Closed interval is covering compact

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Discussion Overview

The discussion revolves around proving that the closed interval [a,b] is covering compact. Participants explore definitions, approaches to the proof, and the relationship between covering compactness and compactness.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests starting with the set of elements in the interval that finitely many members of an open cover U cover, and questions how to use the least upper bound theorem to show that b is included in this set.
  • Another participant asks if covering compact is equivalent to compactness, indicating a potential confusion about the definitions.
  • It is proposed that covering compact implies compactness, but not the other way around, prompting a request for clarification on definitions.
  • A participant emphasizes that the question specifically asks for a proof of covering compactness, not compactness, which leads to a discussion about definitions.
  • One participant mentions that the definition of compactness most commonly used aligns with the concept of covering compactness, noting that it is also referred to as quasi-compact by some.
  • Another participant suggests that the usual proof involves contradiction, proposing a method of dividing the interval into subintervals to demonstrate the existence of a finite subcover.
  • Concerns are raised about the applicability of the proof if the interval is considered over rational numbers, as the properties of least upper bound and greatest lower bound may not hold.

Areas of Agreement / Disagreement

Participants express differing views on the definitions of covering compactness and compactness, and there is no consensus on the best approach to proving the statement. The discussion remains unresolved regarding the specific proof method to be used.

Contextual Notes

There are limitations regarding the definitions of compactness and covering compactness, as well as the assumptions about the nature of the interval (real numbers vs. rational numbers) that are not fully clarified.

Scousergirl
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The question asks to prove directly that the closed interval is covering compact

- U= an open covering of the closed set [a,b]
I started by taking C=the set of elements in the interval that finitely many members of U cover. Now I need to somehow use the least upper bound theorem to show that b is in C?
 
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You want to show that for any arbitrary open cover of [a,b] there is a finite subcover.

Is covering compact the same thing as compact?
 
Right so I assume U is my arbitrary open cover.

Covering compact implies compact but not vice versa right?
 
What are your definitions of compact and covering compact?
 
for any arbitrary open cover of [a,b] there is a finite subcover for covering compact but I the question isn't asking to use the definition of compact just covering compact.
 
That is the definition of compact most people use. You've been given slightly non-standard notation. This is also called quasi-compact by some as well.

You're on the right lines. Show that the set of points you described (as having the finite covering property) is closed, then think again.
 
The question asks to prove directly that the closed interval is covering compact

- U= an open covering of the closed set [a,b]
I started by taking C=the set of elements in the interval that finitely many members of U cover. Now I need to somehow use the least upper bound theorem to show that b is in C? ineed solution
 
What makes you think that you can show that without showing, first, that there exist a finite subcover for the entire interval. That is NOT the usual proof for this statement.

The usual proof is by contradiction. Suppose there is NOT finite subcover for the interval. Now look at the two subintervals, [a, c], [c, b] where c is between a and b (for simplicity, you can choose it half way between- c= (a+b)/2. Since the entire interval cannot be covered by a finite collection of these open sets, at least one of the two subintervals cannot. Cut that interval into two pieces and repeat. Repeat until you get to a single point.

Of course, you have to be able to show that you do, in the limit, get a single point. For that you must specify that your interval is an interval or real numbers so that you can use the "least upper bound" and "greatest lower bound" properties. If you are thinking of [a, b] as an interval of rational numbers, the statement is not true.
 

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