Owlcoffee
  • Owlcoffee
Topology
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Owlcoffee
  • Owlcoffee
Let \(\left\{ x_i \right\}_i \in I \) be a family of topological spaces. The only open ones in \(\prod_{i \in I}^{} X_i \) with the form \(\prod_{i \in I}^{} G_i \neq \emptyset\) are the basic open, in other words, those who also satisfy that any \(G_i \) is open and \(G_i = X_i\) for almost any "i".
Astrophysics
  • Astrophysics
Our topology master @zzr0ck3r
Owlcoffee
  • Owlcoffee
I have to prove that statement, by the way.

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Willie579
  • Willie579
I have absolutely no idea what this is. :|
anonymous
  • anonymous
OwlCoffee can you please help me with my question I just tagged you in? It'll only take a minute..
Owlcoffee
  • Owlcoffee
I think I solved it: Suppose \(\prod_{i \in I}^{} G_i\) is a non-empty space, and I consider a point \(x \in \prod_{i \in I}^{}G_i\) there must exist a \(\prod_{i \in I}^{} H_i \) such that \[x \in \prod_{i \in I}^{}H_i \subset \prod_{i \in I}^{}G_i \] I had to actually prove that to continue this excercise... Anywho... For any index "i" the following must satisfy: \(x \in H_i \subset G_i\) and since almost all \(H_i\) is equal to \(X_i\) we will then have \(G_i = X_i\) for almost any "i". Also we have that \(G_i\) is an enviroment for \(x_i \), but since almost any element \(a \in G_i\) it is possible to form a \(x \in \prod_{i \in I}^{}G_i\) such that \(x_i = a\), this is enough to claim that \(G_i \) is an enviroment for all its points, this means it's open. For the second part I had to take care of the subspaces in a topological space. For any subset of \(N\) of a metric space \(M\) the acotated distance will be \(N \times M\) implies that we will have a topology in \(N\) and yet another in \(M\) (had to prove this, took forever). This wil make evident that we can obtain the topology of \(N\) directly from \(M\) without really having to use the metric space of \(M\).
zzr0ck3r
  • zzr0ck3r
I don't even know what you are asking. What do you mean by open ones? every topological space is open by definition. Also, not every topological space is metrizable. Do you mean neighborhood(environment) ?
Owlcoffee
  • Owlcoffee
Well, yes, any topological space is indeed open by definition, but in a family of topological spaces, it's not quite the situation since in ay prehilbertian space a body of vectors do not necessarily form a topological space, this is due to the restrictions of such spaces. But this is a proof by absurd, so I had to deny something that would potentially throw away the whole excercise, this being "A topological space is closed". Yes, I call "enviroment" to the "surrounding" points to any given point in any given space.

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