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Mathematics 7 Online
OpenStudy (anonymous):

find the set of limit points of

OpenStudy (anonymous):

R={ x/x∈R }

OpenStudy (anonymous):

@Mchilds15

OpenStudy (anonymous):

@zepdrix

OpenStudy (anonymous):

@Callisto

OpenStudy (anonymous):

@texaschic101 @beccaboo333 @mathslover

OpenStudy (anonymous):

@johnweldon1993 @chmvijay @uri

OpenStudy (anonymous):

@Chris911

OpenStudy (anonymous):

@Ashleyisakitty

OpenStudy (anonymous):

@inkyvoyd

OpenStudy (anonymous):

There are no limit points. For any neighborhood of \(x\in R\), you will never find a point that does not lie in \(R\).

OpenStudy (anonymous):

can you tell me stepwise,i have to explain the procedure of how it is solved

OpenStudy (anonymous):

^ that implies that every point is a limit point, not that there are none.

OpenStudy (anonymous):

how to prove it @Jemurray3

OpenStudy (anonymous):

What's the definition of a limit point?

OpenStudy (anonymous):

@Jemurray3, you're right, I had the definition wrong in my mind.

OpenStudy (anonymous):

A point x in X is a limit point of S if every neighbourhood of x contains at least one point of S different from x itself?

OpenStudy (anonymous):

Yes. So, written in more precise language, A point \( x \in S\) is a limit point of \(S\) if, for all \( \epsilon > 0\), there exists a point \(y \in S\) such that \( 0 < |x-y| < \epsilon \). So a proof that every point in the set of real numbers is a limit point might begin like this: Let \( \epsilon > 0\) and \(x \in R\) be given. We seek to demonstrate that there exists some \( y \in R\) such that \( 0 <|x-y| < \epsilon \). [...] So your job now is to pick some real number Y that would make that statement true. Don't think too hard - there are an infinite number of choices, and all of them are fairly obvious.

OpenStudy (anonymous):

oh thanks ^^,

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