Question of connection between $0 < | x - c | < δ$ and $c - δ < x < c + δ$, and $x ≠ c$ For a homework question of mine, it says:
$x$ is a solution of $0 < | x - c | < δ$ if and only if $c - δ < x < c + δ$ and $x ≠ c$.
I'm wondering if $x ≠ c$ creates the $ 0 < $ part, while $c - δ < x < c + δ$ creates the $| x - c | < δ$.
So, if $x = c$, does that mean $x$ cannot be a solution? Please help me with this dissection.

The things you need to know are the following general facts:

*

*$0\leq\lvert x\rvert$

*$\lvert x\rvert < b \iff -b<x<b$
Now, to prove the forward direction, suppose $x$ is a solution of $0<\lvert x-c\rvert<\delta$. We have that
\begin{align*}
\lvert x-c\rvert<\delta&\implies-\delta<x-c<\delta\\
&\implies c-\delta<x<c+\delta,
\end{align*}
which proves the first part. Now, assume, to the contrary, that $x=c$. Then $\lvert x-c\rvert=\lvert 0\rvert=0$. Since $0\not<0$, we have reached a contradiction, and it must be that $x\neq c$.
To prove the backwards direction, assume $c-\delta<x<c+\delta$ and $x\neq c$. We have
\begin{align*}
c-\delta<x<c+\delta&\implies-\delta<x-c<\delta\\
&\implies \lvert x-c\rvert<\delta.
\end{align*}
Clearly, $0\leq \lvert x-c\rvert$. Since $x\neq c$, we have $\lvert x-c\rvert\neq 0$. Hence, $0< \lvert x-c\rvert$. Putting this together gives us $0< \lvert x-c\rvert<\delta$.
