Zero is not a Limit Point of Sequence of Reciprocals and Reciprocals + 1

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:

$\ds a_n$

$=$

$\ds \begin {cases} \dfrac 2 {n + 1} & : \text {$n$ odd} \\ 1 + \dfrac 2 n & : \text {$n$ even} \end {cases}$

$\ds $

$=$

$\ds \sequence {\dfrac 1 1, 1 + \dfrac 1 1, \dfrac 1 2, 1 + \dfrac 1 2, \dfrac 1 3, 1 + \dfrac 1 3, \dotsb}$

Then $0$ is not a limit of $\sequence {a_n}$.

The open interval $\openint {-1} 1$ contains $0$, and also contains all terms of $\sequence {a_n}$ with odd indices greater than $1$.

However, all terms of $\sequence {a_n}$ with even indices are outside $\openint {-1} 1$.

Hence $0$ cannot be a limit of $\sequence {a_n}$.

$\blacksquare$

1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $3$