Subsequence of Real Sequence Diverging to Negative Infinity Diverges to Negative Infinity
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Theorem
Let $\sequence {x_n}_{n \mathop \in \N}$ be a real sequence with $x_n \to -\infty$.
Let $\sequence {x_{n_j} }_{j \mathop \in \N}$ be a subsequence of $\sequence {x_n}_{n \mathop \in \N}$.
Then:
- $x_{n_j} \to -\infty$
Proof
Let $M > 0$ be a real number.
From the definition of divergence to $-\infty$, there exists $N \in \N$ such that:
- $x_n < -M$
for each $n \ge N$.
From Strictly Increasing Sequence of Natural Numbers, we have:
- $n_j \ge j$ for each $j$.
So, we have:
- $x_{n_j} < -M$
for each $j \ge N$.
Since $M$ was arbitrary, we have:
- $x_{n_j} \to -\infty$
by the definition of divergence to $-\infty$.
$\blacksquare$