here

- anonymous

here

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- schrodinger

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- anonymous

If
[xn] be a sequenceinR
endowed with the finite complement topology. If
[xn] converges,In R then

- anonymous

The limit is unique

- anonymous

@zzr0ck3r

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## More answers

- anonymous

(xn)
converges to only two points

- anonymous

(xn)
converges to every element in
R

- anonymous

what do you think sir?

- 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

the last option

- anonymous

(xn)
converges to one point in
R and one point outside R

- zzr0ck3r

there is no outside of R

- anonymous

i have not heard anything outside R

- zzr0ck3r

R is the universal set.

- anonymous

yes that was y i din't put that option first

- 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

actually I think it would be unique in this case.

- anonymous

so, i should go with unique ?

- anonymous

@zzr0ck3r

- zzr0ck3r

well really you should convince yourself...

- zzr0ck3r

what does it mean for a sequence to converge?

- anonymous

A sequence is said to be convergent if it approaches some limit

- zzr0ck3r

lol, what does that mean?

- anonymous

in real analysis , a sequence that has limit is convergent

- anonymous

or explain more

- zzr0ck3r

but what does that mean. If you are going to show that a sequence converges, then what do you need to show?

- anonymous

please explain . thats all i know

- zzr0ck3r

then there is no way you can understand this question

- anonymous

please help me understand

- zzr0ck3r

Just google converging sequences.

- zzr0ck3r

http://wolfweb.unr.edu/homepage/jabuka/Classes/2006_spring/topology/Notes/04%20-%20Congergent%20sequences.pdf

- zzr0ck3r

What is the finite compliment topology?

- anonymous

ok i will do that later but help with more question

- zzr0ck3r

you mean. Will I answer your questions so that you don't have to study ? lol

- zzr0ck3r

ok sure.

- anonymous

a cofinite subset of a set X is a subset A whose complement in X is a finite set.

- anonymous

sir, its because i have less time

- zzr0ck3r

but you are not learning anything...

- anonymous

and you are not always online

- zzr0ck3r

ok, ill stop telling you how to learn. Just keep asking questions.

- anonymous

Let X be a topological space. Then X is regular if

- anonymous

XisbothT1andT2
XisT3only
XisT2only
XisbothT2andT

- zzr0ck3r

read that and tell me what is wrong

- zzr0ck3r

this is a definition. I bet you could figure this out with a look in a book

- anonymous

is it T3?

- anonymous

ok, give me a book to look into

- zzr0ck3r

I already have

- zzr0ck3r

https://en.wikipedia.org/wiki/Regular_space

- anonymous

waw. T3 then?

- 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

Can you rule out any cases?

- anonymous

i think b

- anonymous

what do you think?

- zzr0ck3r

why?

- anonymous

i can't really explain but i fink it is the answer. please correct me if am wrong

- anonymous

@zzr0ck3r

- anonymous

@zzr0ck3r

- zzr0ck3r

But that is the entire point, you need to be able to explain.

- zzr0ck3r

why would you want to know the answer, if you don't understand it.
You are not correct

- zzr0ck3r

I can't tell you why you are wrong unless you tell me why you thought you were right.

- anonymous

first i know what Hausdorff

- anonymous

but i understand Hausdorff more with set than real line

- 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

well to show something is NOT Hausdorff, what do you need to show?

- anonymous

that there intersection does not give empty set

- 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

ok.

- 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

This proves that non infinite set endowed with the finite complement topology can not be Hausdorff.

- anonymous

waw. thank you sir.

- 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

I do not know this terminology but since it is a definition, it could not be any easier...

- anonymous

i think is second

- zzr0ck3r

why?

- anonymous

i have something like that in my book but not sure

- zzr0ck3r

@GIL.ojei why?

- zzr0ck3r

do you think that is a good reason?

- 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

ok. sp am i correct?

- zzr0ck3r

seriously?

- anonymous

ok.so its the second category

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