Convergence of sequences in topological spaces?

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The discussion focuses on the convergence of the sequence (1/n) in different topological spaces. In the discrete topology, the sequence does not converge to any points. In the indiscrete topology, every point in R is a point of convergence for the sequence. For the third topology, defined by sets where the complement is countable or all of X, the sequence also does not converge to any points. The participants confirm the correctness of these conclusions without requiring formal proofs.
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hi
I was having difficulty with this problem in the book

If (1/n) is a sequence in R

which points (if any) will it converge (for every open set there is an integer N such that for all n>N 1/n is in that open set) to using the following topologies

(a) Discrete
(b) Indiscrete
(c) { A in X : A\X is countable or all of X }

For indiscrete I know that any sequence will converge to any point in R

For discrete -the sequence doesn't coverge to any points

and for (c) I am thinking the sequence again doesn't converge to any points

but I am not sure how to prove the last ones ...or if theyre even right

any help?
 
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A\X is countable or all of X }

What do you mean by A\X? Most people write set theoretic subtraction as X\A (since A is a subset of X, you can't subtract X from A meaningfully), is that your intent?
 
Office_Shredder said:
What do you mean by A\X? Most people write set theoretic subtraction as X\A (since A is a subset of X, you can't subtract X from A meaningfully), is that your intent?

yea that's what i meant, sorry -)
 
i don't really require a proof - i just want to know if i got the right answers
that is discrete topology -no point of convergence
indiscrete -all points are points of convergence
and the last one -no point is a point of convergence

thanks
 
Yes, those answers are correct to the best of my knowledge.
 
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