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I asked a probability question here relating to the (topological?) idea of a "box" in $\mathbb{R}^k$. In the comments, user "BGM" said

It just mean a neighborhood / open-ball in $\mathbb{R}^k$,

and the user "NCh" said

Box in $\mathbb{R}^k$ containing the origin is the set of $t_1, \dots, t_k$ such that

$$|t_1|\leq a_1, \; |t_2|\leq a_2, \;\ldots,\; |t_k|\leq a_k.$$

"NCh" then said, put into my own words, that these two views are equivalent, since a box is a rectangle, and every rectangle contains a ball. After thinking about this, I think I had some good thoughts, which I would like to share in order to verify that my understanding is correct.

The volume of a ball will obviously always be less than the volume of a box, given their geometries, right? So, if we take a box $A$ and the largest possible ball $B_1$ that can fit inside said box, then we have that $A - B_1 = R_1 > 0$ remaining volume. Then, we can always take the amount remaining $R_1$, and there is always a ball $B_2$ that we can fit also in that box, so that $R_1 - B_2 = R_2 > 0$, where $R_2 < R_1$. And if we repeat this to infinity, then, at the limit, the sum of the volumes of all the balls approximates the box, right? Is my understanding of the topology here correct? Thank you.

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Dom Fomello
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  • It is an elementary result in functional analisis that in finite dimension all the normes are equivalent, its definissent the same topology. – Piquito Feb 23 '20 at 11:01
  • @Piquito Oh, ok. I haven't studied functional analysis yet, so this is beyond my understanding at the moment. What is the theorem from functional analysis called? And what do you mean by "definissent"? Do you meant "definitely the same topology"? – Dom Fomello Feb 23 '20 at 11:03
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    See at https://math.stackexchange.com/questions/57686/understanding-of-the-theorem-that-all-norms-are-equivalent-in-finite-dimensional – Piquito Feb 23 '20 at 11:18
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    All norms are equivalent in finite dimension. Anyway I think that what "Nch" meant is that if you take a ball or a box, since you can always take a box inside a ball and viceversa, your definition of moment generating function does not change, so you could use ball instead of box, if you prefere.. – Peanojr Feb 23 '20 at 12:40
  • @Peanojr I understand. Thanks. – Dom Fomello Feb 23 '20 at 12:43

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