Zero is Limit Point of Integer Reciprocal Space Union with Closed Interval

From ProofWiki
Jump to navigation Jump to search


Let $A \subseteq \R$ be the set of all points on $\R$ defined as:

$A := \set {\dfrac 1 n : n \in \Z_{>0} }$

Let $T = \struct {A, \tau_d}$ be the integer reciprocal space under the usual (Euclidean) topology.

Let $B$ be the uncountable set:

$B := A \cup \closedint 2 3$

where $\closedint 2 3$ is a closed interval of $\R$.

$2$ and $3$ are to all intents arbitrary, but convenient.

Then $0$ is a limit point of $B$ in $\R$.


Let $U$ be an open set of $\R$ which contains $0$.

From Open Sets in Real Number Line, there exists an open interval $I$ of the form:

$I := \openint {-a} b \subseteq U$

By the Axiom of Archimedes:

$\exists n \in \N: n > \dfrac 1 b$

and so:

$\exists n \in \N: \dfrac 1 n < b$

But $\dfrac 1 n \in B$.

Thus an open set $U$ which contains $0$ contains at least one element of $B$ (distinct from $0$).

Thus, by definition, $0$ is a limit point of $B$ in $\R$.