Question

here

Answer 1

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

Answer 2

The limit is unique

Answer 3

@zzr0ck3r

Answer 4

(xn)
converges to only two points

Answer 5

(xn)
converges to every element in
R

Answer 6

what do you think sir?

Answer 7

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...

Answer 8

the last option

Answer 9

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

Answer 10

there is no outside of R

Answer 11

i have not heard anything outside R

Answer 12

R is the universal set.

Answer 13

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

Answer 14

I think since we are assuming it converges, it must have some repeated element infinite times, and thus will converge to all x

Answer 15

actually I think it would be unique in this case.

Answer 16

so, i should go with unique ?

Answer 17

@zzr0ck3r

Answer 18

well really you should convince yourself...

Answer 19

what does it mean for a sequence to converge?

Answer 20

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

Answer 21

lol, what does that mean?

Answer 22

in real analysis , a sequence that has limit is convergent

Answer 23

or explain more

Answer 24

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

Answer 25

please explain . thats all i know

Answer 26

then there is no way you can understand this question

Answer 27

please help me understand

Answer 28

Just google converging sequences.

Answer 29

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

Answer 30

What is the finite compliment topology?

Answer 31

ok i will do that later but help with more question

Answer 32

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

Answer 33

ok sure.

Answer 34

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

Answer 35

sir, its because i have less time

Answer 36

but you are not learning anything...

Answer 37

and you are not always online

Answer 38

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

Answer 39

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

Answer 40

XisbothT1andT2

XisT3only

XisT2only

XisbothT2andT

Answer 41

read that and tell me what is wrong

Answer 42

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

Answer 43

is it T3?

Answer 44

ok, give me a book to look into

Answer 45

I already have

Answer 46

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

Answer 47

waw. T3 then?

Answer 48

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

Answer 49

Can you rule out any cases?

Answer 50

i think b

Answer 51

what do you think?

Answer 52

why?

Answer 53

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

Answer 54

@zzr0ck3r

Answer 55

@zzr0ck3r

Answer 56

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

Answer 57

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

You are not correct

Answer 58

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

Answer 59

first i know what Hausdorff

Answer 60

but i understand Hausdorff more with set than real line

Answer 61

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

Answer 62

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

Answer 63

that there intersection does not give empty set

Answer 64

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 \(O\cap U=\emptyset\)

Answer 65

ok.

Answer 66

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.

Answer 67

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

Answer 68

waw. thank you sir.

Answer 69

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

Answer 70

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

Answer 71

i think is second

Answer 72

why?

Answer 73

i have something like that in my book but not sure

Answer 74

@GIL.ojei why?

Answer 75

do you think that is a good reason?

Answer 76

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

Answer 77

ok. sp am i correct?

Answer 78

seriously?

Answer 79

ok.so its the second category