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how can it be shown that the ball $B(a,5)$ may be a proper subset of $B(b,3)$ in a metric space, but if $B(a,6)\subseteq B(b,3)$, then they are equal?

Henno Brandsma
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John Stern
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    The only thing you'll use is the triangle inequality. For the second part, use that two sets are the same iff one is contained in the other and vice versa. – Rodrigo Dias Oct 08 '17 at 13:15
  • the admin edited question but in original version, B(a,6) is proper subset of B(b,3), no equality. – John Stern Oct 08 '17 at 14:00
  • @JohnStern If (as in original version) $B(a,6)$ is a proper subset of $B(b,3)$ then still it can be proved that $B(b,3)\subseteq B(a,6)$. But then we find that $B(b,3)$ is a proper subset of itself. This contradiction shows that $B(a,6)$ cannot be a proper subset of $B(b,3)$ and based on the false statement that it is everything can be concluded. – drhab Oct 08 '17 at 14:42

2 Answers2

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If $B(a,6)\subseteq B(b,3)$ and $x\in B(b,3)$ then $d(a,x)\leq d(a,b)+d(b,x)< 3+3=6$ so that $x\in B(a,6)$. This proves that $B(b,3)\subseteq B(a,6)$ and consequently $B(b,3)=B(a,6)$.

Now let set $(0,6)\subseteq\mathbb R$ be the underlying set of the metric space, and let it be equipped with usual metric $d(x,y)=|x-y|$.

Then $B(3,3)=(0,6)$ so it contains properly every proper subset of $(0,6)$.

The ball $B(0.5,5)$ is such a proper subset of $(0,6)$.

drhab
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For the second part suppose that $B(a,6)\subset B(b,3)$, we show that $B(b,3)\subset B(a,6)$. Let $x\in B(b,3)$, $d(a,x)\leq d(a,b)+d(b,x)\leq 3+3=6$.

Tsemo Aristide
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