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I'm trying to show that if a sequence converges it has only one limit point. I can see how to do it by contradiction by assuming that there are 2 limit points and constructing an $\epsilon$.

I was wondering if I could just use the definition of convergence which would make a much more succinct proof.

Let $a_{n} \to a$. Then by definition of convergence, for every $\epsilon > 0$, there exists an integer $N$ such that $|a_{n} - a| < \epsilon$ for every $n \geq N$. Does this imply that $a$ is the only limit point of $\{a_{n}\}$? Since it is true for every $n \geq N$, it must be true that there exists an $n \geq N$ such that $|a_{n} - a| < \epsilon$.

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Are you asking how to prove this using the definition of convergence? Or are you just asking if your last paragraph makes sense? – yunone Mar 21 '12 at 1:40
@yunone: I'm asking if my last paragraph makes sense and can be used to prove "if a sequence converges it has only one limit point." – Student Mar 21 '12 at 1:42
up vote 1 down vote accepted

What you should be trying to prove is this:

Suppose a sequence $a_n$ converges to $L$ and also to $M$ in a metric space $X$. Then $L = M$.

Proof: $a_n$ converging to $L$ means that for all $\epsilon >0$ there exists a natural number $N_1$ such that $|a_n - L| < \epsilon/2$ whenever $n \geq N_1$.

Similarly because $a_n$ converges to $M$, for all $\epsilon > 0$ there exists a natural number $N_2$ such that $|a_n - M| < \epsilon/2$ whenever $n \geq N_2$.

Here is what you should set up. From here I would advise you to consider the fact that

$|M - L| = |-a_n + M + a_n - L| \leq |a_n - M| + |a_n - L|$

Can you finish the problem now? Recall that if given $x$ a non-negative real number such that $x$ is always less than any given $\epsilon > 0$, then $x= 0$.

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Then we can say $|M-L| < \epsilon$. So the distance between $L$ and $M$ can be made as small as possible so $L = M$? – Student Mar 21 '12 at 14:20
@Jon Exactly. In other words, for all $n \geq \max\{N_1,N_2\}$ we have $|M - L| \leq |a_n - M| + |a_n - L| <\epsilon/2 + \epsilon/2 = \epsilon$. Since $\epsilon$ is arbitrary, by my last remark above this means that $|M - L|= 0$ so that $M = L$. – user38268 Mar 21 '12 at 21:25
Thank you that helped alot! – Student Mar 21 '12 at 21:33
@Jon Prove the proposition in the last line that I wrote in my answer, namely prove: If given $x$ a non-negative real number such that $x$ is always less than any given $\epsilon>0$, then $x = 0$. – user38268 Mar 21 '12 at 22:13
How about this? If $x \neq 0$, then we could always make $\epsilon < x$. We know we can do this because there is always a rational between two real numbers. If we did this, then there would be a contradiction because we assumed that $x$ is always less than any given $\epsilon > 0$. So $x$ must be 0. – Student Mar 21 '12 at 22:34

The way you have stated what you are trying to show, it's false. The sequence $1,2,1,3,1,4,1,5,1,6,\dots$ has only one limit point in the reals but it doesn't converge.

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Sorry I meant the converse. – Student Mar 21 '12 at 1:26

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