## Every open set contains both open and closed cell.

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.

 PhysOrg.com science news on PhysOrg.com >> 'Whodunnit' of Irish potato famine solved>> The mammoth's lament: Study shows how cosmic impact sparked devastating climate change>> Curiosity Mars rover drills second rock target
 Recognitions: Homework Help Science Advisor Why not just show that for e.g. a1 and b1 there exists c1 and d1 such that a1
 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.

Recognitions:
Homework Help