Thank you, yes I plan to do more!

Ah good catch, yes it should be that!

Same words in different theories. May have different meanings. But in this case ... First, let's just be clear about technical details. Given a topological space X, the answer to "is X a metric space?" is immediately "No", because the structure of matric space is NOT part of the structure of topological space. But people may ask that meaning "does X admits a structure of metric space compatible with its topology?" That's the technically correct question. In the other hand, "is a ring R an abelian group?" has the answer "yes, because part of the structure of a ring is the structure of abelian group". And even such statement can be pedantically contested, because if a ring is defined as a monoid in the Category of Abelian Groups, then it has a component which is a group, but it is not really a group. Anyways. Second, the Wikipedia definition of pseudo-Riemannian manifold is, in particular, a differentiable manifold. Because of that, it admits a structure of metric space. But this structure may have nothing to do with metric tensor. Actually, the metric tensor is just nondegenerate, doesn't need to be positive-definite. So, for what I know, it doesn't define a metric over the tangent spaces, but, gues what, a pseudometric, which is a metric that allows negative values. Simple like that.

@@samueldeandrade8535 Thank you that clarifies it

@@amarug rigorously no. But if we are practical, yes. That's just technicalities. What happens is that Every metric space has a structure as topological space in a natural way. It is good to remember this when we start working with objects with too much structure. We end it up feeling the need to separate the structures, to understand which ones are based on some other, etc. It's always good to make students understand that the "identity function" X → X x → x may not be continuous, for example. It depends what topologies you are considering. Such things, related to abuse of language and specific possibly confusing cases, inevitably appear.

@@amarug The short answer is no. There is a natural topology induced by the metric, where the open sets are the open balls around each point. But a metric space can be given a different topology, and for a given topology, it is not in general immediately obvious whether that topology is equivalent to the metric topology. As a a trivial example, one may consider the trivial topology on R^n, or for a less trivial example, one may consider the cocountable topology. These topologies are not only not equivalent to the metric topology on R^n, they are not even metrizable.

Sry to hear, will probably remake most of these vids in the future with better set-up and more details

@@coolmaththeorems it is wrong to be intelligent AND cute. HAHAHAHAHA. I was not serious, man. Relax.

@@samueldeandrade8535 hahaahahhaa okay misunderstood lol. Thank you appreciate it!

​@@coolmaththeorems when I think about metric spaces I prefer the short definition d: S×S→ |R such that d(x,y)=0 iff x=y d(x,y) ≤ d(x,z)+d(y,z) Let me check if it is right. By choosing z=x, we get d(x,y) ≤ d(y,x) This implies symmetry. Now, for y=x, d(x,x) ≤ d(x,z)+d(x,z) 0 ≤ 2d(x,z) 0 ≤ d(x,z) So it is positive. Yeah. I don't remember if this is the shortest way to define it. Anyway. Cheers!

@@samueldeandrade8535 I totally agree. And I am 100% serious.