Limit Point of Underlying Set of Sequence of Reciprocals and Reciprocals + 1

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Theorem

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

\(\ds a_n\) \(=\) \(\ds \begin {cases} \dfrac 2 {n + 1} & : \text {$n$ odd} \\ 1 + \dfrac 2 n & : \text {$n$ even} \end {cases}\)
\(\ds \) \(=\) \(\ds \sequence {\dfrac 1 1, 1 + \dfrac 1 1, \dfrac 1 2, 1 + \dfrac 1 2, \dfrac 1 3, 1 + \dfrac 1 3, \dotsb}\)

Let $\struct {\R, \tau}$ denote the real number line under the usual (Euclidean) topology.

Let $S$ denote the set of terms of $\sequence {a_n}$ considered as a subset of $\struct {\R, \tau_d}$.


Then $0$ is a limit point of $S$.


Proof

Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number.

Then the open interval $\openint {-\epsilon} \epsilon$ contains $0$ and all elements $a_m$ of $S$ such that $0 < \dfrac 1 m < \epsilon$.

Hence $\dfrac 1 m \in \openint {-\epsilon} \epsilon$.

Hence the result by definition of limit point of $S$.

$\blacksquare$


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