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I am just learning what a topology is and from what I have understood up till now is that a topological space is nothing but a set with a notion of nearness that is given introducing open sets.

Ok, so, in the definition of manifold that I have seen (in Wald's general relativity book) a manifold is constructed mapping subsets of the manifold to be set with open subsets of $\mathbb{R}^n$.

This way we introduce a topology in our manifold to be, since we can use the notion of open balls in $\mathbb{R}^n$ to define open balls of the manifold and hence we get a topology.

Now, we can add more structure to the manifold endowing it with a metric and making it a metric space.

Now my question. I know that a metric induces naturally a topology. But we already had a topology before itroducing the metric structure, so, are this two topologies the same topology?

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  • $\begingroup$ Doesn't it matter if the metric is a riemannian one or a lorentzian one? $\endgroup$ – PhoenixPerson Aug 8 '14 at 16:06
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    $\begingroup$ Of course, the metric must be Riemannian. You don't get a metric space otherwise. $\endgroup$ – Zhen Lin Aug 8 '14 at 16:06
  • $\begingroup$ in place of a riemmanian metric you could (as is done in general relativity) define a pseudo-riemmanian metric right? $\endgroup$ – PhoenixPerson Aug 8 '14 at 16:08
  • $\begingroup$ No, that wouldn't work. In a pseudo-Riemannian metric there can be distinct points that are distance zero apart, which cannot happen in a metric space. $\endgroup$ – Zhen Lin Aug 8 '14 at 16:10
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    $\begingroup$ In terms of pseudo-Riemannian manifolds, you may be interested in the Alexandrov topology. $\endgroup$ – Willie Wong Aug 21 '14 at 10:44
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For the Riemannian case this is surely true, you may find it in "Foundations of Differential Geometry" by Kobayashi and Nomizu, volume 1, page 166, proposition 3.5.

In the pseudo-Riemannian case this is not true as indicated in a comment above by Zhen Lin (the topology induced by the pseudometric will no longer be Hausdorff, for instance).

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