Exercises for Section 2.3 1. Prove that if f : A −→ Rm and g : f (A) −→ Rp are both continuous, then g ◦ f : A −→ Rp is continuous. Proof. Let a ∈ A, and b = f (a). By continuity of f and g, we have lim f (x) = f (a)

and

x→a

lim g(y) = g(b).

y→b

Then, by the Main Limit Theorem, lim g ◦ f (x) = g(f (a)).

x→a

Thus g ◦ f is continuous at a. But a was chosen arbitrarily, so g ◦ f is continuous on A. 2. Prove that f : A −→ Rm is continuous at a if and only if each f i is. Proof. This follows immediately from Problem 1.4.5. 3. Prove that a linear transformation T : Rn −→ Rm is continuous. Hint: See Problem ??.??. Proof. Let a ∈ Rn . We claim lim T x = T a.

x→a

Let ε > 0. By Problem 1.1.8, there is a real number M > 0 such that kT hk ≤ Mkhk for all h ∈ Rn . Choose δ = ε/M. Then 0 < kx − ak < δ =⇒ kT x − T ak = ≤ < =

kT (x − a)k Mkx − ak Mδ ε

4. Prove that B n (a; r) = {x ∈ Rn : kx − ak < r} is open, by considering the function f : Rn −→ R with f (x) = kx − ak. See Problem ?? from Section ??. 1



Proof. We have shown (Problem ??.??) that lim kg(x)k = lim g(x) . x→a x→a Therefore, if G is continuous, then kgk is continuous. Thus the function f (x) = kx − ak is a continuous function. Moreover, B n (a; r) = f −1 (−1, r) is the pre-image of an open set. Thus, by continuity, the ball is open. 5. A function f : A −→ R is said to be a bounded function if there is R > 0 such that f (x) < R for all x ∈ A. If A ⊂ Rn is not closed, show that there is a continuous function f : A −→ R which is unbounded. Hint: If x ∈ Rn − A but x ∈ / int (Rn − A), let f (y) = 1/ky − xk. You still have to show that such an x exists, and you still have to show that f is unbounded! Proof. Since A is not closed, ∂A 6⊂ A. Therefore, there is some point x ∈ ∂A with x ∈ 6 A. Define f : A −→ R via f (y) = 1/ky − xk. Let R > 0, and consider the open set B(x, 1/R). Then, because x ∈ ∂A, this open set must intersect A. Therefore, there is some y ∈ A with y ∈ B(x, 1/R). Thus, ky − xk < 1/R, and f (y) = 1/ky − xk > R. Hence, f is unbounded. 6. The image of a compact set under a continuous function is compact, hence closed. Is the image of a closed set under a continuous function necessarily closed? Prove that it is, or give a counterexample. Solution. The claim is false. Proof. Define f : R −→ (−π/2, π/2) via f (x) = arctan(x). Then the domain, R, is closed, and f is continuous; but the image is not closed. 7. Is the image of an open set under a continuous function necessarily open? Prove that it is, or give a counterexample. Solution. This claim is false. Proof. Define f : (0, 1) −→ R via f (x) = 17. Then the domain is open, but the image is the closed set {17}. 2

8. Is the sphere S = {(x, y, z) ∈ R3 : x2 + y 2 + z 2 = 1} connected? Why or why not? Solution. In fact it is path-connected. Finding a path between two arbitrary points is straight forward, especially in spherical coordinates. 9. (Extreme Value Theorem). If A is compact, prove that every continuous function f : A −→ R takes on a maximum and a minimum value. Proof. If A is compact and f is continuous, then f (A) is compact in R. But this means that f (A) is the union of finitely many closed intervals [ai , bi ]. Let m = min{ai } and M = max{bi }. Then a and b are the maximum and minimum values of f , respectively. Moreover, since a, b ∈ f (A), then f attains those values for some points in A. 10. Let B be closed in Rn , and f : B −→ Rm . Prove that the following are equivalent: i. f is continuous on B. ii. f −1 (E) is closed in Rn for every closed set E ⊂ Rm . Proof. (⇒). Let f be continuous on B. Then the preimage f −1 (U) of any open set U is open in B (meaning there is an open set V such that f −1 (U) = V ∩ B). Let E ⊂ Rm be closed. Take U = Rm − E, which is open. Then f −1 (U) = {x ∈ B : f (x) ∈ U} = {x ∈ B : f (x) ∈ / E} −1 = = B − f (E) Therefore, f −1 (E) = B − f −1 (U) is the complement of an open set, so it is closed. (⇐). Similarly, Let U be open in Rm , and take E = Rm − U. Then E is closed, as is f −1 (E) = {x ∈ B : f (x) ∈ E} = {x ∈ B : f (x) ∈ / U} −1 = = B − f (U) 3

Therefore, f −1 (U) is open in B, because it is the complement of a closed set. 11. We will now try to finish the proof that a set E ⊂ R is connected if and only if it is an interval. Recall that R is connected. (a) Find a continuous bijective function f : R −→ (a, ∞). Hint: Start with a = 0. Solution. f (t) = et + a. (b) Find a continuous bijective function g : R −→ (−∞, b). Solution. g(t) = b − et . (c) Find a continuous bijective function h : R −→ (a, b). Solution. h(t) = a + 1 +

b−a π

tan−1 (t).

(d) Prove that (a, ∞), (−∞, b), and (a, b) are all connected. Hint: Why did we want f, g, and h? Proof. The three intervals of interest are all the images under continuous functions f, g, and h of the connected set R. (e) Prove that [a, ∞), (−∞, b], and [a, b] are all connected. Hint: See Problem ??.??. Proof. These sets, are the closures of connected sets. They are therefore connected. (f) Prove that [a, b) and (a, b] are connected. Proof. These sets lie between the connected set (a, b) and its closure [a, b], and are therefore connected by Problem ??.??. 12. Prove that the graph y = f (x) of a continuous function f : [a, b] −→ R is compact and connected in R2 . Proof. Define g : [a, b] −→ R2 via g(t) = (t, f (t)). Then g is a continuous function because all of its components are continuous (see Problem 2). Therefore, since [a, b] is compact and connected, then the image of g must also be compact and connected. Moreover, the image of g in R2 is exactly the graph of f . 4

13. (Intermediate Value Theorem) Suppose that A ⊂ Rn is connected, f : A −→ R is continuous, and there are a, b ∈ A with f (a) < 0 and f (b) > 0. Show that there exists c ∈ A with f (c) = 0. Proof. We know already that f (A) is connected. But since f (A) ⊂ R, then it must be an interval. Moreover, f (a) < 0 and f (b) > 0 implies that this interval contains 0. Thus 0 ∈ f (A). In other words, there is some c ∈ A with f (c) = 0. 14. Prove that if f : T −→ R is a continuous real-valued function on the unit circle T , then there is a point p ∈ T such that f (p) = f (−p), where −p is the point diametrically opposed to p. Proof. Define g : T −→ R via g(x) = f (x) − f (−x). So g(x) = 0 if and only if f (x) = f (−x). Choose any point x ∈ T . If g(x) = 0 we are finished. If not, then either g(x) > 0 or g(x) < 0. In either case, note that g(−x) = −g(x), so g(x) and g(−x) have opposite signs. By the intermediate value theorem, then there must be some point p ∈ T with g(p) = 0. 15. Prove that Rn − {0} is connected for n ≥ 2. How is this different from R − {0}? Proof. If n ≥ 2, then given two points, x, y ∈ Rn − {0}, we can always draw a path from x to y that misses 0. Therefore, Rn − {0} is pathconnected, hence connected. Such is not the case for R − {0}. You should be able to figure out which sets in Rn we can delete to get a disconnected set. 16. Find an example of a set A ⊂ R2 that is connected, but not pathconnected. Solution. The most familiar set to you that has this property is called the Topologist’s Sine Curve. Consider the graph of f (x) = sin(1/x) together with the point (0, 0) in R2 . (You should have seen this function quite a bit in Analysis I). Then this set, call it S, is connected, but there is no path from (0, 0) to the point (1, sin 1) (or any other point on the curve, for that matter). 5