# Limit of Decreasing Sequence of Unbounded Below Closed Intervals

## Theorem

Let $x \in \R$.

Let $\sequence {x_n}_{n \mathop \in \N}$ be a decreasing sequence converging to $x$.

Then:

$\ds \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} = \hointl {-\infty} x$

That is:

$\hointl {-\infty} {x_n} \downarrow \hointl {-\infty} x$

where $\downarrow$ denotes the limit of decreasing sequence of sets.

## Proof

We first show that:

$\ds \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} \subseteq \hointl {-\infty} x$

Let:

$\ds t \in \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n}$

Then, from the definition of set intersection, we have:

$t \in \hointl {-\infty} {x_n}$ for each $n \in \N$.

That is:

$t \le x_n$ for each $n$.

Aiming for a contradiction, suppose suppose that:

$t > x$
$x \le x_n$ for each $n$.

Since $x_n \to x$, there exists $N \in \N$ such that:

$x_N - x < t - x$

That is:

$x_N < t$

$t \le x_n$ for each $n$.

So we have $t \le x$, and so $t \in \hointl {-\infty} {x_n}$, giving:

$\ds \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} \subseteq \hointl {-\infty} x$

by the definition of subset.

$\Box$

We now show that:

$\ds \hointl {-\infty} x \subseteq \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n}$

Let:

$\ds t \in \hointl {-\infty} x$

Then we have:

$t \le x$

Since:

$x \le x_n$ for each $n$

we have:

$t \le x_n$ for each $n$.

That is:

$t \in \hointl {-\infty} {x_n}$ for each $n$.

So, from the definition of set intersection, we have:

$\ds t \in \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n}$

so:

$\ds \hointl {-\infty} x \subseteq \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n}$

by the definition of subset.

$\Box$

So, from the definition of set equality, we have:

$\ds \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} = \hointl {-\infty} x$

$\blacksquare$