Exterior and limit points: Quick Question

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The assertion that X\lim(A) equals ext(A) for a subset A of a metric space X is false. The exterior of A is defined as X minus the closure of A, which does not equate to the set of limit points of A. The discussion highlights the importance of considering isolated points, as demonstrated with the example A = [0, 1] ∪ {2} in the real numbers R, where 2 is an isolated point that affects the relationship between limit points and the exterior.

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Is the following true?

X\lim(A) = ext(A), where A is a subset of a metric space X.

I think I've found a proof, but I don't feel very secure about it. So could someone just tell me if this is a correct assertion? What I'm trying to prove is that lim(A) is closed.

Thanks
 
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No, this is false. The exterior of A can be expressed as X minus something, but that "something" is not the set of limit points of A. (As a hint, think about isolated points.)
 
Ahh really? Hmm I'm going to ahve to think about it more then...damnit
 
In R, consider A= [0, 1]\cup \{2\}.
 
Yup, those damn isolated points. Thanks!
 

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