walters
Show that {x} are open sets in X for all points x∈X, then all subsets of X are also open in X.
Delete
Share
This Question is Closed
experimentX
Best Response
You've already chosen the best response.
1
{x} collection of open sets ... open covering of X?
walters
Best Response
You've already chosen the best response.
0
yes
experimentX
Best Response
You've already chosen the best response.
1
looks like i am not understanding the Q ... you sure the Q is right?
walters
Best Response
You've already chosen the best response.
0
yes it is like that
experimentX
Best Response
You've already chosen the best response.
1
any background on the Q? for some reason I find point set topology very hard.
walters
Best Response
You've already chosen the best response.
0
what i know that the set is open when all its point are interior
experimentX
Best Response
You've already chosen the best response.
1
But discrete points are itself closed set.
experimentX
Best Response
You've already chosen the best response.
1
where does this problem come from (book) ??
walters
Best Response
You've already chosen the best response.
0
assignment
experimentX
Best Response
You've already chosen the best response.
1
which book are you using as Text Book ??
walters
Best Response
You've already chosen the best response.
0
RG Bartle and DR Sherbert ,introduction to real analysis john wiley &sons
experimentX
Best Response
You'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
Best Response
You've already chosen the best response.
0
not really but i think his question is open and closed sets
walters
Best Response
You've already chosen the best response.
0
i understand it this way but i can't show it |dw:1361566625628:dw|
walters
Best Response
You've already chosen the best response.
0
|dw:1361567087644:dw|
experimentX
Best Response
You've already chosen the best response.
1
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.
experimentX
Best Response
You've already chosen the best response.
1
is X compact??
walters
Best Response
You've already chosen the best response.
0
"compact "what do u mean
experimentX
Best Response
You've already chosen the best response.
1
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.
experimentX
Best Response
You've already chosen the best response.
1
X-{x}
walters
Best Response
You've already chosen the best response.
0
is what i am thinking "but not quit sure"
experimentX
Best Response
You've already chosen the best response.
1
let me ask experts ... I'll reply you in sometime.
walters
Best Response
You've already chosen the best response.
0
i think since the subset of X are open this also implies that {x}and its complements are also open
experimentX
Best Response
You've already chosen the best response.
1
i got the hint: in the discret topology all points are open
walters
Best Response
You've already chosen the best response.
0
so does this mean that there is no relationship between {x} and its complement
experimentX
Best Response
You've already chosen the best response.
1
no ... use the fact that union of infinite number of open sets is and open set.
experimentX
Best Response
You've already chosen the best response.
1
or simply just union of open sets.
walters
Best Response
You've already chosen the best response.
0
i mean like "ie if {x} is open it does not imply that its complement is also closed "
walters
Best Response
You've already chosen the best response.
0
is it possible ?
experimentX
Best Response
You've already chosen the best response.
1
I guess not in discrete topology
experimentX
Best Response
You've already chosen the best response.
1
|dw:1361569963336:dw|
usually we have neither closed nor open set.
experimentX
Best Response
You've already chosen the best response.
1
the neighborhood of the boundary will not contain any point of that inner set, which makes the compliment open.
walters
Best Response
You've already chosen the best response.
0
ok i get ur statement
experimentX
Best Response
You've already chosen the best response.
1
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.
experimentX
Best Response
You've already chosen the best response.
1
P.S. I am not sure, I don't have experience with Topology.
walters
Best Response
You've already chosen the best response.
0
@phi
phi
Best Response
You've already chosen the best response.
0
I am an engineer, and have not studied analysis.
walters
Best Response
You've already chosen the best response.
0
@JamesJ and @KingGeorge
walters
Best Response
You've already chosen the best response.
0
|dw:1361653147330:dw|
i think they mean it this way because of the question
walters
Best Response
You've already chosen the best response.
0
@jacobian