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

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{x} collection of open sets ... open covering of X?
looks like i am not understanding the Q ... you sure the Q is right?

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Other answers:

yes it is like that
any background on the Q? for some reason I find point set topology very hard.
what i know that the set is open when all its point are interior
But discrete points are itself closed set.
where does this problem come from (book) ??
which book are you using as Text Book ??
RG Bartle and DR Sherbert ,introduction to real analysis john wiley &sons
@walters have you managed to go all though chapters to reach Point Set Topology at the end?
not really but i think his question is open and closed sets
i understand it this way but i can't show it |dw:1361566625628:dw|
Let 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.
is X compact??
"compact "what do u mean
I 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.
is what i am thinking "but not quit sure"
let me ask experts ... I'll reply you in sometime.
i think since the subset of X are open this also implies that {x}and its complements are also open
i got the hint: in the discret topology all points are open
so does this mean that there is no relationship between {x} and its complement
no ... use the fact that union of infinite number of open sets is and open set.
or simply just union of open sets.
i mean like "ie if {x} is open it does not imply that its complement is also closed "
is it possible ?
I guess not in discrete topology
|dw:1361569963336:dw| usually we have neither closed nor open set.
but in discrete topology we seem to have
the neighborhood of the boundary will not contain any point of that inner set, which makes the compliment open.
ok i get ur statement
as 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.
P.S. I am not sure, I don't have experience with Topology.
  • phi
I am an engineer, and have not studied analysis.
|dw:1361653147330:dw| i think they mean it this way because of the question

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