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:
- $t > x$
From Monotone Convergence Theorem (Real Analysis): Decreasing Sequence, we have:
- $x \le x_n$ for each $n$.
Since $x_n \to x$, there exists $N \in \N$ such that:
- $x_N - x < t - x$
That is:
- $x_N < t$
This contradicts:
- $t \le x_n$ 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:
- $\ds \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} = \hointl {-\infty} x$
$\blacksquare$