# Are the topology of a manifold and the topology induced by the metric of a manifold the same?

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?

• Doesn't it matter if the metric is a riemannian one or a lorentzian one? – PhoenixPerson Aug 8 '14 at 16:06
• Of course, the metric must be Riemannian. You don't get a metric space otherwise. – Zhen Lin Aug 8 '14 at 16:06
• in place of a riemmanian metric you could (as is done in general relativity) define a pseudo-riemmanian metric right? – PhoenixPerson Aug 8 '14 at 16:08
• 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. – Zhen Lin Aug 8 '14 at 16:10
• In terms of pseudo-Riemannian manifolds, you may be interested in the Alexandrov topology. – Willie Wong Aug 21 '14 at 10:44