here

- anonymous

here

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- schrodinger

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- anonymous

In the finite complement topology of R, let the sequence [x_{n} be defined by x_{n} = n, for n\epsilon N. If the limit of the sequence is x, then x must be
∞
0
A unique constant
Arbitrary in
R

- anonymous

@zzr0ck3r

- anonymous

what is the complement of R?

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## More answers

- zzr0ck3r

what is the set of all elements that are not in R
For the 6th times, please stop using epsilon. IT DOES NOT MEAN IN

- anonymous

i do not know

- anonymous

i thought there are two types of numbers

- anonymous

the imaginary and the real

- zzr0ck3r

when we talk about the real numbers as the universe, we do not consider them as a subset of the complex numbers.
The compliment of the real numbers is the emptyset

- anonymous

ok. thanks so much

- anonymous

so will that option b empty since is the complement of R?

- zzr0ck3r

In the cofinite topology,
1) a sequence that does not take an infinite many of the same value will converge to every element in the space.
2) A sequence that does(but only one number), will converge to that number
3) A sequence that has more than one number for which it obtains infinite times does not converge

- zzr0ck3r

So what is the answer?

- anonymous

Arbitrary in R

- zzr0ck3r

correct

- anonymous

i wonder how \(\{ x_n,\varnothing ,R
\}\) is considered cofinite :O

- anonymous

what does arbitrary really mean. because i see it as finite or infinite

- zzr0ck3r

um, it means if you close your eye and pick one

- anonymous

Arbitrary = any

- zzr0ck3r

@halmos? \(\{x_n, \emptyset, \mathbb{R}\}\) what is this?

- anonymous

In the finite complement topology of R, let the sequence [x_{n} be defined by x_{n} = n
was trying to interpret in notations

- zzr0ck3r

Arbitrary size means we are not given anything about the set, so as far as order, a set of arbitrary order is either finite or infinite. But we use the term this other way as well :)

- zzr0ck3r

he sucks with latex :)

- zzr0ck3r

he also usees \(\epsilon\) when he means \(\in\) so you got to be careful with that as well :)

- anonymous

oh i see now :)

- zzr0ck3r

Sorry @GIL.ojei I am just playing with you...

- zzr0ck3r

but seriously start using \(\in\) when you mean in. It is coded as `\(\in)`

- zzr0ck3r

\in

- zzr0ck3r

\(\epsilon\), in general, means a positive real number

- anonymous

@GIL.ojei u can use this next time for math notations :)
http://prntscr.com/8e7ekc
and yes epsilon used to denote very small values

- anonymous

thank u sir

- zzr0ck3r

or "arbitrarily small" positive number :)

- anonymous

ok

- anonymous

Let
(X,τ)
be a topological space, and let A be a subset of X. A is dense in X if and only if every non-empty open subset U of X, _________________
A⋂U=0
A⋃U=ϕ
A⋂U=ϕ
A⋂U=X

- zzr0ck3r

What is that weird symbol?

- zzr0ck3r

@GIL.ojei ?

- anonymous

this is just a definition, and i think the notation should be empty set u can use `\varnothing` \( \varnothing \) instead

- anonymous

empty set

- zzr0ck3r

then what is the 0?

- anonymous

i think he copied it from a place which typed it as \(\emptyset \) first (we see it 0) then as \(\phi\)

- anonymous

its just zero

- zzr0ck3r

that makes no sense

- zzr0ck3r

It should be intuitively clear from the definition of a dense set, that if \(U\) is open then \(A\cap U\ne \emptyset\).
But I do not see this option, and I dont know what the weird \(\phi \) looking thing is

- zzr0ck3r

or `\emptyset`

- zzr0ck3r

The union of any non empty set and anything else is non empty

- zzr0ck3r

these options don't make sense.

- zzr0ck3r

@GIL.ojei please reread the question and see if it shuold say \(\neq \emptyset\)

- anonymous

Let (X,τ) be a topological space, and let A be a subset of X. A is dense in X if and only if every non-empty open subset U of X, _________________

- anonymous

A⋂U=0

- anonymous

that is option a

- anonymous

A⋃U=\[ϕ \]

- anonymous

that is option b

- anonymous

A⋂U=\[ϕ \]

- anonymous

thats option C

- anonymous

A⋂U=X

- anonymous

option D

- anonymous

@zzr0ck3r

- anonymous

i think is option c

- anonymous

or option D

- zzr0ck3r

option a makes no sense
the union of subsets of some general topology are not necessarily equal to a number
option b is wrong because U is non empty
option c is wrong because of what I wrote about the non empty intersection
option d is only true sometimes

- zzr0ck3r

\(A\cap U \neq \emptyset\) is always true

- anonymous

Which of these is not true about
T1
- spaces?
a)Every singleton set is closed
b)Every finite set is closed
c)Every Hausdorff space is
T1
d)Y1isinR

- anonymous

this is the last question on topology for today. please, i will tell you what to teach me today . please

- anonymous

i know option A and B are correct

- anonymous

but what about C and D?
i fink C is also correct

- zzr0ck3r

what does Y1isinR mean and why do you keep writing things like this?

- anonymous

that was how i saw it

- anonymous

\[Y_1 \] is in R

- zzr0ck3r

you look at the page and it looks like this ?
Y1isinR
?

- anonymous

i think that was what they wanted to write

- zzr0ck3r

what is \(Y_1\)?

- anonymous

do not have idea but is Every Hausdorff space is T1?

- zzr0ck3r

Tell me the definitions of both

- anonymous

X is a T1 space if any two distinct points in X are separated.

- anonymous

i know that Hausdorff space has to do with intersection. but what is really separated?

- zzr0ck3r

what does separated mean?

- anonymous

i asked the question

- zzr0ck3r

So if it is Hausdorff, then for all \(x,y\) we have open set \(O,U\) such that \(x\in O, y\in U, O\cap U=\emptyset\)
Does this imply that there is a open set \(A\) that contains \(x\) and does not contain \(y\)?
Does this imply that there is a open set \(B\) that contains \(y\) and does not contain \(x\)?
IF the answer is yes, then we must be \(T_1\).

- anonymous

waw. so, T1 space is also Hausdorff space

- zzr0ck3r

we are not showing that T1 is Hausdorff, we are showing that Hausdorff is T1.

- zzr0ck3r

every horse has 4 legs but not every 4 legged animal is a horse
i.e. if a implies b it is not necessarily true that b implies a

- anonymous

ok

- anonymous

i want to close the tab and open another. i think there are thinks i want to understand

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