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Open n-cell is open?

  1. Sep 26, 2007 #1
    1. The problem statement, all variables and given/known data

    An open-n-cell in R^n defined as (a1,b1) x (a2,b2) x ....... (an, bn). Prove that every open n-cell is open.

    2. Relevant equations

    3. The attempt at a solution

    I was thinking of using induction. Clearly the base case n=1 is open as (a1,b1) is open in R1. It is a segment. But my problem is trying to picture an open-n-cell. Some hint would be really good.
  2. jcsd
  3. Sep 26, 2007 #2


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    It's just a box without it's boundary. Just pick x=(x1,x2...xn) in the n-cell. Then x1 is in (a1,b1), x2 in (a2,b2) etc. Doesn't that make it pretty easy to find a neighborhood of x contained in the n-cell?
  4. Oct 1, 2007 #3
    Probably the last thing on this problem. The cell concept is soemwhat not easy for me to grasp at this moment. In order to prove that a given point is internal to a cell, I need to prove that all points in the cell are internal points. In other words, the neighborhood of a point x is contained in the cell (as you mentioned).

    Now, I suppose we are talking of neighborhood defined as Nh(x) , which implies a sphere of radius h center point x. I can prove the following, which is sufficient but I feel it is not elegant. Any other comments would be useful as well.

    Let us assume that the distance between A (a1,a2,..an) and B (b1,b2..bn) is delta. Then the distance between any two points in the cell is less than delta ( in open cell). Let us take a mid point of the cell Q {ai+bi/2} for each i. Then the distance of any point within the cell shall be less than delta/2.

    Now consider any other point x within the cell with a distance delta/2 -h from Q. Hence, Nh(x) around x would be a sphere of radius h with x as center point. Take any other point y on this sphere. Then d(y,Q) < delta/2-h+h < delta/2. Hence, any point y of Nh(x) belongs to the cell. Hence, x is an internal point. So, for all points within the cell we can find a radius, "e" such that the Ne(x) is a subset of the cell. Hence, all points are internal points. So, cell is open.
  5. Oct 1, 2007 #4


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    I don't know what you're supposed to used here, but isn't (x, y) open in |R for all x < y in |R and doesn't that make (a1, b1) x (a2, b2) x ... x (an, bn) a basis element for the product induced topology?
  6. Oct 1, 2007 #5

    matt grime

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    You are supposed to show that the product topology on n is the same as the usual metric topology.

    Try the case of R^2 first. An open cell is the inside of a square. Given a square, and an interior point, you must find an open ball, i.e. a circle around it inside the square. Then generalize this.
  7. Oct 1, 2007 #6


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    "In order to prove that a given point is internal to a cell" you do not have to "prove that all points in the cell are internal points"! For example, the point 1/2 is interior to [0,1] but it is not true that "all points in the cell are internal points"- 0 and 1 are in the set and are not internal points. Perhaps that was just a typo. In order to prove that the set is open you need to "prove that all points in the cell are internal points". If p is a point is (a,b)x(c,d)x(e,f)x... the p is made of "coordinates" (x1,x2,x3,...) such that x1 is in (a,b), x2 is in (c,d), x3 is in (e,f), etc.
  8. Oct 1, 2007 #7
    I understand that to prove a point x as internal point, I do not need to prove all points are internal points. I can just take x and find a neigborhood of x that is contained in the cell. And if all points of cell are internal points, then the set/cell is open. If that was not apparent from my earlier response, then I think I should rewrite it myself.
    Anything else that you would like to add? Thanks!
    Last edited: Oct 1, 2007
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