Kenneth Kuttler

4.2. EXERCISES 101

x ∈ f−1 (c,d). Thus f−1 (c,d)∩ (a,b) must be open after all. Of course, if there is nox ∈ f−1 (c,d)∩ (a,b) , then f−1 (c,d)∩ (a,b) = /0 which is open.

⇐ Let x ∈ (a,b) and suppose f−1 (c,d)∩ (a,b) is always open. Why is f continuousat x? If not, there exists xn → x but f (xn) ↛ f (x) , the symbol ↛ meaning it doesn’tconverge. It follows there exists ε > 0 and a subsequence

{xnk

}such that f

(xnk

)/∈

( f (x)− ε, f (x)+ ε) ≡ Iε . But x ∈ f−1 (Iε)∩ (a,b) and this is open so eventually xnk ∈f−1 (Iε)∩ (a,b) and so f

(xnk

)∈ Iε after all. Thus xn → x ⇒ f (xn) → f (x) and so f is

continuous at x after all.

4.2 Exercises1. Let f (x) = 2x+ 7. Show f is continuous at every point x. Hint: You need to let

ε > 0 be given. In this case, you should try δ ≤ ε/2. Note that if one δ works in thedefinition, then so does any smaller δ .

2. Suppose D( f ) = [0,1]∪{9} and f (x) = x on [0,1] while f (9) = 5. Is f continuousat the point, 9? Use whichever definition of continuity you like.

3. Let f (x) = x2 +1. Show f is continuous at x = 3. Hint:

| f (x)− f (3)|=∣∣x2 +1− (9+1)

∣∣= |x+3| |x−3| .

Thus if |x−3| < 1, it follows from the triangle inequality, |x| < 1+ 3 = 4 and so| f (x)− f (3)| < 4 |x−3| . Now complete the argument by letting δ ≤ min(1,ε/4) .The symbol, min means to take the minimum of the two numbers in the parenthesis.

4. Let f (x) = 2x2 +1. Show f is continuous at x = 1.

5. Let f (x) = x2 + 2x. Show f is continuous at x = 2. Then show it is continuous atevery point.

6. Let f (x) = |2x+3|. Show f is continuous at every point. Hint: Review the twoversions of the triangle inequality for absolute values.

7. Let f (x) = 1x2+1 . Show f is continuous at every value of x.

8. If x ∈R, show there exists a sequence of rational numbers, {xn} such that xn → x anda sequence of irrational numbers, {x′n} such that x′n → x. Now consider the followingfunction.

f (x) =

{1 if x is rational

0 if x is irrational.

Show using the sequential version of continuity in Theorem 4.0.2 that f is discontin-uous at every point.

9. If x ∈R, show there exists a sequence of rational numbers, {xn} such that xn → x anda sequence of irrational numbers, {x′n} such that x′n → x. Now consider the followingfunction.

f (x) =

{x if x is rational

0 if x is irrational.

Show using the sequential version of continuity in Theorem 4.0.2 that f is continuousat 0 and nowhere else.

4.2. EXERCISES 101x € f-!'(c,d). Thus f-!(c,d)M (a,b) must be open after all. Of course, if there is nox € f-!(e,d)N (a,b), then f—! (c,d) M (a,b) = @ which is open.< Let x € (a,b) and suppose f~! (c,d) (a,b) is always open. Why is f continuousat x? If not, there exists x, — x but f(x,) + f(x), the symbol -» meaning it doesn’tconverge. It follows there exists € > 0 and a subsequence {x,,} such that f (%n,) ¢(f (x) —€, f(x) +e) =Ie. But x € f-' (Ie) (a,b) and this is open so eventually x, €f-' (le) A (a,b) and so f (x»,) € Ze after all. Thus x, + x => f (xn) > f (x) and so f iscontinuous at x afterall. JJ4.2 Exercises1. Let f(x) = 2x+7. Show f is continuous at every point x. Hint: You need to let€ > 0 be given. In this case, you should try 6 < €/2. Note that if one 6 works in thedefinition, then so does any smaller 6.2. Suppose D(f) = [0,1] U{9} and f (x) =x on [0,1] while f (9) =5. Is f continuousat the point, 9? Use whichever definition of continuity you like.3. Let f (x) =x +1. Show f is continuous at x = 3. Hint:If (x) — f (3)| = |x? +1-(9+1)| = [x +3] [x3].Thus if |x—3| < 1, it follows from the triangle inequality, |x| < 1-+3 = 4 and so|f (x) — f (3)| < 4|x—3]. Now complete the argument by letting 6 < min(1,¢/4).The symbol, min means to take the minimum of the two numbers in the parenthesis.4. Let f (x) =2x* +1. Show f is continuous at x = 1.5. Let f(x) =x?+2x. Show f is continuous at x = 2. Then show it is continuous atevery point.6. Let f(x) = |2x+3]|. Show f is continuous at every point. Hint: Review the twoversions of the triangle inequality for absolute values.7. Let f (x) = za: Show f is continuous at every value of x.8. Ifx € R, show there exists a sequence of rational numbers, {x, } such that x, — x anda sequence of irrational numbers, {x}, } such that x/, + x. Now consider the followingfunction.1 if x is rationalf(x) = ee0 if x is irrationalShow using the sequential version of continuity in Theorem 4.0.2 that f is discontin-uous at every point.9. If x € R, show there exists a sequence of rational numbers, {x, } such that x, + x anda sequence of irrational numbers, {x/, } such that x/, + x. Now consider the followingfunction.x if x is rationalf(x) = Pe0 if x is irrationalShow using the sequential version of continuity in Theorem 4.0.2 that f is continuousat O and nowhere else.