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EXERCISE 2.3 1. π₯ β π1 arbitrary β π₯ β₯ 0. Therefore, 0 is a lower bound of π1. Also, any negative number π¦ < 0 is a lower bound of π1 : π¦ < 0 β€ π₯, βπ₯ β π1. To show inf π1 = 0, we have to show that 0 β₯ π¦, for any lower bound π¦ of π1. If π¦ > 0, then π¦ is not a lower bound of π1 because
π¦>0β
π¦ π¦ π¦ > 0 β β π1 & < π¦ 2 2 2
Therefore, if π¦ is a lower bound of π1, then π¦ β€ 0. 0 is a lower bound, therefore, 0 = inf π1. For any π₯ β₯ 0, π₯ is an element of π1 and π₯ + 1 > π₯ implies π₯ + 1 is also an element of π1, larger than π₯. Therefore, π₯ is not an upper bound of π1 . 2. Let S2 := { x β β βΆ π₯ > 0} . Does S2 have lower bounds? Does S2 have upper bounds? Does inf S2 exists? Does sup S2 exists? Prove your statements Proof: 1) Let S2 := { x β β βΆ π₯ > 0} Since 0 < x for all x β S1, we have that 0 is a lower bound of S2 and therefore, any real number smaller than 0 is also a lower bound of S2. Suppose now that S2 has at least one upper bound, then there exists u β β such that x β€ π’ for all x β S2. Since, 0 < x and x β€ u for all x β S2 it follows that 0 < u which gives us that 0 < 1 β€ u + 1 and therefore u,u+1 β S2. Thus since u is an upper bound of S2 and u + 1 β S2 we have that u + 1 β€ u which gives us that 1 β€ 0 which we know is not true. Hence, S2 has no upper bounds.
2) We already know that 0 is a lower bound of S2. Suppose there exists a lower bound s of S2 such that 0 < s. Then, we have in particular that s β S2. Observe now that 0 < s/2 < s. Thus, we have that s/2 β S and s/2 < s which contradicts the fact that s is a lower bound of S2. Hence, S2 has no lower bound bigger than 0 and therefore inf S2 = 0.
Since the supremum of a nonempty set is always an upper bound of the set and S2 has no upper bounds, it follows that sup S2 does not exists.
1
3. Let π3 = { π βΆ π β β }. Show that sup π3 = 1 and inf π3 β₯ 1 ( It will follow from the Archimedean Property In section 2.4 that inf π3 β 0. Answer : 1
Let π3 β {π βΆ π β β }. 1
Since 1 β€ π πππ πππ π β β we have that 0 < π β€ 1 for all π β β which gives us that 0 is a lower bound of π3 and 1 is an upper bound of π3 .
Observe also that since
1 1
= 1 we have that 1 β π3 which gives us that given π₯ < 1
there exists π β² β π3 such that π₯ < π β² ( just take π β² = 1 ) which gives us that π₯ is not an upper bound od π3 and then since π₯ is an arbirary number smaller than 1 we have that sup π3 = 1. Since π3 is bounded from below we have that inf π3 exists. Since 0 is a lower bound of π3 it follows from the definition of infimum that 0 β€ inf π3 . 4. Let π4 βΆ= {1 β (β1)π π : π β π} . πΉπππ inf π4 πππ sup π4 Answer : ο Let π4 β {1 β
(β1)π π
: π β π}.
π ππ πππ π€π βππ£π π‘βππ‘
If
(β1)π π
β1
π
1
= 1 β (π) β₯
π
1
= β π πππ 1 β€ π π€βππβ πππ£ππ π’π π‘βππ‘ 1 β
1
=1β(π)=
1
1+π β₯1>2
πΌπ π ππ ππ£ππ π€π βππ£π π‘βππ‘ (β1)π
(β1)π
(β1)π π 1
1
= π πππ 2 β€ π π€βππβ πππ£ππ π’π π‘βππ‘ 1 β 1
1β2=2
πππππ β 1/π < 1/π β€ 1 πππ πππ π β π π€π βππ£π π‘βππ‘ 1 β 1/π < 1 + 1/π β€ 2 πππ πππ π β π. 1
πβπ’π , 2 β€ 1 =
(β1)π π
β€ 2 πππ πππ π β π.
π»ππππ, π4 ππ πππ’ππππ πππππ€ ππ¦
1 πππ π4 ππ πππ’ππππ ππππ£π ππ¦ 2 πππ π‘βπππππππ inf π4 πππ sup π4 πππ‘β ππ₯ππ π‘. 2
ο ππ βππ£π πππ π π‘βππ‘ inf π4 β€ π β€ sup π4 πππ πππ π β π4 . πβπ’π , π ππππ 1 β (β1) = 1 2
(β1)2 2
1
1
1
= 1 β 2 = 2 π€π βππ£π π‘βππ‘ 2 πππ 1 β
(β1)1 π
=1β
1
2 π€π βππ£π π‘βππ‘ 2 , 2 β π4 πππ π‘βπππππππ inf π4 β€
πππ 2 β€ sup π4 .
ππ βππ£π πππ π π‘βππ‘ π’ inf π4 πππ πππ πππ€ππ πππ’ππ π’ ππ π4 πππ sup π4 β€ π’ πππ πππ π’ππππ πππ’ππ π’ ππ π4 . 1 1 πβπ’π , π ππππ ππ π πππ€ππ πππ’ππ ππ π4 π€π βππ£π π‘βππ‘ 2 2 β€ inf π4 πππ π ππππ 2 ππ ππ π’ππππ πππ’ππ ππ π4 π€π βππ£π π‘βππ‘ sup π4 β€ 2. π»ππππ,
1 1 1 β€ inf π4 β€ πππ 2 β€ sup π4 β€ 2 πππ π‘βπππππππ inf π4 = πππ sup π4 = 2. 2 2 2
5. Find the infimum and supremum, if they exist, of each of the following sets. (a) π΄ βΆ= {π₯ β β: 2π₯ + 5 > 0} Answer: π΄ = {π₯ β β: 2π₯ + 5 > 0}, 5
π΄ = {π₯ β β: π₯ > β 2}, 5
5
All x β π΄ are greater that β 2, that is we can say that A is bounded below and β 2 is a lower bound of A. 5
Let suppose infA= β 2 = π€ The condition (1β) Is already statisfied. For condition (2β) lets suppose the 5
opposite, that is that there exist a lower bound of A called t and that π‘ > π€ = β 2, 5
then π‘ + 2 > 0 and π‘ β€ π₯, for all π₯ β π΄. By the Archimedean prperty, βπ β β such that
0
β π 2 1
From there, using the definition of the set A, we can conclude that π‘ β π β π΄. 1
1
π‘ β π < π‘ and π‘ β π β π΄, gives us a contradiction with the hypothesis that t is a lower bound. We found a number that is smaller from t and also inside the set A wich contradicts the definition of the lower bound). From that we conclude that t cannot be a lower bound, that is π‘ β₯ π€ and π€ = β
5 2
5
is the greatest lower bound. Therefore, infA= β 2 (b) Find the infimum and supremum, if they exist, of each of the following sets π΅ βΆ= {π₯ β β, π₯ + 2 β₯ π₯ 2 }. Answer
:
Let : x = -2 then the value of π₯ + 2 β₯ π₯ 2 is 0 β₯ 4. So -2 does not fit to the inequality x = -1 then the value of π₯ + 2 β₯ π₯ 2 is 1 β₯ 1. So -1 fit to the inequality x = 0 then the value of π₯ + 2 β₯ π₯ 2 is 2 β₯ 0. So 0 fit to the inequality x = 1 then the value of π₯ + 2 β₯ π₯ 2 is 3 β₯ 1. So 1 fit to the inequality x = 2 then the value of π₯ + 2 β₯ π₯ 2 is 4 β₯ 4. So 2 fit to the inequality x = 3 then the value of π₯ + 2 β₯ π₯ 2 is 5 β₯ 9. So 3 does not fit to the inequality From 2.3.2 Defenition, suppose that inf B = -1 and sup B = 2. Then we need to do fill all of the conditions from 2.3.2 defenition. ο·
Let infB = w = -1 Then, the first condition is already satisfied To fit the second condition, we know that 0 is another lower bound, let 0 =w Then, π‘ β€ π€ β β1 β€ 0 (True)
ο·
Let supB = u = 1
Then, the first condition is already satisfied To fit the second condition, we know that 2 is another upper bound, let 2 = v Then, π’ β€ π£ β 1 β€ 2 (True) So, we can conclude that the infimum and supremum from π΅ βΆ= {π₯ β β, π₯ + 2 β₯ π₯ 2 } is inf B = -1 and sup B = 2 (c) Lets consider the given set and rewrite the conditions : πΆ ={π₯ βπ
βΆ
={π₯ βπ
βΆ
={π₯ βπ
βΆ
1 < 0} π₯
π₯2 β 1 < 0} π₯
(π₯ β 1)(π₯ + 1) < 0} π₯
There are two cases: 1. (π₯ β 1)(π₯ + 1) > 0 πππ π₯ < 0 1.1 (π₯ β 1) > 0 πππ (π₯ + 1) > 0 πππ π₯ < 0 π₯ > 1 πππ π₯ > β1 πππ π₯ < 0 π₯ > 1 πππ π₯ < 0
(no solution )
1.2 (π₯ β 1) < 0 πππ (π₯ + 1) < 0 πππ π₯ < 0 π₯ < 1 πππ π₯ < β1 πππ π₯ < 0 π₯ < β1 πππ π₯ < 0 π₯ β β©ββ, β1βͺ 2. (π₯ β 1 )(π₯ + 1) < 0 πππ π₯ > 0 2.1 (π₯ β 1) > 0 πππ (π₯ + 1) < 0πππ π₯ > 0 π₯ > 1 πππ π₯ < β1 πππ π₯ > 0 2.2 (π₯ β 1) < 0 πππ (π₯ + 1) > 0 πππ π₯ > 0 π₯ < 1 πππ π₯ > β1 πππ π₯ > 0 β1 < π₯ < 1 πππ π₯ > 0 π₯ β β©0,1βͺ Hance, we have
( no solution)
πΆ = {ββ, β1} βͺ {0,1} It is easy to see that infimum of C does not exist since C is not bounded below. We can see that C is bounded above We soppose that sup πΆ = 1 = π£ The condition (1) is already satistied. For the condition (2) lets suppose the opposite, that there exists upper bound of C called π’ such that π’ < π£ β 1 πππ π’ β₯ π₯ for all π₯ β πΆ. Then π’ β 1 < 0 , that 1 β π’ < 0 by the achimedean property, β π β π such that 0
πΌ πππ π₯ < π½ No solution
2Β°
(π₯ β πΌ) < 0 πππ (π₯ β π½) > 0 π₯ < πΌ πππ π₯ > π½ π½ 0 πππ π‘ β€ π₯,
for all π₯ β π·.
By the Archimedean property, β π β π π π’πβ π‘βππ‘ 1
0π½ From there, using the definition of the set D, we can conclude that 1
π‘βπβπ· 1
1
π‘ β π < π‘ πππ π‘ β π β π·, gives us a contradiction with the hypothesis that t is a lower bound. (We found a number that is smaller from t and also inside the set D which contradicts the definition of the lower bound). From that we conclude that t cannot be a lower bound, that is, π‘ β₯ π€ πππ π€ = π½ is the greatest lower bound. Therefore, ππππ· = π½ = 1 β β6. Next we suppose π π’ππ· = πΌ = π£ The condition (1β) is already satisfied. For the condition (2β) letβs suppose the opposite that There exists an upper bound of D called u such that π’ < π€ = πΌ πππ π’ β₯ π₯, πππ πππ π₯ β π·. By the Archimedean property, β π β π π π’πβ π‘βππ‘ 1
0