In the infinity norm, if three balls intersect pairwise, their total intersection is nonempty.

Question

I am trying to prove that in $\mathbb{R}^n$ equipped with the infinity norm $d_\infty(x,y)=\max_{i=1,...,n}|x_i - y_i|$ if I have three balls $B_1$, $B_2$ and $B_3$ and point $x \in B_1 \cap B_2$ a point $y \in B_1 \cap B_3$ and $z \in B_2 \cap B_3$ then there must exist a point $p \in B_1 \cap B_2 \cap B_3$. With the standard norm this isn't the case, I can draw three balls that overlap pairwise but have no common intersection, but with this infinity norm 'balls' look like 'cubes' and I cannot. I can imagine why this must be true but I was wondering if someone could give me a formal proof, or please correct me if I'm wrong, thanks

[edited to clarify the question]

Answer 1

Original question (paraphrased): Show that if $B_1,B_2,B_3\subset\Bbb{R}^n$ are balls w.r.t. the infinity norm, then if a point lies in the intersection of each pair of balls, then it lies in the intersection of all three.

Proof. If a point lies in the pairwise intersection of any two balls, then it lies in all three balls, hence it lies in their intersection.

Note that this has nothing to do with $\Bbb{R}^n$, the infinity norm or balls. It is an elementary consequence of the definition of the intersection of (finitely many) sets.

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After the question has been edited into a different question:

New question (paraphrased): Show that if $B_1,B_2,B_3\subset\Bbb{R}^n$ are balls w.r.t. the infinity norm, then if their pairwise intersections are nonempty, then their intersection is nonempty.

Proof. In the infinity norm on $\Bbb{R}^n$, for $x,y\in\Bbb{R}^n$ and $r\in\Bbb{R}_{\geq0}$, we have $$y\in B(x,r)\qquad\Leftrightarrow\qquad \forall i:\ y_i\in B(x_i,r).$$ So it suffices to prove the claim for $n=1$. But this is obvious; if three intervals in $\Bbb{R}$ intersect pairwise, then there is a point contained in all three intervals.

Answer 2

I will use the argument used in some exercise related with Helly theorem (cf. Discrete and polyhedral geometry).

We consider first coordinate-orthogonal projections of the balls $B_i$.

That is, we have three intervals s.t. each pair has nonempty intersection. Note that three intervals have nonempty intersection $x=(x_i)$ ( When $B_i =[a^i_1,b^i_1]\times [a^i_2,b^i_2]\times \cdots $, then this means $a^i_1\leq x_1\leq b^i_1$ for all $i$) So we consider second coordinate. In the long run we have $a^i_\alpha\leq x_\alpha\leq b^i_\alpha$ for all $i,\ \alpha$.