Closure of Integer Reciprocal Space
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Theorem
Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
- $A := \set {\dfrac 1 n: n \in \Z_{>0} }$
Let $\struct {A, \tau_d}$ be the integer reciprocal space under the usual (Euclidean) topology.
Then:
- $A^- = A \cup \set 0$
where $A^-$ denotes the closure of $A$ in $\R$.
Proof
By definition, the closure of $A$ is:
- $A \cup A'$
where $A'$ is the derived set of $A$.
By definition of derived set, $A'$ consists of all the limit points of $A$ in $\R$.
From Zero is Limit Point of Integer Reciprocal Space, the only limit point of $A$ is $0$.
Hence the result:
- $A^- = A \cup \set 0$
$\blacksquare$
Sources
- 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: $1 \ \text{(a)}$