Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Calculus and Beyond Homework Help
Understanding Upper and Lower Bounds in Analysis Solutions
Reply to thread
Message
[QUOTE="silimay, post: 1460427, member: 46417"] I was recently looking at the solutions to a problem set I have in analysis and there was something I didn't understand. This is the problem in the problem set: [h2]Homework Statement [/h2] Let A ⊂ R be nonempty. Define −A = {−x / x ∈ A}. Show that sup(−A) = − inf A and inf(−A) = − sup A. So I thought the definition of a, say, least upper bound, was that [tex]\alpha[/tex] is a least upper bound of [I]E[/I] if it is an upper bound of [I]E[/I] and if [tex]\gamma < \alpha[/tex] then [tex]\gamma[/tex] is not an upper bound of [I]E[/I]. But in the solution, they almost seem to be using a different definition of least upper bound / greatest lower bound. [b]3. (part of the solution)[/b] This is just part of the solution. I'm not really confused about the actual answer to the question, I'm just confused about how they define the least upper bound. In the solution, in a case where they assumed A had a lower bound: Let us define r := inf A. Then r ∈ R. Since r is a lower bound for A,we know that x ≥ r, ∀x ∈ A. It follows that: −x ≤ −r, ∀x ∈ A. Hence, −r is an upper bound for A. We must show that it is the least upper bound for A. To see this, let e > 0 be given. Since r = inf A, it follows that ∀ e > 0, ∃a ∈ A such that a < r + e. So, ∀ e > 0, ∃a ∈ A such that −a > −r −e . Hence, since such an element of −A can be found ∀ e > 0, it follows that −r is the least upper bound. ...What they were doing with e seems almost more like something with a limit point...So I got confused... [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Calculus and Beyond Homework Help
Understanding Upper and Lower Bounds in Analysis Solutions
Back
Top