Limit Point of Sequence is Accumulation Point

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Theorem

Let $T = \left({S, \tau}\right)$ be a topological space.

Let $A \subseteq S$.


Let $\left \langle {x_n} \right \rangle$ be a sequence in $A$.

Let $\alpha$ be a limit point of $\left \langle {x_n} \right \rangle$.


Then $\alpha$ is also an accumulation point of $\left \langle {x_n} \right \rangle$.


Proof

Let $\alpha$ be a limit point of $\left \langle {x_n} \right \rangle$.

Then by definition of limit, $\left \langle {x_n} \right \rangle$ converges to $\alpha$.

By definition of convergence that means:

for any open set $U \subseteq T$ such that $\alpha \in U$: $\exists N \in \R: n > N \implies x_n \in U$.

As there is an infinite number of values of $n > N$, there are an infinite number of terms of $\left \langle {x_n} \right \rangle$.

Hence the result, from definition of accumulation point.

$\blacksquare$


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