Chapter 4 1

a) Prove directly that the function 1x2 is continuous on (0,∞).

|f(x) – f(x0)| = | 1x2 - 1

x02 | ≤ |x – x0|

|x| + |x0|x2x0

2 .

Then let δ = min(x02 ,εx0

3

10 ).

If |x – x0| < δ then | 1x2 - 1

x02 | < ε.

b) A constant function f: A → m is a function such that f(x) = f(y) for all x,y ∈ A. Show that f is continuous. If ε > 0, let δ = ε. If |x – y| < δ then |f(x) – f(y)| = |f(x) – f(x)| < ε

c) Is the function f(y) = 1y4+y2+1 continuous. Justify your answer.

Yes, the function is continuous.

|f(y) – f(y0)| = | 1y4 + y2 + 1 - 1

y04 + y0

2 + 1 | = |(y04 + y0

2 + 1) - (y4 + y2 + 1) (y0

4 + y02 + 1)(y4 + y2 + 1) | =

|y – y0||y0

3 + y03y + y0(y2 + 1) + y(y2 + 1)

(y04 + y0

2 + 1)(y4 + y2 + 1) |

Let δ = min(y02 , εy0

5

240 )

Then if |x – x0| < δ then | 1y4 + y2 + 1 - 1

y04 + y0

2 + 1 | < ε.

3 a) If f: → is continuous and K ⊂ is connected, is f-1(K) necessarily connected? No. Let f(x) = sin(x) and K = {1}. b) Show that if f: n → m is continuous on all of n and B ⊂ n is bounded, then f(B) is bounded. If f is continuous on all of n, then f is continuous on the closure of B which is compact. Then f(cl(B)) is compact and is therefore bounded and since f(B) ⊂ f(cl(B)), f(B) is also bounded.

7 Consider a compact set B ⊂ n and let f: B → m be continuous and one-to-one. Then prove that f-1: f(B) → B is continuous. Show by example that this may fail if B is connected but not compact. Consider the closed subset C ⊂ B ⊂ n. Since B is closed and since C is a closed subset of B, the set C is also bounded and therefore compact. Since f is continuous, f(c) is compact and therefore closed. Then, since f is one-to-one, it maps B bijectively onto its image f(B). It may fail for f:[0,2π) → 2 defined by f(x) = (sin(x), cos(x)) 9 Prove the following “gluing lemma”: Let f:[a,b) → m and g:[b,c] → m be continuous. Define h:[a,c] → m by h = f on [a,b) and h = g on [b,c]. If f(b) = g(b), then h is continuous. Generalize this result to sets A,B in a metric space. Extend f by letting f(b) = g(b). Then limx→b

- h(x) = limx→b- f(x) = f(b) = g(b) = limx→b

h(x). Hence, h is continuous. In a metric space, let A,B ⊂ M and let C,D ⊂ A. The define mappings f: C → B and g: D → B so that C ∩ D ∫ ∅.

Then f ∪ g: C ∪ D → B is defined by f ∪ g(t) =

∈∈

D t if g(t)C t if )(tf

.

11 a) For f:(a,b) → , show that f is continuous, then its graph G is path-connected. Argue intuitively that if the graph of f is path-connected, then f is continuous. If any two points (c,f(c)) and (d,f(d)) are on the graph then there exists a function ϕ(t) = (t,f(t)) which is a continuous path from the point (c,f(c)) to the point (d,f(d)). Hence, the graph is path-connected. Similarly, of the graph of f is path connected then f is continuous since there exists a continuous function that maps every point on the graph to a corresponding point in f. b) For f: A → m, A ⊂ n, show that for n ≥ 2, connectedness of the graph does not imply continuity.

Let f(x) = (x, sin1x ). Then f is connected but not continuous.

c) Discuss b) above for m = n = 1. For the case of m = n = 1 the connectedness of the graph does imply continuity. Since if a set in is connected, then it is an interval which is continuous. To show this, assume that A ⊂ is connected but not an interval. Then ∃x,y,z ∈ with x < y < z such that x, z ∈ A and y – A. Then take U = (-∞,y) and V = (y,∞) which are two sets that separate A. This would imply that A is disconnected which is a contradiction. Hence, if a set in is connected, it is a continuous interval. 19 Find a continuous map f:n → n and a closed set A ⊂ n such that f(A) is not closed. In fact, do this when f:2 → is the projection on the x axis. A = {(x,y) ∈ 2 | xy = 1} f(A) = { x ∈ | x ∫ 0 } A is closed but f(a) is not closed.