Accumulation Point of Sequence is not necessarily Limit Point

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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 point of $\sequence {a_n}$.


Proof

Proof by Counterexample:

Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.

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

\(\displaystyle \sequence {a_n}\) \(=\) \(\displaystyle \begin {cases} 1 & : \text {$n$ odd} \\ n / 2 & : \text {$n$ even} \end {cases}\)
\(\displaystyle \) \(=\) \(\displaystyle \sequence {1, 1, 1, 2, 1, 3, 1, 4, \dotsc}\)


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

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

Hence the result.

$\blacksquare$


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