I was reading Is this proof that all metric spaces are Hausdorff spaces correct? to see how the statement is proved, but I have a question. Is it possible to generalize that any ball in metric space is open? How do we know the "open set" in metric space? Is it a convention to say that topology of metric space is topology generated by every ball around every point? I can't think it right now but would there be any topology in metric space that some ball around some point is not open?
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open balls are a base for the topology of a metric space – J. W. Tanner Jan 14 '21 at 01:59
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But maybe that's the one that needs to be proved as well using axioms. right? – able20 Jan 14 '21 at 02:01
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Given a metric space $(X,d)$, the topology induced by the metric $d$, i.e., the topology generated by the open balls: $$ B(x,r)=\{y\in X\mid d(y,x)<r\} $$ is called "metric topology" of $X$.
By definition, any open ball is an open set under this topology.
but would there be any topology in metric space that some ball around some point is not open?
As a set, you can of course define other topology $\tau$ on $X$. An "open ball" of $(X,d)$ is then not necessarily an "open" set in $\tau$.