Limit of Intersection of Closed Intervals from Zero to Positive Integer Reciprocal
Jump to navigation
Jump to search
Theorem
For all (strictly) positive integers $n \in \Z_{>0}$, let $A_n$ be the closed real interval:
- $A_n = \closedint 0 {\dfrac 1 n}$
Let $A \subseteq \R$ be the subset of the real numbers defined as:
- $A = \ds \lim_{n \mathop \to \infty} \bigcap A_n$
Then:
- $A = \set 0$
Proof
First it is noted that:
- $\forall x \in \R_{<0}: x \notin A$
and that by definition of closed real interval:
- $\forall n \in \Z_{>0}: 0 \in A_n$
and so by definition of intersection:
- $0 \in A$
It remains to demonstrate that:
- $\forall x \in \R_{>0}: x \notin A$
Aiming for a contradiction, suppose $\exists x \in \R_{>0}: x \in A$.
By the Axiom of Archimedes:
- $\exists N \in \Z: N > \dfrac 1 x$
and so from Reciprocal Function is Strictly Decreasing:
- $\exists N \in \Z: \dfrac 1 N < x$
Thus:
- $x \notin A_N$
and so by definition of intersection:
- $x \notin A$
This contradicts our supposition that $x \in A$.
Hence the only element of $A$ is $0$, and so:
- $A = \set 0$
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $1$: Sets and Logic: Exercise $10$