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 one year ago
Show that {x} are open sets in X for all points x∈X, then all subsets of X are also open in X.
 one year ago
Show that {x} are open sets in X for all points x∈X, then all subsets of X are also open in X.

This Question is Closed

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1{x} collection of open sets ... open covering of X?

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1looks like i am not understanding the Q ... you sure the Q is right?

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1any background on the Q? for some reason I find point set topology very hard.

walters
 one year ago
Best ResponseYou've already chosen the best response.0what i know that the set is open when all its point are interior

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1But discrete points are itself closed set.

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1where does this problem come from (book) ??

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1which book are you using as Text Book ??

walters
 one year ago
Best ResponseYou've already chosen the best response.0RG Bartle and DR Sherbert ,introduction to real analysis john wiley &sons

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1@walters have you managed to go all though chapters to reach Point Set Topology at the end?

walters
 one year ago
Best ResponseYou've already chosen the best response.0not really but i think his question is open and closed sets

walters
 one year ago
Best ResponseYou've already chosen the best response.0i understand it this way but i can't show it dw:1361566625628:dw

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1Let F be collection of A's ... and \( X = \cup_{A \in F} A\) and \( B(x;r) \subset A_i \forall x \in A, \implies B(x;r) \subset X\) shows that X is open set.

walters
 one year ago
Best ResponseYou've already chosen the best response.0"compact "what do u mean

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1I think what you are doing is Let X be a closed set, and \( x \in X \) be a point in X. and \( X  {x}\) is a subset of X which is open in X. let me ask others.

walters
 one year ago
Best ResponseYou've already chosen the best response.0is what i am thinking "but not quit sure"

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1let me ask experts ... I'll reply you in sometime.

walters
 one year ago
Best ResponseYou've already chosen the best response.0i think since the subset of X are open this also implies that {x}and its complements are also open

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1i got the hint: in the discret topology all points are open

walters
 one year ago
Best ResponseYou've already chosen the best response.0so does this mean that there is no relationship between {x} and its complement

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1no ... use the fact that union of infinite number of open sets is and open set.

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1or simply just union of open sets.

walters
 one year ago
Best ResponseYou've already chosen the best response.0i mean like "ie if {x} is open it does not imply that its complement is also closed "

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1I guess not in discrete topology

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1dw:1361569963336:dw usually we have neither closed nor open set.

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1but in discrete topology we seem to have http://mathworld.wolfram.com/DiscreteSet.html http://mathworld.wolfram.com/DiscreteTopology.html

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1the neighborhood of the boundary will not contain any point of that inner set, which makes the compliment open.

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1as to your question, the subsets of X will be power sets of its elements which are open. And the union of open set is open. Hence all the subsets will be open.

experimentX
 one year ago
Best ResponseYou've already chosen the best response.1P.S. I am not sure, I don't have experience with Topology.

phi
 one year ago
Best ResponseYou've already chosen the best response.0I am an engineer, and have not studied analysis.

walters
 one year ago
Best ResponseYou've already chosen the best response.0@JamesJ and @KingGeorge

walters
 one year ago
Best ResponseYou've already chosen the best response.0dw:1361653147330:dw i think they mean it this way because of the question
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