Exercise 2.3 [PDF]

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EXERCISE 2.3 1. 𝑥 ∈ 𝑆1 arbitrary ⇒ 𝑥 ≥ 0. Therefore, 0 is a lower bound of 𝑆1. Also, any negative number 𝑦 < 0 is a lower bound of 𝑆1 : 𝑦 < 0 ≤ 𝑥, ∀𝑥 ∈ 𝑆1. To show inf 𝑆1 = 0, we have to show that 0 ≥ 𝑦, for any lower bound 𝑦 of 𝑆1. If 𝑦 > 0, then 𝑦 is not a lower bound of 𝑆1 because



𝑦>0⇒



𝑦 𝑦 𝑦 > 0 ⇒ ∈ 𝑆1 & < 𝑦 2 2 2



Therefore, if 𝑦 is a lower bound of 𝑆1, then 𝑦 ≤ 0. 0 is a lower bound, therefore, 0 = inf 𝑆1. For any 𝑥 ≥ 0, 𝑥 is an element of 𝑆1 and 𝑥 + 1 > 𝑥 implies 𝑥 + 1 is also an element of 𝑆1, larger than 𝑥. Therefore, 𝑥 is not an upper bound of 𝑆1 . 2. Let S2 := { x ∈ ℝ ∶ 𝑥 > 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 ∈ ℝ ∶ 𝑥 > 0} Since 0 < x for all x ∈ 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 ∈ ℝ such that x ≤ 𝑢 for all x ∈ S2. Since, 0 < x and x ≤ u for all x ∈ S2 it follows that 0 < u which gives us that 0 < 1 ≤ u + 1 and therefore u,u+1 ∈ S2. Thus since u is an upper bound of S2 and u + 1 ∈ S2 we have that u + 1 ≤ u which gives us that 1 ≤ 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 ∈ S2. Observe now that 0 < s/2 < s. Thus, we have that s/2 ∈ 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 𝑆3 = { 𝑛 ∶ 𝑛 ∈ ℕ }. Show that sup 𝑆3 = 1 and inf 𝑆3 ≥ 1 ( It will follow from the Archimedean Property In section 2.4 that inf 𝑆3 − 0. Answer : 1



Let 𝑆3 ≔ {𝑛 ∶ 𝑛 ∈ ℕ }. 1



Since 1 ≤ 𝑛 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛 ∈ ℕ we have that 0 < 𝑛 ≤ 1 for all 𝑛 ∈ ℕ which gives us that 0 is a lower bound of 𝑆3 and 1 is an upper bound of 𝑆3 .



Observe also that since



1 1



= 1 we have that 1 ∈ 𝑆3 which gives us that given 𝑥 < 1



there exists 𝑆 ′ ∈ 𝑆3 such that 𝑥 < 𝑠′ ( just take 𝑠 ′ = 1 ) which gives us that 𝑥 is not an upper bound od 𝑆3 and then since 𝑥 is an arbirary number smaller than 1 we have that sup 𝑆3 = 1. Since 𝑆3 is bounded from below we have that inf 𝑆3 exists. Since 0 is a lower bound of 𝑆3 it follows from the definition of infimum that 0 ≤ inf 𝑆3 . 4. Let 𝑆4 ∶= {1 − (−1)𝑛 𝑛 : 𝑛 ∈ 𝑁} . 𝐹𝑖𝑛𝑑 inf 𝑆4 𝑎𝑛𝑑 sup 𝑆4 Answer :  Let 𝑆4 ≔ {1 −



(−1)𝑛 𝑛



: 𝑛 ∈ 𝑁}.



𝑛 𝑖𝑠 𝑜𝑑𝑑 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑡ℎ𝑎𝑡



If



(−1)𝑛 𝑛



−1



𝑛



1



= 1 − (𝑛) ≥



𝑛



1



= − 𝑛 𝑎𝑛𝑑 1 ≤ 𝑛 𝑤ℎ𝑖𝑐ℎ 𝑔𝑖𝑣𝑒𝑠 𝑢𝑠 𝑡ℎ𝑎𝑡 1 −



1



=1−(𝑛)=



1



1+𝑛 ≥1>2



𝐼𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑡ℎ𝑎𝑡 (−1)𝑛



(−1)𝑛



(−1)𝑛 𝑛 1



1



= 𝑛 𝑎𝑛𝑑 2 ≤ 𝑛 𝑤ℎ𝑖𝑐ℎ 𝑔𝑖𝑣𝑒𝑠 𝑢𝑠 𝑡ℎ𝑎𝑡 1 − 1



1−2=2



𝑆𝑖𝑛𝑐𝑒 − 1/𝑛 < 1/𝑛 ≤ 1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛 ∈ 𝑁 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑡ℎ𝑎𝑡 1 − 1/𝑛 < 1 + 1/𝑛 ≤ 2 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛 ∈ 𝑁. 1



𝑇ℎ𝑢𝑠, 2 ≤ 1 =



(−1)𝑛 𝑛



≤ 2 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛 ∈ 𝑁.



𝐻𝑒𝑛𝑐𝑒, 𝑆4 𝑖𝑠 𝑏𝑜𝑢𝑛𝑑𝑒𝑑 𝑏𝑒𝑙𝑜𝑤 𝑏𝑦



1 𝑎𝑛𝑑 𝑆4 𝑖𝑠 𝑏𝑜𝑢𝑛𝑑𝑒𝑑 𝑎𝑏𝑜𝑣𝑒 𝑏𝑦 2 𝑎𝑛𝑑 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 inf 𝑆4 𝑎𝑛𝑑 sup 𝑆4 𝑏𝑜𝑡ℎ 𝑒𝑥𝑖𝑠𝑡. 2



 𝑊𝑒 ℎ𝑎𝑣𝑒 𝑎𝑙𝑠𝑜 𝑡ℎ𝑎𝑡 inf 𝑆4 ≤ 𝑠 ≤ sup 𝑆4 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑠 ∈ 𝑆4 . 𝑇ℎ𝑢𝑠, 𝑠𝑖𝑛𝑐𝑒 1 − (−1) = 1 2



(−1)2 2



1



1



1



= 1 − 2 = 2 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑡ℎ𝑎𝑡 2 𝑎𝑛𝑑 1 −



(−1)1 𝑛



=1−



1



2 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑡ℎ𝑎𝑡 2 , 2 ∈ 𝑆4 𝑎𝑛𝑑 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 inf 𝑆4 ≤



𝑎𝑛𝑑 2 ≤ sup 𝑆4 .



𝑊𝑒 ℎ𝑎𝑣𝑒 𝑎𝑙𝑠𝑜 𝑡ℎ𝑎𝑡 𝑢 inf 𝑆4 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑙𝑜𝑤𝑒𝑟 𝑏𝑜𝑢𝑛𝑑 𝑢 𝑜𝑓 𝑆4 𝑎𝑛𝑑 sup 𝑆4 ≤ 𝑢 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑢𝑝𝑝𝑒𝑟 𝑏𝑜𝑢𝑛𝑑 𝑢 𝑜𝑓 𝑆4 . 1 1 𝑇ℎ𝑢𝑠, 𝑠𝑖𝑛𝑐𝑒 𝑖𝑠 𝑎 𝑙𝑜𝑤𝑒𝑟 𝑏𝑜𝑢𝑛𝑑 𝑜𝑓 𝑆4 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑡ℎ𝑎𝑡 2 2 ≤ inf 𝑆4 𝑎𝑛𝑑 𝑠𝑖𝑛𝑐𝑒 2 𝑖𝑠 𝑎𝑛 𝑢𝑝𝑝𝑒𝑟 𝑏𝑜𝑢𝑛𝑑 𝑜𝑓 𝑆4 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑡ℎ𝑎𝑡 sup 𝑆4 ≤ 2. 𝐻𝑒𝑛𝑐𝑒,



1 1 1 ≤ inf 𝑆4 ≤ 𝑎𝑛𝑑 2 ≤ sup 𝑆4 ≤ 2 𝑎𝑛𝑑 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 inf 𝑆4 = 𝑎𝑛𝑑 sup 𝑆4 = 2. 2 2 2



5. Find the infimum and supremum, if they exist, of each of the following sets. (a) 𝐴 ∶= {𝑥 ∈ ℝ: 2𝑥 + 5 > 0} Answer: 𝐴 = {𝑥 ∈ ℝ: 2𝑥 + 5 > 0}, 5



𝐴 = {𝑥 ∈ ℝ: 𝑥 > − 2}, 5



5



All x ∈ 𝐴 are greater that − 2, that is we can say that A is bounded below and − 2 is a lower bound of A. 5



Let suppose infA= − 2 = 𝑤 The condition (1’) Is already statisfied. For condition (2’) lets suppose the 5



opposite, that is that there exist a lower bound of A called t and that 𝑡 > 𝑤 = − 2, 5



then 𝑡 + 2 > 0 and 𝑡 ≤ 𝑥, for all 𝑥 ∈ 𝐴. By the Archimedean prperty, ∃𝑛 ∈ ℕ such that



0
− 𝑛 2 1



From there, using the definition of the set A, we can conclude that 𝑡 − 𝑛 ∈ 𝐴. 1



1



𝑡 − 𝑛 < 𝑡 and 𝑡 − 𝑛 ∈ 𝐴, 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 𝑡 ≥ 𝑤 and 𝑤 = −



5 2



5



is the greatest lower bound. Therefore, infA= − 2 (b) Find the infimum and supremum, if they exist, of each of the following sets 𝐵 ∶= {𝑥 ∈ ℝ, 𝑥 + 2 ≥ 𝑥 2 }. Answer



:



Let : x = -2 then the value of 𝑥 + 2 ≥ 𝑥 2 is 0 ≥ 4. So -2 does not fit to the inequality x = -1 then the value of 𝑥 + 2 ≥ 𝑥 2 is 1 ≥ 1. So -1 fit to the inequality x = 0 then the value of 𝑥 + 2 ≥ 𝑥 2 is 2 ≥ 0. So 0 fit to the inequality x = 1 then the value of 𝑥 + 2 ≥ 𝑥 2 is 3 ≥ 1. So 1 fit to the inequality x = 2 then the value of 𝑥 + 2 ≥ 𝑥 2 is 4 ≥ 4. So 2 fit to the inequality x = 3 then the value of 𝑥 + 2 ≥ 𝑥 2 is 5 ≥ 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. 



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, 𝑡 ≤ 𝑤 → −1 ≤ 0 (True)







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, 𝑢 ≤ 𝑣 → 1 ≤ 2 (True) So, we can conclude that the infimum and supremum from 𝐵 ∶= {𝑥 ∈ ℝ, 𝑥 + 2 ≥ 𝑥 2 } is inf B = -1 and sup B = 2 (c) Lets consider the given set and rewrite the conditions : 𝐶 ={𝑥 ∈𝑅 ∶



={𝑥 ∈𝑅 ∶



={𝑥 ∈𝑅 ∶



1 < 0} 𝑥



𝑥2 − 1 < 0} 𝑥



(𝑥 − 1)(𝑥 + 1) < 0} 𝑥



There are two cases: 1. (𝑥 − 1)(𝑥 + 1) > 0 𝑎𝑛𝑑 𝑥 < 0 1.1 (𝑥 − 1) > 0 𝑎𝑛𝑑 (𝑥 + 1) > 0 𝑎𝑛𝑑 𝑥 < 0 𝑥 > 1 𝑎𝑛𝑑 𝑥 > −1 𝑎𝑛𝑑 𝑥 < 0 𝑥 > 1 𝑎𝑛𝑑 𝑥 < 0



(no solution )



1.2 (𝑥 − 1) < 0 𝑎𝑛𝑑 (𝑥 + 1) < 0 𝑎𝑛𝑑 𝑥 < 0 𝑥 < 1 𝑎𝑛𝑑 𝑥 < −1 𝑎𝑛𝑑 𝑥 < 0 𝑥 < −1 𝑎𝑛𝑑 𝑥 < 0 𝑥 ∈ 〈−∞, −1〉 2. (𝑥 − 1 )(𝑥 + 1) < 0 𝑎𝑛𝑑 𝑥 > 0 2.1 (𝑥 − 1) > 0 𝑎𝑛𝑑 (𝑥 + 1) < 0𝑎𝑛𝑑 𝑥 > 0 𝑥 > 1 𝑎𝑛𝑑 𝑥 < −1 𝑎𝑛𝑑 𝑥 > 0 2.2 (𝑥 − 1) < 0 𝑎𝑛𝑑 (𝑥 + 1) > 0 𝑎𝑛𝑑 𝑥 > 0 𝑥 < 1 𝑎𝑛𝑑 𝑥 > −1 𝑎𝑛𝑑 𝑥 > 0 −1 < 𝑥 < 1 𝑎𝑛𝑑 𝑥 > 0 𝑥 ∈ 〈0,1〉 Hance, we have



( no solution)



𝐶 = {−∞, −1} ∪ {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 𝐶 = 1 = 𝑣 The condition (1) is already satistied. For the condition (2) lets suppose the opposite, that there exists upper bound of C called 𝑢 such that 𝑢 < 𝑣 − 1 𝑎𝑛𝑑 𝑢 ≥ 𝑥 for all 𝑥 ∈ 𝐶. Then 𝑢 − 1 < 0 , that 1 − 𝑢 < 0 by the achimedean property, ∃ 𝑚 ∈ 𝑁 such that 0
𝛼 𝑎𝑛𝑑 𝑥 < 𝛽 No solution



2°



(𝑥 − 𝛼) < 0 𝑎𝑛𝑑 (𝑥 − 𝛽) > 0 𝑥 < 𝛼 𝑎𝑛𝑑 𝑥 > 𝛽 𝛽 0 𝑎𝑛𝑑 𝑡 ≤ 𝑥,



for all 𝑥 ∈ 𝐷.



By the Archimedean property, ∃ 𝑛 ∈ 𝑁 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 1



0𝛽 From there, using the definition of the set D, we can conclude that 1



𝑡−𝑛∈𝐷 1



1



𝑡 − 𝑛 < 𝑡 𝑎𝑛𝑑 𝑡 − 𝑛 ∈ 𝐷, 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, 𝑡 ≥ 𝑤 𝑎𝑛𝑑 𝑤 = 𝛽 is the greatest lower bound. Therefore, 𝑖𝑛𝑓𝐷 = 𝛽 = 1 − √6. Next we suppose 𝑠𝑢𝑝𝐷 = 𝛼 = 𝑣 The condition (1’) is already satisfied. For the condition (2’) let’s suppose the opposite that There exists an upper bound of D called u such that 𝑢 < 𝑤 = 𝛼 𝑎𝑛𝑑 𝑢 ≥ 𝑥, 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ∈ 𝐷. By the Archimedean property, ∃ 𝑛 ∈ 𝑁 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 1



0