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I want to prove that if there is a limit of a sequence $\{a_n\}$ in a metric space $(M,d)$, there is only one limit and that this limit is an accumulation point (i.e. a point where every $\epsilon$-ball around it contains infinite points of $\{a_n\}$).

Proof. Let $a$ and $b$ be two limits ($a \neq b$) and let $\epsilon = \frac{d(a,b)}{2}.$ There exists a $n_0$ such that $d(a,a_n) < \epsilon, \forall n \geq n_0.$ For such $a_n$ it holds that $d(b,a_n)\geq d(a,b)-d(a_n,a) \geq 2\epsilon - \epsilon = \epsilon$, thus there are only finitely many elements in $K(b,\epsilon)$ thus $b$ is not a limit point nor an accumulation point.

The thing I am having trouble is that in the book I'm reading, the proofs are seperate for the 1st part and the 2nd part of the statement and I cannot see a difference. So is this proof 100% correct or am I missing something?

Jake B.
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1 Answers1

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I believe the proof that there is only one limit is wrong because you begin by defining $d(a,b) < 2\varepsilon$. Instead try this.

Proof. Let $b$ and $c$ be limits of sequence $a = (a_1, a_2, \dots)$. By the definition of a limit we can say that for any $\varepsilon > 0$ there exists $n>N$ for all $N$ such that $\dots$

$$d(b, a_n) < \varepsilon \hspace{3mm} and \hspace{3mm} d(a_n, c) < \varepsilon.$$

Since the above is defined for any $\varepsilon > 0$, $\frac{\varepsilon}{2}$ works just as well. By triangle inequality, we can say that $\dots$

$$d(a, b) \leq d(b, a_n) + d(a_n, c)$$ $\dots $ and since $d(b, a_n), d(a_n, c) <\frac{\varepsilon}{2}$ we get $\dots$ $$d(a, b) \leq d(b, a_n) + d(a_n, c) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2}$$ $$d(a, b) < \varepsilon.$$ And two limits that are within $\varepsilon$ of each other are the same limit, we have proven the first part.

As for the second part with the $\varepsilon-ball$, I am not exactly sure how to go about it, hopefully someone else can help.

bkarthik
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