Accumulation Point of Sequence of Reciprocals and Reciprocals + 1

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Theorem

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

Let $\sequence {a_n}$ denote the sequence in $\struct {\R, \tau}$ defined as:

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


Then $0$ is an accumulation point of $\sequence {a_n}$.


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 $0 < \dfrac 1 m < \epsilon$.

We have that:

$\forall n \in \N: n \ge m$

have the property that $0 < \dfrac 1 n < \epsilon$.

Hence there are a countably infinite number of terms of $\sequence {a_n}$ such that $a_n \in \openint {-\epsilon} \epsilon$.

Hence the result by definition of accumulation point of $\sequence {a_n}$.

$\blacksquare$


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