$x$ is a solution of $0<|x-c|<\delta$

Question

$x$ is a solution of $0<|x-c|<\delta$ iff $c-\delta<x<c+\delta$.

The book says that statement is false, but I can't understand why. $0<|x-c|$ by definition of the absolute value and it's not 0 unless $x=c$. $$x-c>0:x-c<\delta\iff x<c+\delta$$ $$x-c<0:c-x<\delta\iff x>c-\delta$$ Doesn't it prove it true?

Answer 1

If $x=c$ then $c-\delta<x<c+\delta$ is true, but $|x-c|=0$ and $0<|x-c|<\delta$ is false. If we changed the first line to $0\le|x-c|<\delta$ then it would be true: $0\le|x-c|$ and not $0<|x-c|$ by the absolute value definition.

Answer 2

$\text{“}0<|x-c|<\delta\text{''}$ is not quite the same as $\text{“} |x-c|< \delta \text{''}.$

The latter is equivalent to $c-\delta<x<c+\delta.$ The former gives a bit of additional information.