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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
what is the complement of R?

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Other answers:

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
i do not know
i thought there are two types of numbers
the imaginary and the real
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
ok. thanks so much
so will that option b empty since is the complement of R?
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
So what is the answer?
Arbitrary in R
correct
i wonder how \(\{ x_n,\varnothing ,R \}\) is considered cofinite :O
what does arbitrary really mean. because i see it as finite or infinite
um, it means if you close your eye and pick one
Arbitrary = any
@halmos? \(\{x_n, \emptyset, \mathbb{R}\}\) what is this?
In the finite complement topology of R, let the sequence [x_{n} be defined by x_{n} = n was trying to interpret in notations
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 :)
he sucks with latex :)
he also usees \(\epsilon\) when he means \(\in\) so you got to be careful with that as well :)
oh i see now :)
Sorry @GIL.ojei I am just playing with you...
but seriously start using \(\in\) when you mean in. It is coded as `\(\in)`
\in
\(\epsilon\), in general, means a positive real number
@GIL.ojei u can use this next time for math notations :) http://prntscr.com/8e7ekc and yes epsilon used to denote very small values
thank u sir
or "arbitrarily small" positive number :)
ok
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
What is that weird symbol?
@GIL.ojei ?
this is just a definition, and i think the notation should be empty set u can use `\varnothing` \( \varnothing \) instead
empty set
then what is the 0?
i think he copied it from a place which typed it as \(\emptyset \) first (we see it 0) then as \(\phi\)
its just zero
that makes no sense
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
or `\emptyset`
The union of any non empty set and anything else is non empty
these options don't make sense.
@GIL.ojei please reread the question and see if it shuold say \(\neq \emptyset\)
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
that is option a
A⋃U=\[ϕ \]
that is option b
A⋂U=\[ϕ \]
thats option C
A⋂U=X
option D
i think is option c
or option D
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
\(A\cap U \neq \emptyset\) is always true
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
this is the last question on topology for today. please, i will tell you what to teach me today . please
i know option A and B are correct
but what about C and D? i fink C is also correct
what does Y1isinR mean and why do you keep writing things like this?
that was how i saw it
\[Y_1 \] is in R
you look at the page and it looks like this ? Y1isinR ?
i think that was what they wanted to write
what is \(Y_1\)?
do not have idea but is Every Hausdorff space is T1?
Tell me the definitions of both
X is a T1 space if any two distinct points in X are separated.
i know that Hausdorff space has to do with intersection. but what is really separated?
what does separated mean?
i asked the question
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 \(T_1\).
waw. so, T1 space is also Hausdorff space
we are not showing that T1 is Hausdorff, we are showing that Hausdorff is T1.
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
ok
i want to close the tab and open another. i think there are thinks i want to understand

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