Limit Point of Sequence is Accumulation Point
From ProofWiki
Theorem
Let $X$ be a topological space.
Let $A \subseteq X$.
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.
$\blacksquare$
Sources
- Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (1970)... (previous)... (next): $\text{I}: \ \S 1$: Limit Points