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Every open set contains both open and closed cell.

  1. Sep 30, 2007 #1
    1. The problem statement, all variables and given/known data

    If an open cell is defined as (a1,b1) X (a2,b2) X .... (an,bn) in R^n and closed cell is defined as [a1,b1] X [a2,b2] X .... [an,bn], then every open set in R^n contains an open-n-cell and a closed-n-cell.

    2. Relevant equations

    Def: An open set is a set which has all points as interior points

    3. The attempt at a solution

    I am a bit unsure about my line of reasoning. Can someone please take a look?

    Let y belong to an n-cell, where y = (y1,...,yn) such that ai<yi<bi, where i = 1,2...n. I can prove that every k cell is compact and that if p,q are two points that belong to the n-cell , then d(p,q) < delta, where delta = sqrt( sum(bi-ai)^2). I will omit the proof for here. But I am going to use this result to solve this problem.

    Now, let us consider a point xo = {(ai+bi/2)} , i= 1,2,....n. Now, the distance from the points s (a1,a2,a3...an) = distance from r (b1,b2,b3...bn) = delta/2. Let x0 belong to the cell. It is easy to see this because each point of x0 is between (ai,bi) for i=1,2....n

    Hence, I can construct a neighborhood N, with x0 as center and radius delta/2+h/2, where h>0 such that the diameter of this neighborhood is delta+h. This will contain points x, such that d(xo,x) < (delta+h)/2. This N will contain both s and r as internal points. Why? Because any neighborhood around s or r with radius h/2 will be inside the sphere N. Hence, s and r shall be internal points of N.

    As a result any point p,q that belong to the cell shall also be internal points of N because the max distance between such points shall be delta. We just saw that the end points r and s are internal points. Hence, if p,q belong to the cell , they shall belong to the N. Note that the greatest distance between two points of a sphere is along a diameter (end points of diameter). That shall be delta + h. So, it will contain any points such that the distance between them is only delta.

    Another way to look at p as internal to the spehere is by taking a point p that belongs to the cell such that d(p,xo) = delta/2. We are justified in assuming this because distance of any two points in the cell is less than delta. Let us have a neighborgood N with radius h/2 around p. Then any point m on this Neighborgood shall be inside the sphere with radius delta/2+ h2/ and center x0. (Using triangular inequality). Hence, p is internal to the sphere. In other words all points that belong to cell shall be internal to the sphere N. And N is open because all neighborhoods are open sets. Hence the open set shall contain the open n-cell and also the closed n-cell.

    I know it does not sound very precise. But any suggestions to make it better? Thank you.
    Last edited: Sep 30, 2007
  2. jcsd
  3. Sep 30, 2007 #2


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    Why not just show that for e.g. a1 and b1 there exists c1 and d1 such that a1<c1<d1<b1. Do this for all i and consider the open cell (c1,d1)x(c2,d2)x..x(cn,dn) and the closed cell [c1,d1]x[c2,d2]x..x[cn,dn]?
  4. Sep 30, 2007 #3
    That makes sense too. I suppose in a different way, I was trying to prove the same. But putting it so neatly sounds lot better.
    Last edited: Oct 1, 2007
  5. Sep 30, 2007 #4


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    Yeah, if you talking cells talk cell language. If you are talking balls talk ball language. They are both the same in the end but shifting back and forth just makes things needlessly complicated.
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