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If [xn] be a sequenceinR endowed with the finite complement topology. If [xn] converges,In R then
The limit is unique
(xn) converges to only two points
(xn) converges to every element in R
what do you think sir?
I am thinking and playing with it. I have only done basic work in that space. But I am sure I can figure it out...
the last option
(xn) converges to one point in R and one point outside R
there is no outside of R
i have not heard anything outside R
R is the universal set.
yes that was y i din't put that option first
I think since we are assuming it converges, it must have some repeated element infinite times, and thus will converge to all x
actually I think it would be unique in this case.
so, i should go with unique ?
well really you should convince yourself...
what does it mean for a sequence to converge?
A sequence is said to be convergent if it approaches some limit
lol, what does that mean?
in real analysis , a sequence that has limit is convergent
or explain more
but what does that mean. If you are going to show that a sequence converges, then what do you need to show?
please explain . thats all i know
then there is no way you can understand this question
please help me understand
Just google converging sequences.
What is the finite compliment topology?
ok i will do that later but help with more question
you mean. Will I answer your questions so that you don't have to study ? lol
a cofinite subset of a set X is a subset A whose complement in X is a finite set.
sir, its because i have less time
but you are not learning anything...
and you are not always online
ok, ill stop telling you how to learn. Just keep asking questions.
Let X be a topological space. Then X is regular if
XisbothT1andT2 XisT3only XisT2only XisbothT2andT
read that and tell me what is wrong
this is a definition. I bet you could figure this out with a look in a book
is it T3?
ok, give me a book to look into
I already have
waw. T3 then?
Which of these spaces is not Hausdorff? R with stanard topology R with lower limit topology R metric toplogy R with the finite complement topology
Can you rule out any cases?
i think b
what do you think?
i can't really explain but i fink it is the answer. please correct me if am wrong
But that is the entire point, you need to be able to explain.
why would you want to know the answer, if you don't understand it. You are not correct
I can't tell you why you are wrong unless you tell me why you thought you were right.
first i know what Hausdorff
but i understand Hausdorff more with set than real line
ok, 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
well to show something is NOT Hausdorff, what do you need to show?
that there intersection does not give empty set
you 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 \(O\cap U=\emptyset\)
Ok, 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.
This proves that non infinite set endowed with the finite complement topology can not be Hausdorff.
waw. thank you sir.
Let 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
I do not know this terminology but since it is a definition, it could not be any easier...
i think is second
i have something like that in my book but not sure
do you think that is a good reason?
the 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
ok. sp am i correct?
ok.so its the second category