k1, k2, k3, ... , then mix ((kn), N ) denotes the number of same-color blocks into which the set of all positive integers is divided. Since only N integers are colored red, this number is at most 2N + 1; in particular, it is finite.
If you think through the example you constructed for part (b), you should find that it implicitly used the fact that for the sequence of integers occurring there, mix ((kn), N ) assumed arbitrarily large values, while the positive result of (c) was based on the fact that mix (( jn), N ) never goes higher than 4.
In fact, prove that the following conditions on a rearrangement (kn) of the positive integers are equivalent:
(i) mix ((kn), N ) is bounded as a function of N.
(ii) For every convergent series Σ an of real numbers, the series Σ ak
n also converges.
Show moreover that if these conditions hold, then for every convergent series Σ an of real numbers, Σ ak
n = Σ an.
3.14:4. Rearrangements and the root test. (d : 3)
Find a series which converges by the root test, and a rearrangement of this series for which the root test gives no information.
However, note a reason why, in a situation of this sort, the rearranged series must still converge.
3.14:5. The subsequential limit set of a rearranged convergent series. (d : 1)
Let Σan, α and β be as in the hypothesis of Theorem 3.54, and let Σa′n and (s′n) be as in the conclusion thereof. Assuming the result of 3.2:6(a) (even if you haven’t proven it), deduce that the subsequential limit set of (s′n) is the interval [α,β].
3.14:6. Rearrangements with alternating signs. (d : 5)
Show that if (an) is a non-absolutely convergent series, and α is a real number, then there exists a rearrangement (ak
n) of (an) such that for all even n, ak
n≥0, for all odd n, ak
n≤0, and Σak
n=α. (One could even get the partial sums to have lim sup any α and lim inf any β such that –
∞
≤α≤β≤+∞
, as in Theorem 3.54. But proving that would just be more work, without a really different idea.)point of E. Show that if for some r > 0, the set of real numbers f (E ∩ Nr( p)) is bounded, and if limx→p g(x) = 0, then limx→p f (x) g(x) = 0. (In words, ‘‘If as x→p, one function remains bounded and the other approaches 0, then their product approaches 0.’’)
4.1:2. Limit points of functions that don’t necessarily approach a limit. (d : 2)
In Chapter 3, we saw that even if a sequence of points of a metric space did not approach a limit, we could still look at its ‘‘subsequential limit points’’. We can associate a similar set of points to a function between metric spaces:
(a) If f : E → Y is a function from a subset of a metric space X to another metric space, and p is a limit point of E in X, show that the following three subsets of Y are the same:
(i) {y∈Y! (∀ε> 0) (∃x∈E ) 0 < d(x, p) <ε and d( f (x), y) <ε} .
(ii) The set of all points limn→∞ f (xn) such that (xn) is a sequence converging to p in E – { p}
and ( f (xn)) converges in Y.
(iii) The union, over all sequences (xn) converging to p in E – { p} , of the set of subsequential limit points of ( f (xn)) in Y.
(b) Show that the set described in three equivalent ways in (a) is closed.
4.1:3. Analog of the Cauchy criterion for functions between metric spaces. (d : 2, 1)
(a) Suppose f : E → Y is a function from a subset E of a metric space X to a complete metric space Y, and let p∈X be a limit-point of E. Formulate and prove a necessary and sufficient condition for limx→p f (x) to exist, analogous to the condition ‘‘(sn) is Cauchy’’ for the limit of a sequence in Y to exist.
(Since ‘‘complete metric space’’ is defined in terms of the convergence of Cauchy sequences, your proof will have to relate the behavior of functions to the behavior of sequences. But the actual formulation of your criterion should be as an ‘‘ε-δ’’ condition, not one about sequences.)
(b) Show that if the assumption that Y is complete is deleted from (a), the condition you have obtained is still necessary, but no longer sufficient, for limx→p f (x) to exist.
4.2. CONTINUOUS FUNCTIONS. (pp.85-89)
Relevant exercises in Rudin:
4:R1. Is this the same as continuity? (d : 1)
4:R2. Continuous maps and closures of subsets. (d : 1) 4:R3. The zero-set of a continuous function is closed. (d : 1)
4:R4. Continuous maps agreeing on a dense subset are equal. (d : 2)
4:R5. Extending a continuous real-valued map from a closed subset to all of R . (d : 3)
In the last sentence of this exercise, Rudin remarks that the result remains true ‘‘if R1 is replaced by any metric space’’. He means R1 as the space containing the closed set E; not R1 as the codomain space of our functions. The result is false if the codomain space is taken, for instance, to be {0,1}; you should not find it hard to get examples showing this.
4:R7. Discontinuous functions on R2 that are continuous on all lines. (d : 2) If we call the exercise as Rudin gives it part ‘‘(a)’’, we can add
(b) Find continuous functions a, b : R→ R with a(0) = b(0) = 0 such that the restriction of the function f of the exercise to the curve {(x, a(x)) ! x∈R } is discontinuous at (0, 0), and the restriction of the function g to the curve {(x, b(x))! x∈R } is unbounded as one approaches (0, 0).
4:R22. Continuous functions that glide between two closed sets. (d : 1)
The function ρ referred to in this exercise is defined in 4:R20, but the main result of 4:R20 (which belongs under the next section) is not needed to do this one. (However, having done that exercise can ...
Answers to True/False question 4.1:0. (a)F. (b)T. (c) F. (d)T. (e)F.
shorten the work of this one.)
An explanation of the parenthetical last sentence of this exercise: The concept of ‘‘normality’’ which Rudin refers to is defined in the general context of topological spaces; some topological spaces are normal and some are not. Rudin translates the result of the exercise to say is that all metric spaces are normal as topological spaces.
4:R23. Convex functions on an interval. (d : 2)
4:R24. A test for convexity, using values at midpoints. (d : 3. > 4:R23) Exercises not in Rudin:
4.2:0. Say whether each of the following statements is true or false.
(a) The function f : Q→ R defined by f (x) = 0 if x2< 2, f (x) = 1 if x2> 2, is continuous.
(b) If p is an isolated point of a metric space X, then every function f from X to a metric space Y is continuous at p.
(c) If f : X → Y and g : Y → Z are functions between metric spaces, and g°f is discontinuous at a point p∈X, then either f is discontinuous at p, or g is discontinuous at f (p).
(d) Under the assumptions of the preceding part, f must be discontinuous at p and g must be discontinuous at f (p).
(e) Suppose a map f : X→ Y of metric spaces is one-to-one and onto, so that there exists an inverse map f–1: Y→ X. Then if f is continuous, so is f–1.
(f ) If X is any metric space, then the identity map idX : X→ X , defined by idX(x) = x for all x∈X, is continuous.
(g) If f : X→ Y is a continuous map of metric spaces and E is a bounded subset of X , then f (E ) is a bounded subset of Y.
(h) A mapping f : X → Y of metric spaces is continuous if and only if for every x∈X and every neighborhood N of f (x) in Y, the subset f–1(N ) contains a neighborhood of x in X.
(i) If f : R2 → R is continuous, then for each a∈R, the functions g, h : R → R defined by g(y) = f (a, y) and h(x) = f (x, a) are continuous.
4.2:1. Limits of sequences characterized in terms of continuity. (d : 1)
Let ( pn) be a sequence of points in a metric space X, and p a point of X. Let K be the set {0} ∪ {1 ⁄ n! n = 1, 2, 3, ... } ⊆ R, and let f : K→ X be the function defined by f (0) = p, f (1 ⁄ n) = pn. Show that limn→∞ pn = p in X if and only if f : K→ X is continuous.
4.2:2. Continuity characterized in terms of limits of sequences. (d : 2)
Let f : X→ Y be a map between metric spaces. Show that the following conditions are equivalent:
(i) f is continuous.
(ii) For every sequence ( pn) which converges in X, one has limn→∞ f ( pn) = f (limn→∞ pn). (Note that the right-hand side of the above equation is defined by assumption; the equation thus means that the left-hand side is defined and equal to the right-hand side.)
(iii) For every sequence ( pn) which converges in X, the sequence ( f ( pn)) converges in Y.
(Hint for proving (iii)⇒(ii): Given a sequence as in the hypothesis of (ii), with limit q, apply (iii) to a sequence formed by alternately using the terms of the given sequence, and the element q.)
4.2:3. Connectedness characterized in terms of continuity. (d : 2, 2, 3. > 4:R22)
(If you don’t do this exercise after reading this section, you might do it after section 4.4.) Let X be a metric space.
(a) Show that if A and B are subsets of X, then A and B are separated (Definition 2.45) if and only if there exists a continuous function f : A∪B→ R such that f (a) = 0 for all a∈A and f (b) = 1 for all b∈B.
(b) Deduce that X is connected if and only if every continuous function X→ {0, 1} is constant.
(c) Show likewise that X is connected if and only if every continuous function X→ Z is constant.
4.2:4. The archimedean property for powers of a continuous increasing function. (d : 4 or 2, 3)
(a) Suppose f : R → R is a continuous function such that f (x) > x for all x. For any positive integer n, let fn denote the n-fold composite f°f°...°f. Show that for all x, y∈R, there exists a positive integer n such that fn(x) > y.
(Part (a) above has difficulty d : 4 if you have not done Exercise 1.4:7, d : 2 if you have. Suggestion on how to use that exercise: For each real number x, consider the statement P (t) about a nonnegative real number t saying that for some positive integer n, fn(x) > x + t.)
(b) Does the conclusion of part (a) remain true if the assumption that f is continuous is removed?
4.2:5. Fixed points for continuous increasing functions. (d : 2)
Let E be a closed bounded subset of R, and f : E → E a continuous strictly monotone increasing function, i.e., a continuous function such that for all x, y∈E, x < y ⇒ f (x) < f (y). Take any x0∈E, and define x1, x2, x3, ...∈E by the rule xn+1 = f (xn).
(a) Show that if x1 > x0, then xn+1 > xn for all n, and sketch why the corresponding implications hold with ‘‘>’’ replaced by ‘‘=’’ and by ‘‘<’’.
(b) Show that limn→∞ xn exists, and is a fixed point of f ; i.e., that denoting this limit by x we have f (x) = x.
(c) Show that if x1 > x0, then the point x found in (b) is the least fixed point of f which is > x0. State the corresponding characterizations of x in the cases x1 = x0 and x1 < x0. (You do not have to write out proofs for these cases.)
(Remarks: The same result can be proved if E is the extended real line. Exercise 4.2:4 is related to that fact, though it is not a special case since the function there is not assumed monotone. Another related exercise is 1.2:4.)
4.2:6. The descendants of an amoeba. (d : 2. > 4.2:5)
Let t∈[0, 1]. Suppose we have a population of amoebas such that every hour on the hour, each amoeba in the population splits in two, and such that over the course of each hour, each amoeba has probability t of surviving, and 1 – t of dying, with the survival of different amoebas in the population independent of one another. Suppose we start with a single living amoeba at hour 0 (the result of a division that has just taken place, so that it will divide again at hour 1 if it survives till then), and for n = 0, 1, 2, ... let pn denote the probability that at least one of its descendants will be alive at hour n.
(We count it as one of its own descendants; thus, p0 = 1.) Since this number depends on the constant t as well as n, let us write it, more precisely, as pn(t).
(a) Find a formula for computing pn+1(t) from pn(t).
(Remark: The hard way to approach this question is to look ahead to hour n and consider the number of descendants alive at that time, and whether any will survive during the next hour. The easy way is to note that the original amoeba has probability t of surviving to hour 1, and that if it does so, it will divide and each of the resulting amoebas will have, independently, probability pn(t) of having at least one living descendant n hours later. Recall also the principle: If two independent events have probabilities x and y of occurring, then the probability that both will occur is x y. To determine the probability that one or both will occur, note that this is the probability that they will not both fail to occur. The probability that they will both fail to occur is, by the preceding principle, (1 – x) (1 – y), so the probability that they will not is 1 – (1 – x) (1 – y).)
(b) Does limn→∞ pn(t) exist? If so, determine its value. (Suggestion: Use 4.2:5.)
Tangential remarks: The assumption of synchronized reproduction is, of course, an absurd ...
Answers to True/False question 4.2:0. (a)T. (b)T. (c) T. (d)F. (e)F. (f )T. (g) F. (h)T. (i)T.
simplification; and one can also see that conditions can’t remain constant when n →
∞
, since anexpanding population will run out of food and space. Nevertheless, the result you will obtain above is probably a reasonable first approximation to the probability that a simple organism entering a new habitat will succeed in proliferating rather than dying out.
Actually, I didn’t come upon this question by thinking about amoebas. Rather, I was considering the algebraic situation where one has a binary operation *, not necessarily associative, on a set X, and one wants to study expressions in which * connects an arbitrary number of elements, for instance, (a * b), ((a * (b * c)) * (d * e)), etc.. To study ‘‘statistical’’ properties of such expressions, I wanted to assume them generated in a random way. In the expressions in question, the elements a , b, c etc. were to appear in fixed order, so when generating the expressions, one could use a place-holder 1 instead of these letters, and work with symbols like ((1 * (1 *1)) * (1 *1)). I decided that the mathematically simplest sort of random generation would be to start with a single symbol 1, and execute a series of steps, at each of which every 1 would either be replaced by (1 * 1), with some probability t, or else be declared
‘‘final’’, and undergo no more changes after that. When all squares have become ‘‘final’’, one has a randomly generated expression. But can one expect all squares to eventually become ‘‘final’’? I worked out the probability that this would happen as a function of t – and found that the resulting computation was an example of something it will be useful to know when we reach Chapter 7 of Rudin. So I reformulated the problem as a more concrete question about amoebas, and have introduced it here in preparation for that later chapter.
Some years later, I learned of still another way of looking at this computation. Consider a system of passageways, beginning at an initial point, where each passageway ends by branching into two further passageways. Suppose each passageway has probability t of being open, and 1 – t of being blocked.
Then the above problem concerns the probability that this system has an infinite unblocked route from the initial point. This is a simple case of the subject of percolation theory, which takes its name from the case where the passageways are pores in a solid. Closer to real percolation problems, and more difficult to analyze, are cases where the passageways form infinite ‘‘checkerboard arrays’’ in 2 or more dimensions.
4.2:7. When can one compose limit-statements? (d : 1; 3)
The principle ‘‘If as x→ p, f (x) → q, and if as y→ q, g(y)→ r, then as x→ p, g( f (x))→ r’’
may appear ‘‘obvious’’. But the following example shows that it is false.
(a) Let f : R→ R be defined by f (x) = x sin x–1 if x ≠ 0, f (0) = 0, and let g : R→ R be defined by g(x) = 0 if x ≠ 0, g(0) = 1. Show that limx→0 f (x) = 0 and limx→0 g(x) = 0, but that it is not true that limx→0 g( f ((x))) = 0.
Why is the apparently obvious principle that we started with false? Because ‘‘→ q’’ has slightly different meanings in the two statements ‘‘as x → p, f (x) → q’’ and ‘‘as y → q, g(y) → r’’! In the first it means that f (x) takes on values that become arbitrarily close to q; but in the second, it means that y takes on values getting arbitrarily close to q, other than the value q itself. In the example above, we see that as x→ p the function y = f (x) has the first of these properties, but not the second.
Obviously, it would be good to know conditions under which we can correctly ‘‘compose’’ limit statements. Necessary and sufficient conditions are given in
(b) Suppose f is a function from a subset E of a metric space X to a metric space Y, and g is a function from a subset F of Y which contains f (E ), to a metric space Z . Let p be a limit point of E in X, q a limit point of F in Y, and r a point of Z , such that limx→p f (x) = q and limy→q g(x) = r. Show that the following conditions are then equivalent:
(i) limx→p g( f ((x))) = r.
(ii) Either q∈F and g is continuous at q, or there exists ε > 0 such that q∈f (Nε(x)∩E ) (i.e., the function f does not take on the value q at points arbitrarily close to p).
4.2:8. Two different meanings of ‘‘neighborhood’’ have similar properties. (d : 3)
What Rudin calls a ‘‘neighborhood’’ of a point p of a metric space is nowadays more often called an
‘‘open ball’’ about the point. Topologists instead generally define a neighborhood of p to mean any set E such that p is an interior point of E . In this exercise, let us call this concept a ‘‘topologist’s neighborhood’’, and let the unmodified word have Rudin’s meaning. We shall see that the topologist’s concept is as convenient as Rudin’s for some typical uses of the concept.
(a) Show that if p is a point of a metric space X, then a subset E ⊆ X is a ‘‘topologist’s neighborhood’’ of p if and only if it contains a neighborhood of p in Rudin’s sense.
(b) If f : X → Y is a map between metric spaces and p a point of X, show that the following conditions are equivalent:
(i) The map f is continuous at p.
(ii) The inverse image under f of every neighborhood of f ( p) contains a neighborhood of p.
(iii) The inverse image under f of every ‘‘topologist’s neighborhood’’ of f ( p) is a ‘‘topologist’s neighborhood’’ of p .
(c) State similarly how the concepts of a limit-point of a set and the limit of a sequence can be formulated in terms of the topologist’s concept of neighborhood. (No proofs asked for.)
4.2:9. At-most-countably-many-to-one continuous maps, and perfect sets. (d : 3. > 2.4:2)
Let us call a function f : X → Y ‘‘at most countably many to one’’ if for every y∈Y, the set f–1(y) ⊆ X is at most countable. Suppose f is a continuous, at most countably many to one map of metric spaces. Show that for every compact perfect subset E ⊆ X, f (E )....
is a perfect subset of Y.
(This exercise might be given with one of the next two sections, since the above result nicely parallels the results of those sections on how continuous maps behave on compact sets and connected sets.)
4.2:10. k-dimensional space-filling curves. (d : 2, 1, 3)
In 7:R14 (p.168), Rudin will show how to construct a continuous map Φ from [0,1] onto [0,1]2 = { (x, y) ! x, y∈[0,1] } . This is called a ‘‘space-filling curve’’ because a continuous function from [0,1]
into the plane can be thought of as a parametrized curve, and this curve fills up all the space inside the square [0,1]2. Assuming this result, we shall show here, using only the methods of this section, that there must also exist curves filling higher-dimensional boxes, and even something that can be looked at as an infinite-dimensional space-filling curve.
Given a continuous map Φ from [0,1] onto [0,1]2, let us write Φ(t) = (a(t), b(t)) for each t∈[0,1]. (Rudin writes (x(t), y(t)), but I will be using x and y for real numbers.) Thus, a and b are continuous functions [0,1] → [0,1] such that for every point (x, y)∈[0,1]2 there exists t∈[0,1] such that (a(t), b(t)) = (x, y).
(a) Assuming such functions a and b given, deduce that for every point (x, y, z)∈[0,1]3 there exists t∈[0,1] such that (a(t), a(b(t)), b(b(t))) = (x, y, z). Conclude that the function taking t to (a(t), a(b(t)), b(b(t))) is a continuous function from [0,1] onto [0,1]3 (a 3-dimensional space-filling curve).
(b) Generalize the argument of part (a) to show that if we write bi for the i-fold composite function b°b°...°b (with i factors), then for every n ≥ 2 the function [0,1] → [0,1]n taking t to (a(t), a(b(t)), ... , a(bn–2(t)), bn–1(t)) is continuous and onto (an ‘‘n-dimensional space-filling curve’’).
(Remark: In Rudin, for f a real-valued function, fn usually means the function defined by fn(x) = f (x)n. Hence that notation holds in this exercise packet unless the contrary is stated. In this exercise, I have stated the contrary!)
(c) Deduce that for every sequence (xn) of elements of [0,1], there exists t∈[0,1] such that a(bn–1(t)) = xn for all n≥ 1.
(Recall that a sequence (xn) involves infinitely many terms, x1, x2, ... . We understand b0 to denote the identity map of [0,1], given by b0(t) = t for all t. Thus, the n = 1 and n = 2 cases of the above equation are a(t) = x1 and a(b(t)) = x2.)
Hint: Deduce from part (b) that for every N, the set of t for which the first N of the above
equations hold is nonempty. Then use compactness.
Exercise 4.2:14 below will show that the above result can be looked at describing an ‘‘infinite-dimensional space-filling curve’’.
4.2:11. Plane-filling curves. (d : 2)
As in 4.2:10 above (see first paragraph thereof ), we will assume here the result of 7:R14. Deduce that there also exists a continuous map from R onto R2. (Suggestion: Let your curve send [0,1] onto [0,1]2 by the map given by Rudin, and send various other intervals [n , n+1] onto larger and larger squares.)
Remarks: As in part (b) of the preceding exercise, one can similarly obtain surjective continuous maps R → Rk for all positive integers k. Since this is straightforward, I don’t make it an exercise. On the other hand, part (c) of the preceding exercise uses the compactness of [0,1], but R is not compact. This leads to an obvious question, which is answered in the next exercise.
4.2:12. No infinite-dimensional analog of preceding exercise. (d : 2)
Show that there does not exist a sequence of continuous functions f1, f2, ... : R→ R with the property that for every sequence (xn) in R there exists t∈R such that fn(t) = xn for all n ≥ 1. (Hint: Given a sequence of continuous functions ( fn), show that if you choose x1 sufficiently large, no t∈[ –1, 1]
can satisfy f1(t) = x1; that if you choose x2 sufficiently large, no t∈[ – 2, 2] can satisfy f2(t) = x2, etc..)
4.2:13. The results of the preceding exercises for R imply the same results for [0,1). (d : 2.
> 4.2:11, 4.2:12)
We start with a fact not based on the preceding exercises:
(a) Show by example that there exist continuous maps from R onto [0,1) and continuous maps from [0,1) onto R.
(b) Deduce from part (a) and the result of 4.2:11 that there exists a continuous function from [0,1) onto [0,1)2.
(c) Deduce from part (a) and the result of 4.2:12 that there does not exist any infinite sequence of continuous functions g1, g2, ... : [0,1) → [0,1) with the property that for every sequence (xi) in [0,1) there exists t∈[0,1) such that gn(t) = xn for all n ≥ 1.
4.2:14. A sequence of real-valued continuous functions is equivalent to one function into the space of sequences. (d : 3)
Rudin shows in Theorem 4.10 (p.87) that a family of k maps, f1, ... , fk, from a metric space X into R are all continuous if and only if single map f : X → Rk given by f (x) = ( f1(x), ... , fk(x)) is continuous. We shall describe here a similar result for infinite sequences of maps ( fi).
For simplicity, we will begin with the case of [0,1]-valued functions.
Let [0,1]J denote the set of all sequences (xi) of elements of [0,1]. Given (xi), (yi)∈[0,1]J, define
d&((xi), (yi)') = Σi | xi– yi| ⁄ 2i. (a) Show that this function d is a metric on [0,1]J.
(b) Show that if X is any metric space and ( fi) is a sequence of functions X→ [0,1], then each fi is continuous if and only if the map f : X → [0,1]J defined by f (x) = ( f1(x), f2(x), ... ) is continuous, relative to the above metric on [0,1]J.
(c) Why can the above formula not be used to define a metric on RJ? Show, however, that by replacing
| xi– yi| ⁄ 2i with min ( | xi– yi| , 2– i), one can get a metric on RJ, and a result analogous to (b) above.
4.2:15. The set where two continuous functions agree is closed. (d : 2,1,1,1)
(a) Let X and Y be metric spaces, and let f : X→ Y and g : X → Y be continuous functions. Show that the set E = {x∈X | f (x) = g(x)} is closed in X. (This can be done either by verifying directly that E satisfies the definition of a closed set, or by showing that the complement of E is open, or if one has
done 4:R3, by showing that the function taking x∈X to d ( f (x), g(x))∈R is continuous, and noting that E is the zero-set of this function.)
(b) Show that (a) implies the result of 4:R3.
(c) Deduce the main result of 4:R4 – the final non-parenthetical statement, beginning ‘‘If g( p) = f ( p) ...’’
– from (a).
(d) Assuming the results of 4:R20 (whether or not you have done that exercise), prove the following converse to (a) above: For every closed subset E of a metric space X, there exist continuous functions f and g from X to another metric space Y such that {x∈X | f (x) = g(x)} = E.
4.2:16. Functions which approach a limit everywhere. (d : 3)
Suppose f : X → Y is a function between metric spaces (not assumed continuous) such that for every p∈X, limx→p f (x) exists. Define g : X → Y by g( p) = limx→p f (x) for all p∈X. Show that g is continuous.
(There are several ways this result can be strengthened. The above assumption that the limit of f is defined at every p∈X necessitates that X have no isolated points. One can weaken this to say that the limit of f exists at every non-isolated point of X, defining g as above at those points, while making it agree with f at the isolated points; one can then establish the same conclusion as above. One can also merely assume f to be defined on a dense subset E of X; this still allows it to have a limit at every non-isolated point of X, so that one can define g on all of X as above, and again get the same conclusion.)
4.3. CONTINUITY AND COMPACTNESS (and uniform continuity). (pp.89-93) Relevant exercises in Rudin:
4:R6. A function on a compact space is continuous if and only if its graph is compact. (d : 2) 4:R8. A uniformly continuous function on a bounded subset of Rn is bounded. (d : 3)
4:R9. Uniform continuity in terms of diameters of sets. (d : 1)
In this exercise, for Rudin’s phrase ‘‘the requirement in the definition of uniform continuity’’ simply read ‘‘the condition that f : E→ Y be uniformly continuous’’.
4:R10. Alternative proof of Theorem 4.19. (d : 2)
4:R11 and 4:R13. Extension of real-valued functions from dense subsets. (d : 1, 3 and 3)
These two exercises ask you to prove the same result, but by different methods. The statement of this result is given in the later exercise, to which the second sentence of the earlier exercise refers you. The first sentence of the earlier exercise is an easy but instructive result on Cauchy sequences.
4:R12. A uniformly continuous function of a uniformly continuous function is uniformly continuous. (d : 1) 4:R20. The distance from a set is a uniformly continuous function. (d : 2)
4:R21. The distance from a compact set to a disjoint closed set is bounded below. (d : 3. > 4:R20)
In the last sentence, ‘‘Show that the conclusion may fail’’ means, of course, give an example where the conclusion fails.
4:R25. The ‘‘sum’’ of a compact set and a closed set is closed. (d : 3. > 4:R21)
4:R26. Properties of a map which factors through a continuous one-to-one map on a compact set. (d : 3) Exercises not in Rudin:
4.3:0. Say whether each of the following statements is true or false.
(a) A function f from a subset E of a metric space X to Rk is bounded if and only if f (E ) is a bounded subset of Rk.
(b) If f is a continuous function from R to a metric space X, then for every positive integer n, f ( [– n , n] ) is a compact subset of X.
(c) Let f be a continuous real-valued function on a metric space X. If there exists a point p∈X such that f ( p) = supx∈X f (x), then X is compact.
(d) The function f : ( 0 , 1) → R defined by f (x) = 1 ⁄ x is uniformly continuous.
(e) The function f : (1, +
∞
)→ R defined by f (x) = 1 ⁄ x is uniformly continuous.(f ) The function f : (0, +
∞
)→ R defined by f (x) = 1 ⁄ x is uniformly continuous.(g) If f : R → R is uniformly continuous, then f assumes a maximum value, i.e., there exists a real number a such that f (a) = supx∈R f (x).
(h) If X is a noncompact metric space, then there exists a real-valued function f on X which is uniformly continuous but not continuous.
(i) A subset E ⊆ R is compact if and only if every continuous real-valued function on E is bounded.
4.3:1. On what metric spaces is every function continuous, respectively uniformly continuous? (d : 3) (a) What metric spaces X have the property that every function from X to any metric space Y is continuous?
(b) What metric spaces X have the property that every function from X to any metric space Y is uniformly continuous?
4.3:2. Continuous periodic functions on R are uniformly continuous. (d : 2)
(a) Let c be a positive real number, and f : R→ X a continuous function from R to a metric space X which has period c, i.e., such that f (r + c) = f (r) for all r∈R. Show that f is uniformly continuous.
So, for instance, the function sin x is uniformly continuous. However
(b) Show that the function sin (x2) is not uniformly continuous. (Rudin has not developed the sine function as of this point. However, all you need to know to do (b) is that sin x is a nonconstant periodic continuous real-valued function on R.)
4.3:3. A continuous map on Rk takes bounded sets to bounded sets. (d : 2)
Let X be a metric space and f : Rk→ X a continuous function. Show that if E is a bounded subset of Rk, then f (E ) is a bounded subset of X.
4.3:4. A shrinking map on a compact space has a fixed point. (d : 4, 2, 2, 3, 4, 4)
(a) Suppose K is a compact metric space, and f : K→ K is a map with the property that for every pair of distinct points x, y∈K one has d( f (x), f (y)) < d(x, y). Show that there exists a unique point p∈K such that f (p) = p.
(The remaining parts, though of interest, can be omitted without detracting from the interest or challenge of the above problem.)
(b) Show by example that if the ‘‘<’’ is weakened to ‘‘≤’’ in part (a), the map need not have any fixed point.
(c) Show by examples that the result of (a) is also false if we put any of the noncompact metric spaces [0,1), [0, +
∞
) or R in place of the compact space K.(d) Deduce from (a) that for K a compact metric space, there cannot exist a continuous function f : K→ K such that for all distinct points x, y∈K one has d( f (x), f (y)) > d(x, y).
(e) Show that in the situation of (a), for every x∈K one has limn→∞ fn(x) = p.
(f ) Suppose K is a compact subset of a metric space X, and g : K→ X is a map such that g(K ) ⊇ K, and such for every pair of distinct points x, y∈K one has d(g(x), g(y)) > d(x, y). Show that there exists a unique point p∈K such that g( p) = p.
(For some further related results, see 4.3:8.)
4.3:5. Uniform ‘‘either/or continuity’’ of two or more functions. (d : 3)
(a) Let X, Y and Z be metric spaces, with X compact, and let f : X → Y, g : X → Z be functions.
We shall not assume f and g are continuous, but let us assume that for each p∈X, at least one of f
and g is continuous at p. Show that for every ε > 0 there exists a δ > 0 such that whenever d( p, q) < δ in X, one has either d( f (p), f (q)) < ε or d(g (p), g(q)) < ε.
(b) Does a similar result hold for an infinite family of maps fi : X → Yi (i∈I ) ? More generally, if we have such a family of maps, and also an arbitrary family of positive real numbers εi (i∈I ), can we find a single δ such that whenever d( p, q) < δ in X, there exists some i∈I such that d( fi(p), fi(q)) < εi? 4.3:6. A weak condition that implies boundedness. (d : 3,1,2)
(a) Suppose E is a subset of a compact metric space X, and f is a function from E to a metric space Y which satisfies the following condition much weaker than uniform continuity: For some ε> 0 there exists δ> 0 such that all p, q∈X that satisfy dX( p, q) <δ also satisfy dY( f ( p), f (q)) <ε. Show that f must be bounded. (Note: f is not assumed continuous. Hint: Construct a certain open covering of E..
.) (b) Deduce 4:R8 from (a).
(c) Find a bounded (but, necessarily, non-compact) metric space X and a uniformly continuous function f from X to another metric space Y such that f is not bounded.
4.3:7. More on functions which approach a limit everywhere. (d : 4)
Suppose f : X → Y is a function between metric spaces (not assumed continuous) such that for every p∈X, limx→p f (x) exists. (This condition was considered in 4.2:16, but the present exercise does not depend on that one.) Let D ⊆ X be the set of points where f is discontinuous. Show that for every compact subset K⊆ X, the set D∩K is countable.
(Suggestion: For p∈X, let w( p) = limε→0diam( f (Nε( p))). Show that for every positive constant c, if { p∈X | w( p) > c} had a limit point q∈X, then limx→q f (x) could not exist, a contradiction.
Deduce that for each positive integer n there are only finitely many p∈K with w( p) > 1 ⁄ n.) 4.3:8. Weakly shrinking maps on compact spaces. (d : 5)
(a) Suppose K is a compact metric space, and f : K→ K is a map with the property that for all x, y∈K one has d( f (x), f (y)) ≤ d(x, y). (Cf. 4.3:4.) Show that f is surjective if and only if for all x, y∈K one has d( f (x), f (y)) = d(x, y).
(b) Deduce from (a) that if K is a compact metric space and g : K → K a continuous map with the property that for all x, y∈K one has d( f (x), f (y)) ≥ d(x, y), then for all x, y one has d( f (x), f (y)) = d(x, y).
4.4. CONTINUITY AND CONNECTEDNESS. (p.93)
Relevant exercises in Rudin:
4:R14. Every continuous map [0,1]→ [0,1] has a fixed point. (d : 3)
(A very brief hint will turn the difficulty-rating of the above exercise from d : 3 to d : 1.) 4:R19. When the intermediate-value property implies continuity. (d : 3)
Exercises not in Rudin:
4.4:0. Say whether each of the following statements is true or false.
(a) Every continuous map R→ Q is constant.
(b) Every continuous map Q→ R is constant.