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The following two statements are equivalent:

A) $\forall \epsilon>0$, the set of $n \in \mathbb{N}$ such that $|x_n-L| \geq \epsilon$ is finite.

B) $\forall \epsilon>0$, $\exists N \in \mathbb{N}$ such that $n \geq N \rightarrow |x_n-L|<\epsilon$.

This is how I proved it:

To prove A $\rightarrow$ B, we will use proof by contradiction. Suppose A is true, but B is still false. The negation of the statement B is, $\exists \epsilon>0$, $\forall N\in\mathbb{N}$ such that $n\geq N \rightarrow |x_n-L| \geq \epsilon$. This contradicts our assumption that for any $\epsilon >0$, we only have finite set of $n \in \mathbb{N}$ such that $|x_n-L| \geq \epsilon$.

To prove B $\rightarrow$ A, suppose B is true. We will also use proof by contradiction. Suppose B is true, but A is false. The negation of the statement A is, for some $\epsilon >0$, the set of $n\in \mathbb{N}$ such that $|x_n-L| \geq \epsilon$ is infinite. But this statement states that we can't find some finite $N\in\mathbb{N}$ such that $n\geq N \rightarrow |x_n-L|<\epsilon$. Hence, it contradicts our assumption of statement B being true. $\blacksquare$

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You got the negation of statement B) wrong in first argument. Your negation basically says that $|x_n-L|\ge \epsilon$ for all $n$. Negation should be "there is $\epsilon>0$ such that there is no $N$ with the property that $n\ge N \implies |x_n-L|<\epsilon$" or, in other words, there is a subsequence $x_{n_k}$ of $x_n$ such that $|x_{n_k}-L|\ge \epsilon$ for all $k\ge 1$. Hence your conclusion doesn't really follow.

Your second argument is fine.

But notice that you don't really need any of contradiction proof;

For $A\implies B$, since the set $S=\{n:|x_n-L|\ge \epsilon\}$ is finite, we can find $M_\epsilon=\max S$ and then $|x_n-L|<\epsilon$ for all $n\ge M_\epsilon+1$ which is bascially B)

For $B\implies A$, if there is $N$ such that $|x_n-L|<\epsilon$ for all $n\ge N$, then the set $S=\{n:|x_n-L|\ge \epsilon\}$ is contained in the finite set $\{1,2,\cdots,N-1\}$ and so is finite.

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