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$(a)$Suppose that ${x^{(n)}}^{(infty)}_{(n=m)}$ is a sequence of points in a metric space $(X,d)$, and let $L in X$. We say that $L$ is a limit point of ${x^{(n)}}^{(infty)}_{(n=m)}$ iff for every $N ge m$ and $epsilon gt 0$ there exists an $n ge N$ such that $d(x^{(n)}, L) le epsilon$.

$(b)$Suppose that ${x^{(n)}}^{(infty)}_{(n=m)}$ is a sequence of points in a metric space $(X,d)$, and let $L in X$. We say that $L$ is a limit point of ${x^{(n)}}^{(infty)}_{(n=m)}$ iff for every $epsilon gt 0$ there exists an $n$ such that $d(x^{(n)}, L) le epsilon$. ?

Is $(a)$ equivalent to $(b)$? Obviously $(a) Rightarrow (b)$, and for the converse let $N in mathbb Z$ and let $d_0 = min{d(x^{(i)}, L), i lt N}$, then there exist $m$ such that $d(x^{(m)}, L) lt d_0/2$.

Is this reasoning correct? (feel free to edit the format or to ask for clarifications and thank you for your time)

Suppose there is an n where x(n)=L. This would make L a limit point by (b), but not by (a)

(b) is not correct, we need infinitely many points of the sequence as close to $L$ as we like , not just one or even finitely many (otherwise $x^{(1)} = L$ would be enough to make $L$ a "(b)-limit point", e.g.) and this can only be ensured by asking for indices beyond any finite $N$. What sequences do "at infinity" (for larger and larger indices) is what matters, not incidental stuff like some finitely many values or some such.