Notices
Results 1 to 6 of 6

Thread: Convex sets

  1. #1 Convex sets 
    Moderator Moderator
    Join Date
    Jun 2005
    Posts
    1,620
    Let me apologize straight away for my prolixity - I have been playing village cricket this afternoon which inevitably involves a few beers after. Also, Mrs. G. is at a concert tonight, so I am slightly bored.

    Right. It is no secret that I dislike pure analysis - as a hobby mathematician I am free to pick-and-choose what I study, though I would be the first to admit that neglecting the "fundamentals" may well be unwise. This neglect has tripped me up more than once. Ho hum.

    For reasons nobody here needs to know, I decided to study functional analysis, and I work from a text by Walter Rudin, again admitting I have a weak grounding in pure analysis.

    Anyway, on first reading Rudin's opening chapter (intro to topological vector spaces etc) I am arrogantly thinking, like, yeah, yeah, I know all this stuff and consequently skimmed. But then a coupla chapters down the line I got confused by notation and terminology, so went back to the Intro.

    Which brings me finally, and somewhat boringly, to my question, and I use Rudin's notation exactly.

    Suppose that is a vector space and that . Then is said to be convex if for all . Note that this is shorthand for the following:

    Whenever then for all . The intuitive meaning is clear, provided only that we are prepared to bend the "rules" more than a little. It says that the "further you move incrementally from "x" the closer you get to "y" and still stay in the set".

    Yeah, I know, you're right to say "Ugh"!

    So Rudin now defines a topology on in the obvious way.This seems to suggest to my pea brain that all the above can be written as follows:

    Define a curve in by such that . Then if, for all that for any
    defines a path-connected topological space.

    Or have I taken a wrong turn early on?


    Reply With Quote  
     

  2.  
     

  3. #2  
    Forum Professor river_rat's Avatar
    Join Date
    Jun 2006
    Location
    South Africa
    Posts
    1,517
    Hi Guitarist

    I'm a bit confused, are you asking if convex topological space => path-connected topological space then yes that is true.

    If you are asking if path-connected topological space => convex topological space then this is not generally true.

    If you are asking neither question then please ignore this post


    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
    Reply With Quote  
     

  4. #3  
    Forum Professor
    Join Date
    Jul 2008
    Location
    New York State
    Posts
    1,101
    The definition of convex is simpler than you make it out to be. In plain English. a set is convex if given any pair of points in a set, all points on the straight line segment connecting them are also in the set.

    The description involving arbitrary curves is something different.
    Reply With Quote  
     

  5. #4 Re: Convex sets 
    . DrRocket's Avatar
    Join Date
    Aug 2008
    Posts
    5,486
    Quote Originally Posted by Guitarist
    Suppose that is a vector space and that . Then is said to be convex if for all . =Whenever then for all .
    The idea here is that if x any belong to C then so does the line segment joining them.

    Quote Originally Posted by Guitarist

    Define a curve in by such that . Then if, for all that for any
    defines a path-connected topological space.

    Or have I taken a wrong turn early on?
    Yes you took a wrong turn. This is merely path connected. By replacing a line segment by a curve you allow all sorts of convoluted shapes.

    You are headed for theorems that require convexity, like the Krein-Millman theorem and the theory of locally nconvex spaces. The simplex algorithm is dependent on convexity for the minimization of linear functionals.
    Reply With Quote  
     

  6. #5  
    Moderator Moderator
    Join Date
    Jun 2005
    Posts
    1,620
    I thank you all for your replies. Yeah, I know it sounds churlish, but you guys are giving me definitions, not explanations.

    I admit I may not be the sharpest knife in the drawer, but it is not obvious to me that the definition I copied into the OP requires that the path connecting any 2 points in a convex set must be a line segment, still less a straight line, whatever that means in this context.

    Grrr
    Reply With Quote  
     

  7. #6  
    . DrRocket's Avatar
    Join Date
    Aug 2008
    Posts
    5,486
    Quote Originally Posted by Guitarist
    I thank you all for your replies. Yeah, I know it sounds churlish, but you guys are giving me definitions, not explanations.

    I admit I may not be the sharpest knife in the drawer, but it is not obvious to me that the definition I copied into the OP requires that the path connecting any 2 points in a convex set must be a line segment, still less a straight line, whatever that means in this context.

    Grrr
    Two points: 1) The only finite dimensional real and complex vector spaces are the usual suspects, so when you consider a finite-dimensional subspace of any topological vector space the simple model is accurate. 2) Yes, it is just a line segment. A straight line.
    Reply With Quote  
     

Bookmarks
Bookmarks
Posting Permissions
  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •