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Let ##(x_n)_{n=1}^\infty## be a real sequence and ##L \in \mathbb{R}##. Consider the following conditions on

##(x_n)_{n=1}^\infty## and ##L##. $$\forall \varepsilon > 0,~ \forall n_0 \in \mathbb{N},~\exists n \in \mathbb{N}\mbox{ so that } (n \ge n_0 \mbox{ and } |x_n - L| < \varepsilon) \cdots\cdots(1)$$ $$\exists \varepsilon >0, \exists n_0 \in \mathbb{N},\mbox{ so that }\forall n \in \mathbb{N},(n \ge n_0 \Longrightarrow |x_n-L| < \varepsilon)\cdots\cdots(2)$$ $$\exists n_0 \in \mathbb{N}\mbox{ so that }\forall \varepsilon >0, \forall n\in \mathbb{N},(n \ge n_0 \Longrightarrow |x_n-L| < \varepsilon)\cdots\cdots(3)$$ Which one of the following is equivalent to ##(x_n)_{n=1}^\infty## being bounded ? $$\exists L\in\mathbb{R}\mbox{ such that (1) holds}$$ $$\exists L\in\mathbb{R}\mbox{ such that (2) holds}$$ $$\exists L\in\mathbb{R}\mbox{ such that (3) holds}$$ Now when ##(x_n)_{n=1}^\infty## is bounded, ##\exists M >0## such that ##\forall n \in \mathbb{N}## we have ##|x_n| \leqslant M##. I can this why this leads to the first condition, but I am trying to prove this. I have tried negating the goal but all these quantifiers are causing lot of confusion and I am lost. Any guidance will be helpful.

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# I Proving equivalence between statements about a sequence

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