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 $n$.

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.

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.

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


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