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Sequences and Limits on Epsilon

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1.Let A be subset of R^p and x belongs to R^p. Then x is a boundary point of A if and only if there is a sequence {a_n} of elements in A and a sequence of {b_n}elements of C(A) (complement of A) such that lim(a_n)=lim(b_n).

2.Prove if{x_n} (n=1..infinity) is a sequence in R^p, x in R^p then TFAE:
a. x_n does not converge to x
b.There is some epsilon > 0 such that for every natural number n there is some natural number m for which x_m does not belong to B(x,epsilon)
c.For every epsilon > 0, there is some subsequence {x_n} (n=1..infinity) such that x_n_m does not belong to B(x,epsilon) for all m.

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By definition, x isa boundary point of A if x belongs to the intersection of the closure A' of A and the closure [C(A)]' of the complements C(A). If x belongs to the closure of A, then for every open ball B_e(x) of radius e centered at x, the intersection of B_e(x) and A is not empty. Respectively, x belongs to the closure of the complement of A, if the intersection of every open ball B_e(x) with the complement C(A) is not empty.
Consider a sequence B_{1/n}(x) of open balls around x of radius 1/n. For every n, he intersection of B_{1/n} with A is not empty, so it contains at least one point, call it x_n. Also, the intersection of B_{1/n}(x) with the complement C(A) is also non-empty, and so contains at least one point. Call it y_n.
We have constructed two sequences in R^p: {x_n} and {y_n} both of these sequences converge to x. Indeed, ...

Solution Summary

Sequences and limits are evaluated in this solution.

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