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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

$||t| -|s|| \leq |t - s|$

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By the reversed triangle inequality –  Chasky Oct 25 '12 at 3:16

2 Answers 2

up vote 2 down vote accepted

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.

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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 $$ \big||t|-|s|\big| = |t| - |s| \le |t - s| \quad \mbox{if}\quad |s| \le |t| $$ $$ \big||t|-|s|\big| = |s| - |t| \le |s - t| = |t - s| \quad \mbox{if}\quad |t| \le |s| $$

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