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 $|x_n -x|<r$.

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.

  • $\begingroup$ 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. $\endgroup$ – 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.

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

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