October 19th, 2016, 11:12 AM  #1 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 965 Thanks: 78  Nested open intervals
Why does the nested interval theorem require closed intervals? Let ai and bi converge to a. Lim (ai,bi)=(a,a) doesn't exist: a<x<a Lim (ai,bi]=(a,a] doesn't exist: a$\displaystyle \leq$x<a Lim [ai,bi]=[a,a] exists: a$\displaystyle \leq$x$\displaystyle \leq$a, x=a 
October 19th, 2016, 01:33 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,214 Thanks: 491 
Since ai and bi both converge to a, then a is in the limit interval. (a,a) and (a,a] are both empty (do not contain a), so they are wrong.

February 27th, 2017, 08:32 AM  #3  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 965 Thanks: 78  Quote:
For nested sets in Rk, $\displaystyle C1 \sqsupset C2 \sqsupset C3 \sqsupset$......., $\displaystyle Cn = [(a_{n1},b_{n1})],[(a_{n2},b_{n2})],....,[(a_{nk},b_{nk})]$, [(..)] means open or closed, the previous results apply to the component intervals which can be all open or all closed, unless Lim of a particular interval is 0 (Lim a$\displaystyle _{np}$=a and Lim b$\displaystyle _{np}$=a} in which case they must be closed for that component interval to have a nonempty intersection. Rudin, Theorem 2.39, considers only closed component intervals without specifying a 0 limit. https://en.wikipedia.org/wiki/Nested_intervals also specifies closed intervals for the nested interval theorem (the intersection of a closed nest is nonempty). The motivation for this post is the general confusion about whether a nested set has to be closed to have a nonzero intersection.  
February 27th, 2017, 10:08 AM  #4  
Senior Member Joined: Aug 2012 Posts: 1,144 Thanks: 249  Quote:
For clarity  your own as well as everyone else's  can you please write down the exact question you are asking? Be specific and clear. For example what's a nested set? No such thing. Perhaps you mean a sequence of sets nested downward. Clarity and precision please. General confusion? What general confusion? You answered your own question in your first post, where you showed that if the sets aren't closed then their intersection may be empty. Also by the way the finite intersection property is the logical dual of the definition of compactness. You might give that some thought, it's very englightening. Here's the Wiki page you should be looking at. https://en.wikipedia.org/wiki/Finite...ction_property Last edited by Maschke; February 27th, 2017 at 10:45 AM.  
March 1st, 2017, 10:28 AM  #5 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 965 Thanks: 78 
What I have described are nested ncells, along with references. If one doesn't understand nested ncells, it does not follow that I am confused. Example: Consider a nest of squares (Cn) whose vertical edges (left and right), including corners, belong to the square, and whose horizontal edges (top and bottom) do not. Let the diameter of the squares (the diagonal) go to 0. Let Pn be the midpoint of the nth square. Clearly P=Lim Pn exists. Does Pn belong to all squares, ie, is P=$\displaystyle \cap$Cn? It would if the squares were closed? Ref https://en.wikipedia.org/wiki/Cantor...ection_theorem 
March 2nd, 2017, 08:49 AM  #6 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 965 Thanks: 78 
The intersection of a nest of nonclosed sets whose diameter goes to zero is empty because a nonclosed set cannot consist of a single point. If the diameter limit is finite, the intersection exists. 

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