TITLE: metric spaces open set
QUESTION [0 upvotes]: 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

REPLY [2 votes]: 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 $z \in B(x,\delta)$.
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