned in class yesterday and which some of you may have proved as part of a problem on the homework, says that the inverse image of an open set in a metric space under a continuous function is open, and likewise the inverse image of a closed set under a continuous function is closed. Proposition. Let (X, d X ) and (Y, d Y ) be metric spaces and f : X Y be continuous. Then if U, C Y with U open and C closed, we have that f 1 (U) X is open and f 1 (C) X is closed. Proof. First suppose that U Y is open. We want to show that f 1 (U) is open, i.e. that for all a f 1 (U) there exists a ρ > 0 such that B ρ (a) f 1 (U). So let a f 1 (U). Then there exists an ε > 0 such that B ε (f(a)) U since U is open and f(a) U by the definition of inverse image. Now since f is continuous, there exists a δ > 0 such that for all x B δ (a), f(x) B ε (f(a)). But B ε (f(a)) U, so this says that if x B δ (a), then f(x) U, i.e. that x f 1 (U). So B δ (a) f 1 (U). Thus, since a f 1 (U) was arbitrary, f 1 (U) X is open. Now suppose that C Y is closed. Then Y \ C is open, and by the above proof we have that f 1 (Y \ C) is open in X. But f 1 (Y \ C) = f 1 (Y ) \ f 1 (C) = X \ f 1 (C). This, the complement of f 1 (C) is open in X, which implies that f 1 (C) is closed in X. 5

Now we ll show some applications of this result. The first application is another proof that an open ball in a metric space is indeed an open set. First we will need a lemma. Lemma. Let (X, d) be a metric space and a X. Define a function D a : X R by D a (x) = d(x, a). Then D a is continuous on X. Proof. First, we ll prove the reverse triangle inequality, which follows from the usual triangle inequality. Suppose that x, y, z X. Then the reverse triangle inequality is d(x, y) d(y, z) d(x, z). To see this, note that by the usual triangle inequality we have d(x, y) d(x, z) + d(z, y) = d(x, z) + d(y, z) and hence, subtracting d(y, z) from both sides, d(x, y) d(y, z) d(x, z). Similarly, applying the triangle inequality for d(y, z) now, d(y, z) d(y, x) + d(x, z) = d(x, y) + d(x, z) and hence, subtracting d(x, y) from both sides, [d(x, y) d(y, z)] = d(y, z) d(x, y) d(x, z). Putting it all together, we get the reverse triangle inequality. Now let b X and ε > 0. Then if d(x, b) < ε (so here we take δ = ε in the definition of continuity), then by the reverse triangle inequality, D a (x) D a (b) = d(x, a) d(b, a) = d(x, a) d(a, b) d(x, b) < ε. Thus, since b X and ε > 0 were arbitrary, D a is continuous on X. Proposition. Let (X, d) be a metric space, a X, and ρ > 0. Then B ρ (a) is open. Proof. By definition, B ρ (a) = {x X : d(x, a) < ρ} = {x X : D a (x) < ρ} and {x X : D a (x) < ρ} = Da 1 (, ρ). This is the inverse image of an open set by a continuous function (since we showed that D a is continuous in the previous proposition.) Thus, B ρ (a) is open. As a more fun application of our result on inverse images of open and closed sets under continuous maps, we ll show that certain special sets of matrices are open or closed in the set of 2 2 matrices. Define ( ) a b M 2 (R) := { : a, b, c, d R}, 6

and ( ) a b GL 2 (R) := { M 2 (R) : ad bc 0}, ( ) a b SL 2 (R) := { M 2 (R) : ad bc = 1}. So we have SL 2 (R) GL 2 (R) M 2 (R). Recall that all 2 2 matrices can be viewed as vectors in R 4 = R 2 2, and thus they have a natural norm ( ) a b = a 2 + b 2 + c 2 + d 2, which we called the matrix norm in class. Also recall that given any norm defined on a vector space, we get a natural metri(x, y) = x y. So this matrix norm gives a metric on M 2 (R). Proposition. In the topology induced by the matrix norm, GL 2 (R) is an open subset of M 2 (R) and SL 2 (R) is a closed subset of M 2 (R). Proof. First, we ll show that the function det : M 2 (R) R defined by (( )) a b det = ad bc is continuous. Since M 2 (R) is really the same as R 4, but with the components of a 4-tuple arranged in a different way, it is still true that the projection maps from each component are continuous. Thus, the functions (( )) a b f 1 = a, (( )) a b f 2 = b, (( )) a b f 3 = c, and (( a f 4 c )) b = d d are all continuous on M 2 (R). Since products and differences of continuous functions are continuous, this shows that det is continuous as well. Now we just need to note that and GL 2 (R) = {M M 2 (R) : det M 0} = det 1 (R \ {0}) SL 2 (R) = {M M 2 (R) : det M = 1} = det 1 ({1}). Since R {0} R is open (being the complement of the closed set {0}) and {1} R is closed, we conclude that GL 2 (R) is open and SL 2 (R) is closed in M 2 (R). 7

Note that the sets GL 2 (R) and SL 2 (R) are also groups under matrix multiplication. You can also verify that the operation of matrix multiplication GL 2 (R) GL 2 (R) GL 2 (R) and the taking of matrix inverses GL 2 (R) GL 2 (R) are continuous maps. So these sets are groups with topology (i.e. a notion of open sets) such that the group structure interacts well with the topology. They are examples of what are called topological groups. 3 l (N) and more examples of closed and not closed sets A more interesting example of a metric space than the various finite-dimension spaces R n is the sequence space l (N) defined as l (N) := {(a n ) n=1 : a n R for all n N and (a n ) is bounded}. This set is a vector space over R under sequence addition and scalar multiplication: (a n ) n=1+ (b n ) n=1 = (a n + b n ) n=1 and c(a n ) n=1 = (ca n ) n=1. We define a norm on l (N) by (a n ) n=1 = sup a n. n N Let s check that this function is indeed a norm. Let (a n ), (b n ) l (N) and c R. Positive definiteness: Since a n 0 for all n N, we see that (a n ) n=1 0. If (a n ) is not the zero sequence, then there exists an m N such that a m 0, which implies that a n > 0. Then (a n ) n=1 = sup n N a n a n > 0 since the supremum is an upper bound for the set {a n : n N}. Absolute homogeneity: We have (ca n ) n=1 = sup n N ca n = sup n N c a n = c sup a n = c (a n ). n N Triangle inequality: We have (a n ) n=1 + (b n ) n=1 = sup n N a n + b n sup n N ( a n + b n ) sup n N a n + sup n N b n = (a n ) n=1 + (b n ) n=1 by the triangle inequality in R. Then (a n ) n=1 (b n ) n=1 defines a metric on l, and we can talk about open and closed sets in l. Proposition. Define F l to be the subspace of sequences which are eventually zero, i.e. F = {(a n ) n=1 l : there exists an N N such that n N a n = 0}. Then F is not closed in l. Proof. Recall from the definition of a closed set in a metric space that to show that a set is not closed, we must show that it does not contain all of its limit points. That is, we just have to show that there is one limit point that F does not contain. I claim that the following 8

sequence is a limit point of F : (1/n) n N. To see this, consider the sequence of elements of l ((a k n) n=1) k=1 defined by { 1 a k n k n n = 0 n > k. So our sequence in F is (1, 0, 0, 0, 0, 0,... ) (1, 1/2, 0, 0, 0, 0,... ) (1, 1/2, 1/3, 0, 0, 0,... ) (1, 1/2, 1/3, 1/4, 0, 0,... ) (1, 1/2, 1/3, 1/4, 1/5, 0,... ) and so on. Let s show that this sequence of sequences converges to the sequence (1/n) n N. We compute the distance between (a k n) n=1 for any k N: (a k n) n=1 (1/n) n=1 = (1, 1/2,..., 1/k, 0,... ) (1, 1/2,..., 1/n,... ) = (0, 0,..., 1/(k + 1), 1/(k + 2),... ) 1 = sup n k+1 k + 1 = 1 k + 1. 1 Now, let ε > 0. Let K N be large enough so that < ε. Then by our computation above, K+1 for all k K, (a k n) n=1 (1/n) n=1 = 1 1 < ε. Thus, lim k+1 K+1 k (a k n) n=1 = (1/n) n=1. But (1/n) n=1 / F. So this shows that there is a limit point of F which is not in F. Thus, F is not closed in l (N). 9