Every open ball $D_r(p)$ in a metric space $(M,d)$ is open in $\mathcal{T}?$

Question

For any metric space $(M,d),$ whose underlying topology is $(M,\mathcal{T}),$ it follows that for all $r>0,p \in M,$ an open ball $D_r(p)$ is open in $\mathcal{T}.$

Does this follows follows from inheritance? On the other hand, $\mathcal{T}$ is a set of open sets, in which any open ball $D_r(p)$ of a is a singleton. A singleton is open.

I'm not certain as to either.

Thanks :)

Answer 1

It depends on the topology $\mathcal{T}$ you are referring to. Usually by 'the underlying topology' it is meant the topology generated by the open balls of $M$, so the latter will be trivially open (by construction!). Singletons can be open, and such points are usually called isolated. Take for example any set with the discrete metric.

Answer 2

Given a metric $d$ on $X$, you either define the induced metric topology by declaring that it is the smallest topology that includes all balls $B_d(x,r) = \{y \in X: d(x,y) < r\}$, $x, \in X; r>0$ and then indeed all open balls are open by definition.

Or equivalently you directly define an open set in the metric topology as

$$O \subseteq X \text{ open iff } \forall x\in O: \exists r>0: B_d(x,r) \subseteq O\tag 1$$

In that case you need to prove that all sets of the form $B_d(x,s)$ are open: if $y \in O=B_d(x,s)$ check that we can take $r=s-d(x,y)$ in the definition (1), i.e. $B_d(y,r) \subseteq B_d(x,s)=O$ etc.

Both definitions amount to the same topology but the fact that open balls are indeed open is seen differently.