anonymous
  • anonymous
here
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
If [xn] be a sequenceinR endowed with the finite complement topology. If [xn] converges,In R then
anonymous
  • anonymous
The limit is unique
anonymous
  • anonymous
@zzr0ck3r

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anonymous
  • anonymous
(xn) converges to only two points
anonymous
  • anonymous
(xn) converges to every element in R
anonymous
  • anonymous
what do you think sir?
zzr0ck3r
  • zzr0ck3r
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...
anonymous
  • anonymous
the last option
anonymous
  • anonymous
(xn) converges to one point in R and one point outside R
zzr0ck3r
  • zzr0ck3r
there is no outside of R
anonymous
  • anonymous
i have not heard anything outside R
zzr0ck3r
  • zzr0ck3r
R is the universal set.
anonymous
  • anonymous
yes that was y i din't put that option first
zzr0ck3r
  • zzr0ck3r
I think since we are assuming it converges, it must have some repeated element infinite times, and thus will converge to all x
zzr0ck3r
  • zzr0ck3r
actually I think it would be unique in this case.
anonymous
  • anonymous
so, i should go with unique ?
anonymous
  • anonymous
@zzr0ck3r
zzr0ck3r
  • zzr0ck3r
well really you should convince yourself...
zzr0ck3r
  • zzr0ck3r
what does it mean for a sequence to converge?
anonymous
  • anonymous
A sequence is said to be convergent if it approaches some limit
zzr0ck3r
  • zzr0ck3r
lol, what does that mean?
anonymous
  • anonymous
in real analysis , a sequence that has limit is convergent
anonymous
  • anonymous
or explain more
zzr0ck3r
  • zzr0ck3r
but what does that mean. If you are going to show that a sequence converges, then what do you need to show?
anonymous
  • anonymous
please explain . thats all i know
zzr0ck3r
  • zzr0ck3r
then there is no way you can understand this question
anonymous
  • anonymous
please help me understand
zzr0ck3r
  • zzr0ck3r
Just google converging sequences.
zzr0ck3r
  • zzr0ck3r
http://wolfweb.unr.edu/homepage/jabuka/Classes/2006_spring/topology/Notes/04%20-%20Congergent%20sequences.pdf
zzr0ck3r
  • zzr0ck3r
What is the finite compliment topology?
anonymous
  • anonymous
ok i will do that later but help with more question
zzr0ck3r
  • zzr0ck3r
you mean. Will I answer your questions so that you don't have to study ? lol
zzr0ck3r
  • zzr0ck3r
ok sure.
anonymous
  • anonymous
a cofinite subset of a set X is a subset A whose complement in X is a finite set.
anonymous
  • anonymous
sir, its because i have less time
zzr0ck3r
  • zzr0ck3r
but you are not learning anything...
anonymous
  • anonymous
and you are not always online
zzr0ck3r
  • zzr0ck3r
ok, ill stop telling you how to learn. Just keep asking questions.
anonymous
  • anonymous
Let X be a topological space. Then X is regular if
anonymous
  • anonymous
XisbothT1andT2 XisT3only XisT2only XisbothT2andT
zzr0ck3r
  • zzr0ck3r
read that and tell me what is wrong
zzr0ck3r
  • zzr0ck3r
this is a definition. I bet you could figure this out with a look in a book
anonymous
  • anonymous
is it T3?
anonymous
  • anonymous
ok, give me a book to look into
zzr0ck3r
  • zzr0ck3r
I already have
zzr0ck3r
  • zzr0ck3r
https://en.wikipedia.org/wiki/Regular_space
anonymous
  • anonymous
waw. T3 then?
anonymous
  • anonymous
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
zzr0ck3r
  • zzr0ck3r
Can you rule out any cases?
anonymous
  • anonymous
i think b
anonymous
  • anonymous
what do you think?
zzr0ck3r
  • zzr0ck3r
why?
anonymous
  • anonymous
i can't really explain but i fink it is the answer. please correct me if am wrong
anonymous
  • anonymous
@zzr0ck3r
anonymous
  • anonymous
@zzr0ck3r
zzr0ck3r
  • zzr0ck3r
But that is the entire point, you need to be able to explain.
zzr0ck3r
  • zzr0ck3r
why would you want to know the answer, if you don't understand it. You are not correct
zzr0ck3r
  • zzr0ck3r
I can't tell you why you are wrong unless you tell me why you thought you were right.
anonymous
  • anonymous
first i know what Hausdorff
anonymous
  • anonymous
but i understand Hausdorff more with set than real line
zzr0ck3r
  • zzr0ck3r
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
zzr0ck3r
  • zzr0ck3r
well to show something is NOT Hausdorff, what do you need to show?
anonymous
  • anonymous
that there intersection does not give empty set
zzr0ck3r
  • zzr0ck3r
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 \(O\cap U=\emptyset\)
anonymous
  • anonymous
ok.
zzr0ck3r
  • zzr0ck3r
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.
zzr0ck3r
  • zzr0ck3r
This proves that non infinite set endowed with the finite complement topology can not be Hausdorff.
anonymous
  • anonymous
waw. thank you sir.
anonymous
  • anonymous
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
zzr0ck3r
  • zzr0ck3r
I do not know this terminology but since it is a definition, it could not be any easier...
anonymous
  • anonymous
i think is second
zzr0ck3r
  • zzr0ck3r
why?
anonymous
  • anonymous
i have something like that in my book but not sure
zzr0ck3r
  • zzr0ck3r
@GIL.ojei why?
zzr0ck3r
  • zzr0ck3r
do you think that is a good reason?
zzr0ck3r
  • zzr0ck3r
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
anonymous
  • anonymous
ok. sp am i correct?
zzr0ck3r
  • zzr0ck3r
seriously?
anonymous
  • anonymous
ok.so its the second category

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