Accumulation Point of Sequence is not necessarily Limit Point

Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $\sequence {a_n}$ be a sequence in $T$.

Let $q \in S$ be an accumulation point of $\sequence {a_n}$.

Then it is not necessarily the case that $q$ is also a limit of $\sequence {a_n}$.

Let $\sequence {a_n}$ be the sequence defined as:

$\ds \sequence {a_n}$

$=$

$\ds \begin {cases} 1 & : \text {$n$ odd} \\ n / 2 & : \text {$n$ even} \end {cases}$

$\ds $

$=$

$\ds \sequence {1, 1, 1, 2, 1, 3, 1, 4, \dotsc}$

Then $\sequence {a_n}$ has exactly one accumulation point, that is $1$.

However, $1$ is not a limit of $\sequence {a_n}$, as $\sequence {a_n}$ has no limit.

Hence the result.

$\blacksquare$