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In a metric space $(X,P)$

$$x\in X$$ and

and $$\delta >0$$ are fixed

Show that $B(x,\delta)$ is open set.

I am not entirely sure how to do this problem. The first step I did

is write down the definition

$B(x,\delta)=[y\in X: P(x,y)<\delta$

But now I am not certain what to do.

So then I have $y\in X$

and now I must chose an $r>0$

so that $B(y,r)$$\subset B(x,\delta)$

I think but I am not sure how

Fernando Martinez
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2 Answers2

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To show that $B(x,\delta)$ is open, it's enough to show that it contains an open ball around every one of its elements — that is,

For any $y \in B(x,\delta)$, there is $\xi > 0$ such that $B(y, \xi) \subseteq B(x,\delta)$.

This follows from the triangle inequality. For $y \in B(x,\delta)$, let $d_{xy} = P(x,y)$, and let $\xi = \delta - d_{xy}$. If $z \in B(y, \xi)$, then $$ \begin{align} P(x,z) &\le P(x,y) + P(y,z) \\ &< d_{xy} + \xi \\ &= d_{xy} + \delta - d_{xy} \\ &=\delta \text{,} \\ \end{align} $$ so $z \in B(x,\delta)$.

BrianO
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  • I see you use the assumption that $z$ $\in B(y,that letter)$ and then use triangle inequality that $P(x,z)$ is less than $\delta$ thus showing z element is in both sets. Yes that makes sense. – Fernando Martinez Oct 29 '15 at 20:38
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    Haha, yes. That letter is \xi. "ksee". – BrianO Oct 29 '15 at 20:42
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Let point $x\in B_\delta(c)$ be given, then it has a distance of $r$ from c, we can then define $\epsilon < \delta - r$ and we have then that $B_\epsilon(x)\subset B_\delta(c)$. We know this is inside because $\epsilon$ is less than the remaining distance to the "edge" of the first sphere so it must be contained as any point within this new ball wil be a shorter distance than $\delta$ from our $c$ Which means we can always find these sets within each other and therefore they make up open sets.

Zelos Malum
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