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Homework Statement
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.
Homework Equations
Def: An open set is a set which has all points as interior points
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.
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