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A topological space is defined as a non-empty set $X$ together with a given collection of subsets $T$ (topology) of $X$, such that,

(i) any union of these subsets is one of the subsets.

(ii) any finite intersection of subsets is one of the subsets.

(iii) these collection of subsets contain $\emptyset$ and $X$.

Thus the pair $(X,T)$ together defines a topological space.

Now my question is, if a manifold $M$ is defined as a topological space then what are the collection of subsets of $M$ that specifies the generic topology $T$ in a generic $M$?

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    $\begingroup$ This doesn't make sense! The manifold $M$ comes with a topology i.e. with the collection of subsets $T$. $\endgroup$ – MBN Apr 20 '14 at 13:50
  • $\begingroup$ @MBN- In the definition of a manifold M there are collection open sets that "looks locally" Euclidean and cover M. But that cannot exhaust the topology. Right? Because the topology must also contain finite unions and intersections of these sets, the empty set and M as well right? $\endgroup$ – SRS Apr 20 '14 at 14:49
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    $\begingroup$ @Roopam: they don't exhaust the topology, but generate it $\endgroup$ – Christoph Apr 20 '14 at 15:09
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    $\begingroup$ The topology is the collection of open sets of the space (by definition, a member of the topology is called an "open set"). But when a manifold "locally looks Euclidean", you're talking about charts...the image of a chart is itself an open set in the manifold, which requires a topology to talk about... $\endgroup$ – Alex Nelson Apr 20 '14 at 16:10
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The topology of the manifold is given by unions of pre-images of open subsets of $\mathbb R^n$ under the charts of an atlas.

That is, instead of the requirement that $M$ is a priori a topological space covered by homeomorphisms to $\mathbb R^n$, we can use an atlas of bijective charts with homeomorphic transition maps to define its topology.

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