boundary of a boundary

[lethe]

i was getting into an argument at the bar with one of my buddies. the boundary of a boundary is always zero. or, if that doesn t make sense to you, it is always the empty set. i claimed that this statement is not obvious, and he claimed it was. i m really not sure, so what do you guys think?

[hlreed]

If a boundary is like a sack, then a sack of a sack contains a sack.

A sack is a bag or other container or set.

A box full of boxes is a box.

I don't know what a mathematical boundary is here.

[James R]

Can you give us a mathematical definition of "boundary"? If you can do that, then I imagine the answer will become obvious.

[lethe]

boundaries.

the boundary of a disk is a circle.

the boundary of a line segment is the two endpoints (B-A, actually, if A and B are the endpoints)

the boundary of a filled in square is a the outline of the square. what is the boundary of the outline of a square? (A-B) + (B-C) + (C-D) + (D-A) = 0. so the boundary of the boundary of a filled in sqaure is zero.

the boundary of a circle is also zero. to see that easiest, just recognize that a circle is homeomorphic to a square.

here is the definition of a boundary. if A is a subspace of X, the the boundary is Cl[A] intersect Cl[X-A], where Cl indicates closure (usually with a bar above). and X-A is just the complement of X. i.e. the whole space, but with X removed.

recall what closure means: the closure of the open interval (0,1) is the closed interval [0,1].

let s do the boundary of (0,1). the closure is, as i said, [0,1]. X-A is the two intervals (-inf,0], [1,inf), which is already closed, so that s also its closure.

the intersection of those two spaces is simply {0,1}, which is what you should expect for a boundary.

to prove that the boundary of a boundary is always the empty set is not too hard. let s try it.

let Y be the boundary of X. so Y = Cl[X] intersect Cl[A-X]. let s check the boundary of Y. first notice that since Cl[X] and Cl[A-X] are both closed sets, their intersect is also closed. so Cl[Y] = Y. also, Cl[A-Y] = Cl[A]-Cl[Y]. A is the whole space, so it is its own closure, and Y is closed, so it is also its own closure. thus the boundary of the boundary of X is Y intersect (A-Y). and obviously the intersection of any set and its complement is the empty set.

so it s an easy enough set theoretic proof. but what i really wanted to know was whether it is obvious from a visual geometric standpoint. can you imagine a set in some space? is it obvious to you that the boundary of that set must itself have no boundary?

another example.

the boundary of the ball in R³ is the 2-sphere. the 2-sphere has no edges. no boundaries. so the boundary of the boundary of the ball is the empty set. can you convince yourself, without recourse to any set theory, that this is always true?

i cannot, but i am not sure that such an argument does not exist, and that is what i m asking here.

[James R]

I'm happy with the proof from set theory. I don't think I really grasp why you would think differently in a real situation.

Can you give me any example of a situation where the boundary of a boundary would not be empty?

[lethe]

Originally posted by James R
I'm happy with the proof from set theory. I don't think I really grasp why you would think differently in a real situation.

Can you give me any example of a situation where the boundary of a boundary would not be empty?

no i cannot give you any such examples. in fact, i can sleep at night quite sure in the knowledge that no such example exists, since i have a proof.

and yes, i am also happy with the proof. i have no questions that the statement is true.

what i m grasping at is: how believable is the statement that the boundary of a boundary is always zero?

i told my friend this fact, and he immediately said "oh that s obvious. of course the boundary of a set cannot itself have a boundary". it s a true statement, but i think it is not an obvious statement. but perhaps if you have enough insight, you can "just see" it, as my friend did.

that is what i m trying to discover. can you see that it is impossible? i m really just after you guys gut feelings about it. well, those of you that have at least a passing intuitive notion of what a boundary is. so far the few people who have been inclined to vote in the poll seem to be in agreement with me, that the statement is not obvious.

[James R]

As a physicist, I have a number of intuitive ideas about what a boundary is. Those ideas don't really mesh with the set theory definition of "boundary" as you've presented it. So, for me anyway, I have no particular gut feeling about what should be right one way or another. My only option is to fall back on the maths, which allows me to see the truth of the statement from the given definition of the term.

The real question here may be: to what extent does your mathematical definition of a "boundary" correspond to anything physical? This is relevant because much of our intuition is based in the physical.

[everneo]

what is the boundary of a circle with increasing radius :
r = 0.9+0.09+0.009+0.0009+...

[HallsofIvy]

"A" circle doesn't have increasing radius- the definition of "circle" is that it has a constant radius. In any case, you have given r= .9+ .09+ ... which means r= 1.

The boundary of any circle, whatever the radius is empty. Of course, the boundary of a DISK of radius r is a circle of radius r.
(And, as we all know, the boundary of a boundary is empty!)

I remember seeing the proof that the "boundary of a boundary" is empty in algebraic topology. No, it's not trivial. Whether a person would say it is "obvious" or not depends strongly on that person.

[everneo]

The closure of (0,1) is [0,1] as lethe explained. In maths applying limits is equivalent to quantising in real world. Can u quantize space or time. Whatever limits in space is better approximation suitable for the purpose at hand. The concept of boundary (in the sense it is unique) itself is an approximation, may be the best one for the purpose - but not exact one in real world.

[Crisp]

Hi lethe,

I don't think it is obvious at all that the boundary of a boundary is the empty set... Like you said, the boundary of a disc is the circle surrounding it, but since this is a 1D object, there could have been another boundary (this is not the case for a circle). I could imagine that in some abstract space, the boundary of a n-dim object reduces the dimension to n-1, where another boundary could be possible.

So for your "visual" or "geometric" interpretation... I don't know if it is that obvious at all...

Your proof seems ok btw, and I also think I once saw a proof in a topology course that boundary(boundary(A)) = 0, at least in all "normal" situations... Perhaps if you equip your space with some strange metric that it no longer holds...

Bye!

Crisp