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I have a definition similar to the definition of convergence of a sequence which has been slightly altered and want to prove something that must be true but am not entirely sure if my proof is correct;

Suppose $(a_n)$ is a sequence. Then $\exists N\in\mathbb{N}$ such that $\forall \epsilon >0, \forall n\ge N$, $|a_n-a|< \epsilon$

I am unsure on how to show that $a_n=a$ must be true if this condition holds

ie $\exists N\in\mathbb{N}$ such that $\forall \epsilon >0, \forall n\ge N$, $|a_n-a|< \epsilon\Rightarrow a_n=a$.

I started by supposing for a contradiction than $a_n \neq a$. This implies $\exists$ $a_n=a+ \epsilon$ for some $\epsilon>0$ (taking the positive case first, the negative case becomes easy after).

Then $|a_n-a|=|a+ \epsilon -a|= \epsilon < \epsilon$, a contradiction.

Does this seem reasonable or have I gone wrong somewhere?

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    "**Slightly** altered"? You definition means $\;a_n=a\;$ for all $\;n\ge N\;$ ...! – 2017-02-02
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    That was in reference to the fact that definition has been altered with respect to the definition of convergence of a sequence! – 2017-02-02
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    I know, yet that alteration is anything but "slight" as it yields something completely different from the usual definition of limit. – 2017-02-02
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    The alteration is a slight change yet the effect is more than a slight effect but i am referring to the alteration, not the effect which comes down to english, not maths – 2017-02-02

2 Answers 2

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I don't think this should hold for all $n$, but rather, just for all $n \geq N$. By the given statement, \begin{align*} |a_n - a| < \epsilon \end{align*} for any $\epsilon > 0$ and $n \geq N$. Taking the inf on both sides over the set of $\epsilon > 0$, we conclude \begin{align*} 0 \leq |a_n - a| \leq 0 \Rightarrow |a_n - a| = 0 \Rightarrow a_n = a \end{align*} for all $n \geq N$.

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    Yes it doesnt hold for all n that is correct – 2017-02-02
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Let $n \in \mathbb{N}$. Your condition requires that $\left| a_n - a \right| < \varepsilon$ for all $\varepsilon > 0$. Let $\varepsilon_0 = \frac{\left| a_n - a \right|}{2}  \geq 0$ and note that $\left| a_n - a \right| > 0$ and $$ \left| a_n -a \right| < \varepsilon_0 = \frac{\left| a_n - a \right|}{2}$$ lead to a contradiction $2 < 1$. Hence $\varepsilon_0$ has to be zero or $a_n = a$. Either one yields $a_n = a$.