anonymous
  • anonymous
here
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • 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
  • anonymous
@zzr0ck3r
anonymous
  • anonymous
what is the complement of R?

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zzr0ck3r
  • 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
  • anonymous
i do not know
anonymous
  • anonymous
i thought there are two types of numbers
anonymous
  • anonymous
the imaginary and the real
zzr0ck3r
  • 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
  • anonymous
ok. thanks so much
anonymous
  • anonymous
so will that option b empty since is the complement of R?
zzr0ck3r
  • 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
  • zzr0ck3r
So what is the answer?
anonymous
  • anonymous
Arbitrary in R
zzr0ck3r
  • zzr0ck3r
correct
anonymous
  • anonymous
i wonder how \(\{ x_n,\varnothing ,R \}\) is considered cofinite :O
anonymous
  • anonymous
what does arbitrary really mean. because i see it as finite or infinite
zzr0ck3r
  • zzr0ck3r
um, it means if you close your eye and pick one
anonymous
  • anonymous
Arbitrary = any
zzr0ck3r
  • zzr0ck3r
@halmos? \(\{x_n, \emptyset, \mathbb{R}\}\) what is this?
anonymous
  • 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
  • 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
  • zzr0ck3r
he sucks with latex :)
zzr0ck3r
  • zzr0ck3r
he also usees \(\epsilon\) when he means \(\in\) so you got to be careful with that as well :)
anonymous
  • anonymous
oh i see now :)
zzr0ck3r
  • zzr0ck3r
Sorry @GIL.ojei I am just playing with you...
zzr0ck3r
  • zzr0ck3r
but seriously start using \(\in\) when you mean in. It is coded as `\(\in)`
zzr0ck3r
  • zzr0ck3r
\in
zzr0ck3r
  • zzr0ck3r
\(\epsilon\), in general, means a positive real number
anonymous
  • 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
  • anonymous
thank u sir
zzr0ck3r
  • zzr0ck3r
or "arbitrarily small" positive number :)
anonymous
  • anonymous
ok
anonymous
  • 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
  • zzr0ck3r
What is that weird symbol?
zzr0ck3r
  • zzr0ck3r
@GIL.ojei ?
anonymous
  • anonymous
this is just a definition, and i think the notation should be empty set u can use `\varnothing` \( \varnothing \) instead
anonymous
  • anonymous
empty set
zzr0ck3r
  • zzr0ck3r
then what is the 0?
anonymous
  • anonymous
i think he copied it from a place which typed it as \(\emptyset \) first (we see it 0) then as \(\phi\)
anonymous
  • anonymous
its just zero
zzr0ck3r
  • zzr0ck3r
that makes no sense
zzr0ck3r
  • 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
  • zzr0ck3r
or `\emptyset`
zzr0ck3r
  • zzr0ck3r
The union of any non empty set and anything else is non empty
zzr0ck3r
  • zzr0ck3r
these options don't make sense.
zzr0ck3r
  • zzr0ck3r
@GIL.ojei please reread the question and see if it shuold say \(\neq \emptyset\)
anonymous
  • 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
  • anonymous
A⋂U=0
anonymous
  • anonymous
that is option a
anonymous
  • anonymous
A⋃U=\[ϕ \]
anonymous
  • anonymous
that is option b
anonymous
  • anonymous
A⋂U=\[ϕ \]
anonymous
  • anonymous
thats option C
anonymous
  • anonymous
A⋂U=X
anonymous
  • anonymous
option D
anonymous
  • anonymous
@zzr0ck3r
anonymous
  • anonymous
i think is option c
anonymous
  • anonymous
or option D
zzr0ck3r
  • 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
  • zzr0ck3r
\(A\cap U \neq \emptyset\) is always true
anonymous
  • 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
  • anonymous
this is the last question on topology for today. please, i will tell you what to teach me today . please
anonymous
  • anonymous
i know option A and B are correct
anonymous
  • anonymous
but what about C and D? i fink C is also correct
zzr0ck3r
  • zzr0ck3r
what does Y1isinR mean and why do you keep writing things like this?
anonymous
  • anonymous
that was how i saw it
anonymous
  • anonymous
\[Y_1 \] is in R
zzr0ck3r
  • zzr0ck3r
you look at the page and it looks like this ? Y1isinR ?
anonymous
  • anonymous
i think that was what they wanted to write
zzr0ck3r
  • zzr0ck3r
what is \(Y_1\)?
anonymous
  • anonymous
do not have idea but is Every Hausdorff space is T1?
zzr0ck3r
  • zzr0ck3r
Tell me the definitions of both
anonymous
  • anonymous
X is a T1 space if any two distinct points in X are separated.
anonymous
  • anonymous
i know that Hausdorff space has to do with intersection. but what is really separated?
zzr0ck3r
  • zzr0ck3r
what does separated mean?
anonymous
  • anonymous
i asked the question
zzr0ck3r
  • 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
  • anonymous
waw. so, T1 space is also Hausdorff space
zzr0ck3r
  • zzr0ck3r
we are not showing that T1 is Hausdorff, we are showing that Hausdorff is T1.
zzr0ck3r
  • 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
  • anonymous
ok
anonymous
  • anonymous
i want to close the tab and open another. i think there are thinks i want to understand

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