# Intersection of nested closed bounded convex sets in Euclidean space

I read that in a complete Euclidean space - i.e. a normed real space with the norm induced by the scalar product - any sequence of nested bounded non-empty closed convex sets has a non-empty intersection, but I can't manage to prove it to myself.

Has anybody any ideas or links to online proofs?

### Context

The space is not assumed finite dimensional. At the point where I am in the text, Kolmogorov and Fomin's, they haven't defined a Hilbert space yet.

I have not yet learned about the weak topology and reflexivity.

• "Euclidean space" normally refers to only finite-dimensional spaces (these are always complete). Since you explicitly mention completeness, are you considering infinite-dimensional spaces? (Then the term would be "Hilbert space".) – Daniel Fischer Aug 17 '14 at 12:45
• Have you already learned about the weak topology and reflexivity? – Daniel Fischer Aug 17 '14 at 13:06
• Pity. That would have made it easy. (These sets are all compact in the weak topology, and the intersection property for compact sets is direct.) Let me think if I see a nice way without using that. – Daniel Fischer Aug 17 '14 at 15:26
• math.stackexchange.com/questions/79628/… – Jonas Meyer Aug 17 '14 at 22:09
• @JonasMeyer Thanks for linking; the answer there is based on weak topology, though. I think my answer works... – user147263 Aug 17 '14 at 22:12

## 1 Answer

Let $E_n$, $n\ge 1$, be these nonempty closed convex sets. We want to pick a point $x_n$ in each set so that they form a Cauchy sequence. If the diameter of $E_n$ tends to zero as $n\to\infty$, the Cauchy property comes automatically. Otherwise, we have to make some intelligent choices of $x_n$. I present two versions; the second, pointed out by Daniel Fischer, is more slick.

### Close to the "center of the set"

One idea is to pick $x_n$ close to the "center" of each set. To make this precise, let $$r_n = \inf\{r>0: \exists x\in E_n \text{ such that } E_n\subset B(x,r)\}\tag1$$ Here $B(x,r)$ is closed ball of radius $r$ centered at $x$. (The number $r_n$ is sometimes called the Chebyshev radius of $E_n$.) Observe that $r_n$ is a decreasing sequence of positive numbers, so it has a limit, $r_n\to r_*$.

As usual in infinite dimensions, we don't know if the infimum (1) is actually attained. So we must provide some slack: pick $x_n\in E_n$ such that $E_n\subset B(x_n, r_n+2^{-n})$.

I claim that the sequence $(x_n)$ is Cauchy. Indeed, suppose there is $\epsilon>0$ such that there are arbitrarily large indices $n<m$ for which $\|x_n-x_{m}\|\ge \epsilon$. Then $$E_m \subset B(x_n,r_n+2^{-n})\cap B(x_m,r_m+2^{-m})$$ Observe that both radii here can be made arbitrarily close to $r_*$ by choosing $n,m$ large. Using the parallelogram law, you can show that $$B(x_n,r_n+2^{-n})\cap B(x_m,r_m+2^{-m})\subset B\left(\frac12 (x_n+x_m),\rho\right)$$ for $\rho<r_*$, thus arriving at a contradiction.

(Draw a picture of two intersecting balls of nearly the same radius: if the distance between their radii is substantial, the intersection is contained in a ball of smaller radius).

### Close to the origin of the space

Another idea is to pick $x_n$ of small norm. Let $$R_n = \inf\{\|x\| : x\in E_n \}\tag2$$ Observe that $R_n$ is an increasing sequence of positive numbers, so it has a limit, $R_n\to R_*$.

Pick $x_n\in E_n$ such that $\|x_n\| < R_n+2^{-n}$.

I claim that the sequence $(x_n)$ is Cauchy. Indeed, suppose there is $\epsilon>0$ such that there are arbitrarily large indices $n<m$ for which $\|x_n-x_{m}\|\ge \epsilon$. Then $$\frac{x_n+x_m}{2}\in E_m$$ by convexity. Using the parallelogram law, you can show that $$\left\|\frac{x_n+x_m}{2}\right\| <R$$ when $m,n$ are sufficiently large, thus arriving at a contradiction.

• You can also take the element of smallest norm in $E_n$ as the point $x_n$, that is a Cauchy sequence by pretty much the same argument as the existence. Perhaps that's easier to see. (And I'm happy to note that you heeded my advice regarding the lone upvote ;) – Daniel Fischer Aug 17 '14 at 22:30
• @DanielFischer Indeed, that was my first impulse but I somehow decided it would not work; now I have no idea why I thought so. – user147263 Aug 17 '14 at 22:32
• (For each $n$, a unique $x_n\in E_n$ exists such that $\|x_n\|=R_n$, not that that is needed here.) – Jonas Meyer Aug 18 '14 at 1:04
• @DavideZena Draw the parallelogram with center $(x_n+x_m)/2$ in which the points $0, x_m, x_n$ are three of the vertices. The fourth vertex will be $(x_n+x_m)$. Write down what the parallelogram law says. – user147263 Aug 18 '14 at 5:56
• @DavideZena If $\|x_n-x_m\|\ge \epsilon$, it follows that $$\left\|\frac{x_n+x_m}{2}\right\|^2\le \frac{\|x_n\|^2+\|x_m\|^2}{2} - \frac{\epsilon^2}{4}$$ The expression on the right is less than $R$ when $m,n$ are large enough. – user147263 Aug 18 '14 at 12:36