Limit of Decreasing Sequence of Unbounded Below Closed Intervals with Endpoint Tending to Negative Infinity

Theorem
Let $\sequence {x_n}_{n \mathop \in \N}$ be a decreasing sequence with $x_n \to -\infty$.

Then:


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

That is:


 * $\hointl {-\infty} {x_n} \downarrow \O$

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

Proof
suppose that:


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

Let:


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

Then:


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

From the definition of a sequence diverging to $-\infty$:


 * there exists $N \in \N$ such that $x_N < x$.

But then:


 * $x \not \in \hointl {-\infty} {x_N}$

contradicting that:


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

So we have reached a contradiction, and we have:


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