Say $t,s$ are non-zero real numbers and $\epsilon > 0$ is a positive real number where,
$|t - s| < \epsilon$
Then is is true that
$||t| -|s|| < \epsilon$.
If not what are the conditions required for
$||t| -|s|| \leq |t - s|$
Say $t,s$ are non-zero real numbers and $\epsilon > 0$ is a positive real number where,
$|t - s| < \epsilon$
Then is is true that
$||t| -|s|| < \epsilon$.
If not what are the conditions required for
$||t| -|s|| \leq |t - s|$
Yes. It's true.
\begin{equation} |t| = |(t - s) + s| \leq |t - s| + |s|. \end{equation}
Hence
\begin{equation} |t| - |s| \leq |t - s|. \end{equation}
Swapping the roles of $t$ and $s$ gives $|s| - |t| \leq |t - s|$, whence
\begin{equation} ||t| - |s|| \leq |t - s| \leq \varepsilon. \end{equation}
Where the last inequality is by assumption.
Using absolute value definition:
$ \big||t|-|s|\big| = \begin{cases} |t| - |s| & \mbox{if } |s| \le |t|\\ |s| - |t| & \mbox{if } |t| < |s| \end{cases} $
Then $ \big||t|-|s|\big| = |t| - |s| \le |t - s| \quad \mbox{if}\quad |s| \le |t| $ $ \big||t|-|s|\big| = |s| - |t| \le |s - t| = |t - s| \quad \mbox{if}\quad |t| \le |s| $