# Limit of Intersection of Closed Intervals from Zero to Positive Integer Reciprocal

## 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:

- $A = \displaystyle \lim_{n \mathop \to \infty} \bigcap A_n$

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 Archimedean Principle:

- $\exists N \in \Z: N > \dfrac 1 x$

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 $10$