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

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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:

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


$A = \set 0$


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:

$\exists N \in \Z: \dfrac 1 N < x$


$x \notin A_N$

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:

$A = \set 0$