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Show that unbounded open ball $$B=\{y:d(x,y)>r\}$$ is an open set.

In the case of bounded open ball $$B=\{y:d(x,y)<r\}$$ using triangle inequality we can easily show that, it is indeed an open set. But I left helpless, when I try to use the same technique for unbounded open ball.

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marya
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2 Answers2

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Since you didn't state which definition of openness you're referring to, i'll use this one: A set $A$ is open if for all $a \in A$ there exists an open (bounded) ball $B$ such that $a \in B \subseteq A$.

Now, let $B' = \{ y : d(x,y) > r\}$ and $b \in B'$. Let $r' := d(x,b) -r > 0$. Then we claim that $\{y: d(b,y) < r'\} \subseteq B'$. Take $y$ such that $d(b,y) < r'$. Then

$$d(x,y) + r' > d(x,y) + d(y,b) \geq d(x,b) = r + r'$$ Hence, $d(x,y) > r$, i.e. $y \in B'$.

j4GGy
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Let $(M, d)$ be a metric space; let $x \in M$. Then $d(x, \cdot): M \to \Bbb{R}$ is continuous on $M$. If $r \geq 0$, then $]r, \infty[$ is open in $\Bbb{R}$; hence the preimage of $]r, \infty[$ under $d(x, \cdot)$ is open in $M$.

Yes
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  • To be honest, your answer is complicated than the question(to me). Could you give me an explanation please? – marya Nov 24 '15 at 10:21
  • @Sucre It can be proved that every metric defined on a set $M$ is continuous on $M \times M$. – Yes Nov 24 '15 at 12:40
  • @Sucre I believe you can some proof of that claim at this S.E.. :) – Yes Nov 24 '15 at 13:49