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walters
 2 years ago
Show that {x} are open sets in X for all points x∈X, then all subsets of X are also open in X.
walters
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1{x} collection of open sets ... open covering of X?

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

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

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

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

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

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

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

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

experimentX
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0"compact "what do u mean

experimentX
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0is what i am thinking "but not quit sure"

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

walters
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1i got the hint: in the discret topology all points are open

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

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

walters
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1I guess not in discrete topology

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

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

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

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

walters
 2 years 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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