Limit of Decreasing Sequence of Unbounded Below Closed Intervals with Endpoint Tending to Negative Infinity
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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
Aiming for a contradiction, suppose 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:
- $\ds \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} = \O$
$\blacksquare$