Topology induced by a metric?

[St41n]

When we say that a metric space (X,d) induces a topology or "every metric space is a topological space in a natural manner" we mean that:
A metric space (X,d) can be seen as a topological space (X,τ) where the topology τ consists of all the open sets in the metric space?
Which means that all possible open sets (or open balls) in a metric space (X,d) will form the topology τ of the induced topological space?
Is that correct?

 

[quasar987]

Well, yes, but what are the open sets in (X,d)?

They are those sets A such that for every a in A, there is a r>0 such that B(a;r)={x in X : d(x,a)<r} is entirely contained in A.

You can verify that these sets form a topology on X.

 

[rasmhop]

Which means that all possible open sets (or open balls) in a metric space (X,d) will form the topology τ of the induced topological space?
Is that correct?

Saying: "or open balls" is incorrect, the rest is correct. We say that a topology T on a space X is induced by a metric d on X iff the open balls generated by d forms a BASIS for the topology T (i.e. a set U is open iff it's a union of open balls).

 

[St41n]

Oh I see, that's because the collection of open balls is a subset of the collection of all open sets.
It makes sense, thank you both for your time!