# Is the Open Box Open?

A quick proof of something that bothered me in basic topology. Assume the standard topology on â„n based on open balls. What about an open box? I.e. all points in â„n such that a1 < x1 < b1; a2 < x2 < b2;…;an < xn < bn. Is this an open set? I.e. can you build it up out of a union of open balls? Or, more colloquially, can you pack a square hole with round pegs without leaving any gaps?

Short answer: yes, if the balls can overlap and you have infinitely many of them. Long answer:

In â„1 this is obvious because the open cube is an open ball. I.e. they’re both just open intervals on the real line.

In â„2 or higher we need to generate a set (possibly infinite) of open balls that contains every point of the open box. This sounds like calculus in reverse (fill the rectangular space with progressively smaller curved figures and take the limit) but in fact it’s easier.

All we need to do is establish that every point in the box is contained within some open ball that is also contained in the open box. Then the union of all these open balls is the open box.

So given an arbitrary point in the open box, just form the open ball centered on the point with radius equal to the distance to the nearest edge of the box. By construction this is contained within the box, and we can construct such a ball for every point in the box. Take the union of all these balls. By definition of topology, the union of open sets is open, so the open box is open. QED.

Bonus: this argument isn’t “tight”. I.e. I didn’t actually use the fact that the region in question was a box. The same argument applies to any space in â„n that does not contain its boundary points.

This construction does need an uncountably infinite number of open balls, and in fact every single point in the box is contained in uncountably many open balls. Pure math is rarely economical, but this does seem excessively profligate. I wonder if it’s possible to do this with only countably many balls? Perhaps you could generate a sequence of balls that progressively fill the space, and show that any given point is enclosed within a ball after only a finite number of steps? The geometry seems quite involved, but it might be doable if you first proved it in â„2, and then expanded the proof from there.

One other question: can you do this without intersecting balls? The answer to this is no, some balls have to overlap. To see this just consider a point on the boundary of any given ball B. This is still in the box, but because B is open the point is not in B, so we have to put it in some other open ball. However any neighborhood of the point will inevitably intersect with B.

### One Response to “Is the Open Box Open?”

1. Oisin McGuinness Says:

Hi Rusty,

>> if itâ€™s possible to do this with only countably many balls

The set of balls with centers in Q^n (Q = rational numbers) and rational radii is countable. And they will do the job for you.

Oisin