Help two questions in math. analysis .

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Discussion Overview

The discussion revolves around two questions in mathematical analysis: the proof that every bounded open interval (a,b) is an open set, and the relationship between the supremum of a set and the supremum of that set minus an element that is not its maximum.

Discussion Character

  • Homework-related
  • Conceptual clarification
  • Technical explanation

Main Points Raised

  • One participant suggests that if x is in (a,b), then a
  • Another participant questions the understanding of what constitutes an open set, implying that if x is contained in (a,b) without intersection with the closed interval [a,b], it may indicate that it is an open set.
  • A further contribution defines an open set as one where every point in the neighborhood is included in the set.
  • One participant advises the original poster to read about open intervals and open sets on Wikipedia, suggesting that providing a complete solution may not aid in understanding.
  • There is a discussion about the implications of the statement that c=supA and a is not the maximum of A, with one participant noting that a must be in A and is not an upper bound, leading to the existence of another element b in A such that b>a.

Areas of Agreement / Disagreement

The discussion does not appear to reach a consensus, as participants express varying levels of understanding and propose different approaches to the questions posed.

Contextual Notes

Some statements rely on specific definitions of open sets and supremum, which may not be universally agreed upon. The discussion includes assumptions about the elements of sets and their properties that are not fully unpacked.

ShengyaoLiang
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help...two questions in math. analysis...

1) prove : every bounded open interval(a,b) where a less than b and a,b in R , then is an open set.

2) If c = supA and a is not the maximum of A, then sup(A\{a})= c.

I really don't know how to prove those two...thanks for helping me...really...
 
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if x is in (a,b) then a<x<b, so what can you say about b-x and x-a? Choose a epsilon neighborhood of x that is contained in (a,b) appropriately.

we have that c=supA and that a is not the max of A. The latter statement needs some unraveling - I presume a IS inside A so then it means that a is NOT an upper bound of A, i.e. there exists a b in A s.t. b>a. Take epsilon to be small and positive, then look at c minus epsilon. Remember there is a b in A with b>a, which also means that b IS not a.
 
Do you know what is an open set ?
 
if x is contained in (a,b) and there is no intersection with ]a,b[ , then it is an open set?
 
A better way to define open set would be to say every point in the neighborhood is in the set.
 
Just read the reply by SidharthM, and read about open interval, open set, etc. on wikipedia. Giving the whole solution would not help in your understanding.
 

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