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anonymous
 one year ago
here
anonymous
 one year ago
here

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If [xn] be a sequenceinR endowed with the finite complement topology. If [xn] converges,In R then

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(xn) converges to only two points

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(xn) converges to every element in R

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what do you think sir?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0I am thinking and playing with it. I have only done basic work in that space. But I am sure I can figure it out...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(xn) converges to one point in R and one point outside R

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0there is no outside of R

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i have not heard anything outside R

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0R is the universal set.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes that was y i din't put that option first

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0I think since we are assuming it converges, it must have some repeated element infinite times, and thus will converge to all x

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0actually I think it would be unique in this case.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so, i should go with unique ?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0well really you should convince yourself...

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0what does it mean for a sequence to converge?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A sequence is said to be convergent if it approaches some limit

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0lol, what does that mean?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in real analysis , a sequence that has limit is convergent

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0but what does that mean. If you are going to show that a sequence converges, then what do you need to show?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0please explain . thats all i know

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0then there is no way you can understand this question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0please help me understand

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0Just google converging sequences.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0What is the finite compliment topology?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok i will do that later but help with more question

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0you mean. Will I answer your questions so that you don't have to study ? lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a cofinite subset of a set X is a subset A whose complement in X is a finite set.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sir, its because i have less time

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0but you are not learning anything...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and you are not always online

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0ok, ill stop telling you how to learn. Just keep asking questions.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let X be a topological space. Then X is regular if

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0XisbothT1andT2 XisT3only XisT2only XisbothT2andT

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0read that and tell me what is wrong

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0this is a definition. I bet you could figure this out with a look in a book

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, give me a book to look into

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Which of these spaces is not Hausdorff? R with stanard topology R with lower limit topology R metric toplogy R with the finite complement topology

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0Can you rule out any cases?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i can't really explain but i fink it is the answer. please correct me if am wrong

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0But that is the entire point, you need to be able to explain.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0why would you want to know the answer, if you don't understand it. You are not correct

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0I can't tell you why you are wrong unless you tell me why you thought you were right.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0first i know what Hausdorff

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but i understand Hausdorff more with set than real line

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0ok, the lower limit topology has open sets in the form [a,b) Given two points \(x,y\) we can always find disjoint sets around each even if one of the points is a

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0well to show something is NOT Hausdorff, what do you need to show?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that there intersection does not give empty set

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0you are almost correct You need to show that there is at least two points \(x,y\) where \(x\ne y\) such that there are no open sets \(O\) and \(U\) where \(x\in O\) and \(y\in U\) and \(O\cap U=\emptyset\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0Ok, let \(R\) be endowed with the finite compliment topology. Suppose there exists \(x,y\) where \(x\ne y\) and there is open sets \(O,U\) with \(x\in O\) and \(y\in Y\) where \(U\cap O=\emptyset\). Since \(U\) is open, we have that \(U^C\) is finite and since \(U\cap O=\emptyset\) it must be that \(O\subset U^C\) this means that \(O\) must be finite, but every finite set in \(R\) has uncountably infinite compliment, and thus \(O\) can not be open and thus a contradiction. It is obvious that neither \(U\) or \(O\) is the emptyset.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0This proves that non infinite set endowed with the finite complement topology can not be Hausdorff.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let Y be a subset of a topological space (X,τ) . If Y is a union of a countable number of nowhere dense subset of X, then Y is called a set of the first category, otherwise it is said to be ________ Meager Third Category fourth Category Second Category

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0I do not know this terminology but since it is a definition, it could not be any easier...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i have something like that in my book but not sure

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0do you think that is a good reason?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0the first thing I get when I google This observation motivates the introduction of a larger class of sets: A subset A ⊆ X is called meager (or of first category) in X if it can be written as a countable union of nowhere dense sets. Any set that is not meager is said to be nonmeager (or of second category). The complement of a meager set is called residual

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok. sp am i correct?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok.so its the second category
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