Let SC R. Show that z E Siff Please follow steps: inf {|r- s | s € S} = 0. Let x € 3 and let € > 0. Explain why for € > 0, D(x, €) Ns 0. Take s E D (1,€) S and explain why 0 ≤ inf{|rs|s € S} < €. Why can we conclude that inf {|rs|s € S} = 0.
Let SC R. Show that z E Siff Please follow steps: inf {|r- s | s € S} = 0. Let x € 3 and let € > 0. Explain why for € > 0, D(x, €) Ns 0. Take s E D (1,€) S and explain why 0 ≤ inf{|rs|s € S} < €. Why can we conclude that inf {|rs|s € S} = 0.
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.2: Graphs Of Equations
Problem 23E
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Question
Please follow the steps to answer the question and fill in all the required spaces. Thanks
![Let S CR. Show that Siff
Please follow steps:
inf {|rs| s € S} = 0.
Let S and let e > 0. Explain why for € > 0,
D(x, €) ns +0.
Take s E D (x, €) nS and explain why
0 ≤ inf {|rs| | 8 € S} < €.
-
Why can we conclude that
inf {|xs| 8 € S} = 0.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa8d46346-fe9f-45aa-a3a0-df12b7cae379%2F7ca3fb23-08ab-49c1-aa2e-a0a7b76e9c32%2F5ju585_processed.png&w=3840&q=75)
Transcribed Image Text:Let S CR. Show that Siff
Please follow steps:
inf {|rs| s € S} = 0.
Let S and let e > 0. Explain why for € > 0,
D(x, €) ns +0.
Take s E D (x, €) nS and explain why
0 ≤ inf {|rs| | 8 € S} < €.
-
Why can we conclude that
inf {|xs| 8 € S} = 0.
![Why can we conclude that
Conversely, assume that
inf {|rs|s € S} = 0.
inf {|rs|s € S} = 0.
Take € > 0 and explain why there is s ES, such that
|xs| < €.
Conclude that
Explain why we can find s € SnD (™, €).
D(x, €) ns 0.
Explain why we can say that ze S.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa8d46346-fe9f-45aa-a3a0-df12b7cae379%2F7ca3fb23-08ab-49c1-aa2e-a0a7b76e9c32%2Fuy64m4_processed.png&w=3840&q=75)
Transcribed Image Text:Why can we conclude that
Conversely, assume that
inf {|rs|s € S} = 0.
inf {|rs|s € S} = 0.
Take € > 0 and explain why there is s ES, such that
|xs| < €.
Conclude that
Explain why we can find s € SnD (™, €).
D(x, €) ns 0.
Explain why we can say that ze S.
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