Let us develop the definition of limit point of a sequence
Nathan Kamgang
As a recap, the definition of a limit point of a set states that:
A point $x\in\mathbb{R}$ is a limit point of a set $A\subset\mathbb{R}$ if
$$ \forall \varepsilon>0\ \exists y\in A\setminus{x}\quad\text{such that}\quad |y-x|<\varepsilon. $$Now, consider a sequence of real numbers:
$$ a_1, a_2, a_3, a_4, a_5, a_6, a_7, a_8, a_9, a_{10}, \dots $$Adapting the set definition, one might naively say: for all $\varepsilon>0$, there exists an index $i$ such that $a_i$ is very close to the proposed limit point. Formally, the naive definition would be:
$$ c \text{ is a limit point of } (a_n) \iff \forall \varepsilon>0,\ \exists n\in\mathbb{N} \text{ such that } |a_n-c|<\varepsilon. $$At first glance, this seems like a direct parallel to the set definition. But consider the sequence:
$$ a_1 = 2, \quad a_2 = 100, \quad a_3 = 100, \quad a_4 = 100, \dots $$or more generally,
$$ (a_n) = (2, 100, 100, 100, 100, 100, 100, 100, 100, 100, \dots) $$Intuitively, $100$ should be a limit point. But according to the naive definition:
- For $c=2$: take $\varepsilon < 2$. Then $a_1$ satisfies $|a_1-2|<\varepsilon$, so $2$ would be declared a limit point. This is wrong, because no other terms approach $2$.
- For $c=100$: $a_2, a_3, a_4, \dots$ satisfy the condition, so $100$ is a limit point.
The problem is that the naive definition allows a single element (here $a_1$) to continuously satisfy the definition, which does not capture the intuitive idea of accumulation.
Another natural idea is to forbid using terms equal to the candidate limit point $c$: we require $a_n$ to satisfy $a_n\neq c$, just like the set definition did with $y\in A\setminus{x}$.
$$ c \text{ is a limit point of }(a_n)\iff \forall\varepsilon>0\ \exists n\in\mathbb{N}\ \text{with}\ a_n\neq c\ \text{and}\ |a_n-c|<\varepsilon. $$This copies the set rule “don’t use the point itself” by forbidding $a_n=c$. At first this looks promising: it prevents a single occurrence $a_k=c$ from trivially satisfying every $\varepsilon$-condition. But this attempt also breaks correct cases. For example, take the sequence
$$ (a_n)=(2,100,100,100,100,100,100,100,100,100,\dots). $$Test $c=100$ under the attempted fix. Take $\varepsilon = \tfrac12$. Every term $a_n$ with $n\ge2$ equals $100$, so the condition $a_n\neq 100$ has no element that is close in the sequence. Therefore, this attempted fix fails: it would incorrectly rule out $100$ as a limit point, even though $100$ is clearly an accumulation value of the sequence (indeed, there are infinitely many terms equal to $100$).
In short: excluding the candidate value itself (as we do for sets) does not translate cleanly to sequences. Sequences can repeat values, and repetition can create accumulation even when there are no distinct nearby values. (In sets, excluding $x$ removes the trivial element because sets do not record multiplicity; sequences record multiplicity through indices.)
We therefore need a definition that respects the sequence’s order, for example, requiring an element that is close in every tail or the existence of a convergent subsequence. The formal definitions follow.
Illustration with tails
Start with the sequence:
$$ (a_1, a_2, a_3, a_4, a_5, a_6, a_7, a_8, a_9, a_{10}, \dots) $$- Chop off the first term: $$ (a_2, a_3, a_4, a_5, a_6, a_7, a_8, a_9, a_{10}, \dots) $$
- Chop off the first two terms: $$ (a_3, a_4, a_5, a_6, a_7, a_8, a_9, a_{10}, \dots) $$
- Chop off the first three terms: $$ (a_4, a_5, a_6, a_7, a_8, a_9, a_{10}, \dots) $$
- And so on, ad infinitum.
For a given $\varepsilon>0$, we ask whether there is a term in each tail within distance $\varepsilon$ of the proposed limit point. This ensures no single finite term can satisfy the condition indefinitely.
Apply to our example:
Sequence $(2, 100, 100, 100, 100, 100, 100, \dots)$:
- For $c=2$: choose $\varepsilon = 1/2$. Any tail starting from $N\ge 1$ contains only $100$’s, so there is no term within $\varepsilon$ of $2$. Hence $2$ is not a limit point.
- For $c=100$: any tail contains $100$, so the condition is satisfied. Hence $100$ is a limit point.
This aligns with our intuition.
Formal solution
Using the insight from tails, the correct formal definition for sequences is:
Tail version:
$$ c \text{ is a limit point of } (a_n) \iff \forall \varepsilon>0\ \forall N\in\mathbb{N},\ \exists n > N \text{ such that } |a_n-c|<\varepsilon. $$Subsequence version (equivalent):
There exists a subsequence $(a_{n_k})$ with $n_1