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anonymous
 one year ago
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anonymous
 one year ago
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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In 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
 one year ago
Best ResponseYou've already chosen the best response.0what is the complement of R?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2what 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
 one year ago
Best ResponseYou've already chosen the best response.0i thought there are two types of numbers

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the imaginary and the real

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2when 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
 one year ago
Best ResponseYou've already chosen the best response.0so will that option b empty since is the complement of R?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2In 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
 one year ago
Best ResponseYou've already chosen the best response.2So what is the answer?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i wonder how \(\{ x_n,\varnothing ,R \}\) is considered cofinite :O

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what does arbitrary really mean. because i see it as finite or infinite

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2um, it means if you close your eye and pick one

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2@halmos? \(\{x_n, \emptyset, \mathbb{R}\}\) what is this?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In the finite complement topology of R, let the sequence [x_{n} be defined by x_{n} = n was trying to interpret in notations

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2Arbitrary 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
 one year ago
Best ResponseYou've already chosen the best response.2he sucks with latex :)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2he also usees \(\epsilon\) when he means \(\in\) so you got to be careful with that as well :)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2Sorry @GIL.ojei I am just playing with you...

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2but seriously start using \(\in\) when you mean in. It is coded as `\(\in)`

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2\(\epsilon\), in general, means a positive real number

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@GIL.ojei u can use this next time for math notations :) http://prntscr.com/8e7ekc and yes epsilon used to denote very small values

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2or "arbitrarily small" positive number :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let (X,τ) be a topological space, and let A be a subset of X. A is dense in X if and only if every nonempty open subset U of X, _________________ A⋂U=0 A⋃U=ϕ A⋂U=ϕ A⋂U=X

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2What is that weird symbol?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is just a definition, and i think the notation should be empty set u can use `\varnothing` \( \varnothing \) instead

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think he copied it from a place which typed it as \(\emptyset \) first (we see it 0) then as \(\phi\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2It 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
 one year ago
Best ResponseYou've already chosen the best response.2The union of any non empty set and anything else is non empty

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2these options don't make sense.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2@GIL.ojei please reread the question and see if it shuold say \(\neq \emptyset\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let (X,τ) be a topological space, and let A be a subset of X. A is dense in X if and only if every nonempty open subset U of X, _________________

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2option 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
 one year ago
Best ResponseYou've already chosen the best response.2\(A\cap U \neq \emptyset\) is always true

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Which 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
 one year ago
Best ResponseYou've already chosen the best response.0this is the last question on topology for today. please, i will tell you what to teach me today . please

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i know option A and B are correct

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but what about C and D? i fink C is also correct

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2what does Y1isinR mean and why do you keep writing things like this?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that was how i saw it

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2you look at the page and it looks like this ? Y1isinR ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think that was what they wanted to write

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do not have idea but is Every Hausdorff space is T1?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2Tell me the definitions of both

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0X is a T1 space if any two distinct points in X are separated.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i know that Hausdorff space has to do with intersection. but what is really separated?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2what does separated mean?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i asked the question

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2So 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
 one year ago
Best ResponseYou've already chosen the best response.0waw. so, T1 space is also Hausdorff space

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2we are not showing that T1 is Hausdorff, we are showing that Hausdorff is T1.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2every 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
 one year ago
Best ResponseYou've already chosen the best response.0i want to close the tab and open another. i think there are thinks i want to understand
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