**Proposition**: If a sequence $(x_n)$ converges in $\mathbb{R}$, then the sequence of absolute values $(|x_n|)$ also converges.

**Proof**: Let $\epsilon>0$ be arbitrary. Since $(x_n)$ converges, there exists a number *N* such that if $n>N$ then $|x_n-x|<\epsilon$, where *x* is the limit of the sequence $(x_n)$. We will show that for the same values of $\epsilon$ and *N*, if $n>N$ then $\big||x_n|-|x|\big|<\epsilon$, which implies the convergence of the sequence of absolute values.

By the triangle inequality, $|x_n-x|+|x| \geq |x_n|$, which implies $|x_n-x|\geq |x_n|-|x|$. Also by the triangle inequality, $|x-x_n|+|x_n| \geq |x|$, which implies $|x_n-x|\geq |x|-|x_n|$.

Since $|x_n-x|\geq |x_n|-|x|$ and $|x_n-x|\geq -\big(|x_n|-|x|\big)$, it follows that $|x_n-x|\geq\big||x_n|-|x|\big|$.

Therefore, with the same values of $\epsilon$ and *N* pertaining to the convergence of the sequence $(x_n)$, if $n>N$ then

This implies, by the definition of convergence, that the sequence $(|x_n|)$ converges to $|x|$. $\blacksquare$