# A convergent sequence has precisely one accumulation point

As a step in a proof I've been trying to show that convergence of a sequence implies it must have precisely one accumulation point. This is the definition we use for an accumulation point

Let $$S$$ be a set of real numbers. A real number is an accumulation point $$s_0$$ of $$S$$ if and only if for any $$\epsilon > 0$$, there exists at least one point $$t$$ of $$S$$ such that $$0 < |t-s_0| < \epsilon$$.

I wish to prove:

Lemma: A convergent sequence has precisely one accumulation point.

My thoughts: Since the limit of the sequence exists we know that the set cannot have 2 accumulation points or more, we will show by contradiction.

If we would have multiple accumulation points the limit does not exist because we can never get arbitrarily close to a single point (within ϵ), because there exist certain subsequences that each get arbitrarily close to at least two accumulation points, which determine the minimum distance a sequence can be from an accumulation point (so we get a lower bound and therefore we do not get convergence). There must be precisely one accumulation point.

I do not know how to make this more precise of a statement, there are a lot of words and it's not really structured. I'm looking for some help making a more rigourous argument.

Suppose the sequence converges to $$x$$. Clearly $$x$$ is an accumulation point.

Suppose on the contrary that we have a second accumulation point, $$y$$. Let $$r = \frac{|x-y|}{2}$$.

Notice that there exists $$N>0$$, such that for all $$n>N$$, then $$|x_n -x|.

That is when $$n>N$$, we have $$|x_n - y|=|x_n-x+x-y| \ge ||x_n-x|-|x-y||=||x_n-x|-2r|=2r-|x_n-x|>r$$

Hence we cannot have infinitely many points that get arbitrarily close to $$y$$, hence $$y$$ can't be an accumulation point.

• Beautiful. You always come up with such nice and pretty arguments, this is precisely the technical argument I was thinking of but I was not sure how to formulate it so impeccably, thank you. – Wesley Strik Jan 8 at 14:51

An accumulation point of a sequence $$(a_n)_n$$ is not the same as the accumulation point of its set of values $$\{a_n : n \in \mathbb{N}\}$$.

Indeed, consider the constant sequence $$a_n = a, \forall n \in \mathbb{N}$$. Clearly $$(a_n)_n$$ converges to $$a$$, but the set $$\{a_n : n \in \mathbb{N}\} = \{a\}$$ has no accumulation points by your definition.

The proper definition is:

$$x \in \mathbb{R}$$ is an accumulation point of a sequence $$(a_n)_n$$ if for every $$\varepsilon > 0$$ the interval $$\langle x-\varepsilon, x+\varepsilon\rangle$$ contains infinitely many terms of the sequence $$(a_n)_n$$, i.e. for every $$n \in \mathbb{N}$$ there exists $$m \in \mathbb{N}, m > n$$ such that $$|x-a_m| < \varepsilon$$.

Try to show your lemma now.

• Ah, very useful answer indeed. That's a very good point you make. I din't know that actually, thank you. – Wesley Strik Jan 8 at 19:59