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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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,

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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well its not clear. similar problem
\[x_{n}\rightarrow x \epsilon, as, n\rightarrow N \]

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thats option one . which i fink is correct
one min
please stop using epsilon lol
\[\forall U \in N(x), we have N \in \mathbb N \forall n\geqslant N, x_{n}\epsilon U \] option B
I 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
ok sir
back
hi
hi sir
Let 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
please they are two but help with this sir
What does complete mean, and what does dense mean?
and what does countable mean?
a countable set is a set with the same cardinality
the same carnality as what?
some subset of the set
this 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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