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anonymous
 one year ago
Let X be a topological space, Let {x_{n} be a sequence of elements in X. Then x_{n} is said to converge to x\epsilon X if \forall nbds U of x, there existsN\epsilon\mathbb Nsuch that \forall n\geslant N, x_{n}\epsilon U. i.e one of these conditions does not hold
x_{n}\rightarrowx\epsilon as n\rightarrow N
\forall U\epsilon N(x), we have N\epsilon \epsilon \mathbb N \forall n\geqslantN, x_{n}\epsilon U
x_{n}\rightarrowx\epsilon as n\righ = N
\forall U\epsilon N(x), we have N\epsilon \epsilon \mathbb N \forall n\geqslantN,
anonymous
 one year ago
Let X be a topological space, Let {x_{n} be a sequence of elements in X. Then x_{n} is said to converge to x\epsilon X if \forall nbds U of x, there existsN\epsilon\mathbb Nsuch that \forall n\geslant N, x_{n}\epsilon U. i.e one of these conditions does not hold x_{n}\rightarrowx\epsilon as n\rightarrow N \forall U\epsilon N(x), we have N\epsilon \epsilon \mathbb N \forall n\geqslantN, x_{n}\epsilon U x_{n}\rightarrowx\epsilon as n\righ = N \forall U\epsilon N(x), we have N\epsilon \epsilon \mathbb N \forall n\geqslantN,

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well its not clear. similar problem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[x_{n}\rightarrow x \epsilon, as, n\rightarrow N \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thats option one . which i fink is correct

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0please stop using epsilon lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\forall U \in N(x), we have N \in \mathbb N \forall n\geqslant N, x_{n}\epsilon U \] option B

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0I can read this... I will be back in a bit. I have no idea what this means \[x_{n}\rightarrow x\epsilon\text{ as } n\rightarrow N\] Please, STOP USING EPSILON, and STOP PUTTING TEXT IN LINE WITH MATH it makes this impossible. bbl

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let X be a complete metric space and {On} is countable collection of dense open subset of X. Show that \[\cup O_n \] is not empty

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0please they are two but help with this sir

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0What does complete mean, and what does dense mean?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0and what does countable mean?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a countable set is a set with the same cardinality

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0the same carnality as what?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0some subset of the set

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0this is not correct(not even close). You are jumping way to far ahead. I am not trying to be rude, but I don't to waste time doing this if you will not understand
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