How does $\exists N\; \forall\varepsilon>0 \;\forall k > N \; (d(x_n,x) < \varepsilon)$ differ from definition of convergence?

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$\exists N\; \forall\varepsilon>0 \;\forall k > N \;(d(x_n,x) < \varepsilon) \qquad \implies \qquad\forall\varepsilon>0\;\exists N\;\forall k > N \;(d(x_n,x) < \varepsilon)$

I get that $\forall\varepsilon>0\;\exists N \;\forall k > N \;(d(x_n,x) < \varepsilon)$ is the definition of convergence, and I have no problem proving it either, however I am unsure as to how $\exists N \;\forall \varepsilon >0 \;\forall k > N \;(d(x_n,x) < \varepsilon)$ differs.

I am told that $\exists N \; \forall\varepsilon>0 \;\forall k > N \;(d(x_n,x) < \varepsilon)$ is a stronger statement and that it implies the other, but I am unsure as to why. Any help would be appreciated.

Answer 1

In the first version your N depends on your $\epsilon$ and it might change from $\epsilon $ to $\epsilon.$

The second case is stronger because it provides an $N$ which is good for every $\epsilon$

The second case implies the first case because your $N$ is already provided and you do not have to solve an equation to find it.

Answer 2

The convergence definition says that once $\epsilon$ is picked, you can find an index N such that all the terms after the index is in the $\epsilon$-neighborhood of the limit.

The "if" part of your implication states that you can find an index such that all the terms after are in every $\epsilon$-neighborhood of the limit. That could only be true if the tail of the sequence was constant. Otherwise, we could always find $\epsilon$ to make the statement false.