6.2. EXERCISES 103

Proof: Continuous⇒ 1. Let x ∈ f−1 (V ) so f (x) ∈ B( f (x) ,εx)⊆V for some εx > 0.Then by continuity, there is δ x > 0 such that f (B(x,δ x)∩D( f )) ⊆ B( f (x) ,εx) ⊆ V andso, letting U ≡ ∪x∈ f−1(V )B(x,δ x) , it follows that f−1 (V ) =U ∩D( f ) .

1. ⇒ Continuous. You could pick a particular open set B( f (x) ,ε) ≡ V and thenf−1 (V ) =U ∩D( f ) for open U and so if x∈ f−1 (V ) , then there is δ x such that B(x,δ x)⊆U and so if y ∈ B(x,δ x)∩D( f ) , then f (y) ∈ B( f (x) ,ε) which is the standard definitionof continuity at x.

1.⇒ 2. Let H be closed. Then HC is open and so f−1(HC)= U ∩D( f ) for U open.

But then f−1 (H) =UC ∩D( f ) where UC is closed. (D( f ) = f−1 (H)∪ f−1(HC))

2.⇒ 1. Let U be open. Then UC is closed and so f−1(UC)= C∩D( f ) for a closed

set C. But then CC ∩D( f ) = f−1 (U) where CC is open.This suggests the following definition.

Definition 6.1.3 Let S be a nonempty set. Then one can define relatively open andrelatively closed subsets of S as follows. A set O⊆ S is relatively open if O =U ∩S whereU is open. A set K ⊆ S is relatively closed if there is a closed set C such that K = S∩C.

In words, the above theorem says that a function is continuous at every point of itsdomain if and only if inverse images of open sets are relatively open if and only if inverseimages of closed sets are relatively closed.

6.2 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: Consider the follow-ing which comes from algebra. | 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| . Complete the argument by letting δ ≤ min(1,ε/4) . Thesymbol, 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 rational0 if x is irrational .

6.2. EXERCISES 103Proof: Continuous > 1. Let x € f~!(V) so f(x) € B(f (x) ,€x) CThen by continuity, there is 6, > 0 such that f (B(x, 6.) D(f)) CBso, letting U = Uye p-1(y)B (x, 5x) , it follows that f-'(V) =UND(f).1. => Continuous. "You could pick a particular open set B(f(x),€) = V and thenf-!(V) =UND(f) for open U and so if x € f~! (V), then there is 6, such that B(x, 5,) CU and so if y € B(x,6,) ND(f), then f(y) € B(f (x) ,€) which is the standard definitionof continuity at x.1. = 2. Let H be closed. Then HC is open and so f~! (H©) =UMD(f) for U open.But then f~' (H) =U©ND(f) where UC is closed. (D(f) = f~! (H) Uf! (H%))2. = 1. Let U be open. Then UC is closed and so f~! (U°) =CND(f) for a closedset C. But then CCN D(f) = f-! (U) where C© is open.This suggests the following definition.V for some €, > 0.(f(x), €x) CV andDefinition 6.1.3 Les be a nonempty set. Then one can define relatively open andrelatively closed subsets of S as follows. A set O C S is relatively open if O = UMS whereU is open. A set K CS is relatively closed if there is a closed set C such that K = SOC.In words, the above theorem says that a function is continuous at every point of itsdomain if and only if inverse images of open sets are relatively open if and only if inverseimages of closed sets are relatively closed.6.2 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) =~ 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: Consider the follow-ing which comes from algebra. | f (x) — f (3)| = |x? +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]. Complete the argument by letting 6 < min(1,¢/4). Thesymbol, 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 vane’ inequality for absolute values.7. Let f (x) = . 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, {xj, } such that x/, + x. Now consider the followingfunction.flx)= 1 if x is rational~ | Oif xis irrational °