Nested Interval Property

The first application of the axiom of completeness is the following that says that there are no gaps in the real number line:

For each $n \in N$ assume we are given a closed interval (non-empty) $I_{n} = [a_{n}, b_{n}] = {x \in R | a_{n} \leq x \leq b_{n}} \subseteq R$ . Assume also that each $I_{n} \supseteq I_{n + 1}$ . Then the resulting:

I_{1} \supseteq I_{2} \supseteq \dots

has a nonempty intersection; ie $⋂_{n = 1}^{\infty} I_{n} \neq \emptyset$

Proof
To show this, we are going to use the axiom of completeness (AoC) to produce a single real $x \in R$ for each $n \in N$ . Notice that the AoC is a statement about bounded sets, and the one we want to consider is the set:

A = {a_{n} | n \in N}

of left-hand endpoints of the intervals.

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Because the intervals are nested (given by $\supseteq$ condition), then any $b_{n}$ (choose $b_{1}$ for example) serves as an upper bound for $A$ . Thus, we are justified in choosing $x$ such that:

x = sup (A)

Now consider a particular $I_{n} = [a_{n}, b_{n}]$ . Because $x$ is an upper bound for $A$ , we have $a_{n} \leq x$ for all $n \in N$ . The fact that each $b_{n}$ is an upper bound for $A$ and that $x$ is the least upper bound implies that $x \leq b_{n}$ .

With everything, we have $a_{n} \leq x \leq b_{n}$ , so then $x \in I_{n}$ for every choice of $n \in N$ . Hence $x \in ⋂_{n = 1}^{\infty} I_{n}$ so then the intersection isn't empty.
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