We have in $\mathbb R$ an axiom (sometimes a theorem if we start from another system of axioms) that:

If $(I_n)_{n=1,2,...}$ is a sequence of nested closed intervals such that $l(I_n) \to 0$ when $n \to + \infty$ then their intersection is non-empty (really, the intersection is only one point). Here, $l$ is the length of an interval.

However, if the intervals are not closed, then this need not to be true, as an example $n \to (0, \dfrac {1}{n})$ shows. Here, the intersection is empty.

There is virtually no reason why we shouldn´t consider sequences of nested sets in $\mathbb R^n$, and, most probably, so much is already known about them.

I guess that straightforward generalization of above stated theorem (axiom) also is true, that is:

If $(C_n)_{n=1,2,..}$ is a sequence of nested closed convex sets in $\mathbb R^n$ such that $V(C_n) \to 0$ when $n \to + \infty$ then their intersection is non-empty (it should be one point, right?) Here, $V$ is the volume (or, would calling it $n$-volume be more appropriate?) of a closed convex set.

So, is this true? Is the intersection of nested closed convex sets in $\mathbb R^n$ such that $V(C_n) \to 0$ when $n \to + \infty$ non-empty? Is it one point?

You can start with whatever legal definition of a convex set you want, although, I think that definition:

A set is convex if for every two points $x,y$ from that set we have that line segment that connects (joins) those two points is contained in that set.

is the simplest one.

Notify me if this question can be improved, because, it may well be that there are some issues that I am not aware of, and, I am not completely sure that all is right with definitions I gave.

Also, I take as an implicit assumption that every closed convex set in $\mathbb R^n$ has an $n$-volume.

If this is not true (and I think it is) then suppose that we are dealing only with closed convex sets in $\mathbb R^n$ that have an $n$-volume.

Also, in our considerations, all closed convex sets are bounded.

Edit: By the comment of Emma, it seems clear that the intersection is only-non empty, it need not be one point, because, if closed convex sets are of dimension $n-k$, where $1\leq k <n$ in $\mathbb R^n$ then their $n$-volume is $0$.

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    $\begingroup$ Wouldn't a line segment in $$\mathbb{R}^2$ be an example of a closed bounded convex set with no "2-volume"? $\endgroup$ – emma Jun 21 '19 at 4:43
  • $\begingroup$ @emma Isn´t $2$-volume of such a set equal to zero? $\endgroup$ – Grešnik Jun 21 '19 at 4:44
  • $\begingroup$ Ah, sorry, I interpreted the condition on having volume as having non-zero volume. $\endgroup$ – emma Jun 21 '19 at 4:45
  • $\begingroup$ @emma No problem, I meant that $n$-volume must only exist. Thank you. $\endgroup$ – Grešnik Jun 21 '19 at 4:46
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    $\begingroup$ @emma Actually, you helped me to improve the question, see an edit. $\endgroup$ – Grešnik Jun 21 '19 at 4:53

You seem to be barreling towards Cantor's Intersection Theorem. There are two variants: the compact version and the complete metric space version. Here's the compact version:

Theorem: Suppose $X$ is a topological space, and $C_n$ is a sequence of non-empty closed, compact sets such that $C_{n+1} \subseteq C_n$ for all $n$. Then $$\bigcap_{n=1}^\infty C_n \neq \emptyset.$$

In the setting of $\Bbb{R}^n$, compact sets are simply closed and bounded sets. There's no need for convexity, and there's no need for length/area/volume/Lebesgue measure to enter into it. However, you don't get the uniqueness of the intersection.

On the other hand, we have the complete metric space version:

Theorem: Suppose $X$ is a complete metric space (i.e. Cauchy sequences converge), and $C_n$ is a sequence of non-empty, closed and bounded sets such that $C_{n+1} \subseteq C_n$ for all $n$. Moreover, assume $$\operatorname{diam} C_n := \sup_{x, y \in C_n} d(x, y) \to 0 \text{ as } n \to \infty.$$ Then $$\left|\bigcap_{n=1}^\infty C_n\right| = 1.$$

If you're not familiar with metric spaces, $d(x, y)$ refers to the distance from $x$ to $y$. In the context of $\Bbb{R}$, this refers to the quantity $|x - y|$. In $\Bbb{R}^n$, it (usually) refers to the Euclidean distance between $x$ and $y$ (though other distances are sometimes used).

Note the latter has a stronger conclusion: the sets have a unique point of intersection. In $\Bbb{R}^n$, the condition of being closed and bounded is equivalent to being compact, so the only real difference is the presence of the diameter tending to $0$.

To answer your actual question, if we assume instead that $V(C_n) \to 0$ instead of $\operatorname{diam} C_n \to 0$, then unfortunately, you cannot even guarantee that the intersection is non-empty. For example, for each $n$, take the half-lines $L_n = \{(x, 0) : x \ge n\}$. Note that such sets are not compact, but have zero area/volume/measure in two or more dimensional space. Their intersection is empty.

  • $\begingroup$ Yes, I am aware that intersection need not be a singleton. I made an edit, maybe at the time when you were writing your answer. Good answer, thank you. +1 $\endgroup$ – Grešnik Jun 21 '19 at 5:02
  • $\begingroup$ It seems that intersection, generally, is $n-k$-dimensional closed convex set, where $1\leq k<n$. $\endgroup$ – Grešnik Jun 21 '19 at 5:04
  • $\begingroup$ @Grešnik It's certainly true that intersections of closed and convex sets are closed and convex. If they are all non-empty, then the intersection is non-empty (by CIT). If the intersection had an affine hull of dimension $n$, then this could only occur if it had non-empty interior (needs proof, but it is true in $\Bbb{R}^n$), i.e. only if it contains a ball. Such a ball has strictly positive measure and is contained in each $C_n$, so $V(C_n) \to 0$ would be impossible. $\endgroup$ – Theo Bendit Jun 21 '19 at 5:10
  • $\begingroup$ Ah, I do not know what is an affine hull, I am just an amateur with not-so-good questions, I even do not study math or any math-related college, maybe I shouldn`t participate here with my non-expert questions. $\endgroup$ – Grešnik Jun 21 '19 at 5:14
  • $\begingroup$ @Grešnik The affine hull is the smallest affine subspace of $\Bbb{R}^n$ that contains the convex set. Affine sets are things like points, lines, planes, etc. Basically, the dimension of the affine hull is the only way I know how to talk about the "dimension" of a convex set. For example, a line segment's affine hull is a line, which is $1$-dimensional, so I would say the line segment is $1$-dimensional. $\endgroup$ – Theo Bendit Jun 21 '19 at 5:18

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