Subsequence of Real Sequence Diverging to Negative Infinity Diverges to Negative Infinity

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 $-\infty$.