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**lllll** I'm having some difficulty comprehending the following Lemma:

let $\displaystyle f: \ D \mapsto \mathbb{R}$ be continuous at a point $\displaystyle c$ in $\displaystyle D$. If $\displaystyle f(c) \neq 0$ then there exist an $\displaystyle \varepsilon > 0$ and a neighborhood U of c such that $\displaystyle |f(x)|>\varepsilon \ \forall \ x \ \in \ U \cap D$.

now the proof that is given is:

let $\displaystyle {\color{blue} \varepsilon = \frac{|f(c)|}{2} >0}$. since $\displaystyle f$ os continuous at $\displaystyle c$, there there is an $\displaystyle \varepsilon$ with a corresponding $\displaystyle \delta>0$ such that $\displaystyle |x-c|<\delta$, then $\displaystyle |f(x)-f(c)|<\varepsilon$, therefore $\displaystyle f(c)-\varepsilon < f(x) <f(c)+\varepsilon$. If $\displaystyle f(c) > 0$, then $\displaystyle f(c)-\varepsilon < f(x) \longrightarrow \frac{f(c)}{2}<f(x)$. If $\displaystyle f(c) <0$ then $\displaystyle f(x) <f(c)+\varepsilon \longrightarrow f(x) < \frac{f(c)}{2}$, hence $\displaystyle {\color{red} |f(x)| > \frac{|f(c)|}{2} = \varepsilon} \ \forall \ x \ \in \ U \cap D$. where $\displaystyle U=(c-\delta,c+\delta) $

I don't comprehend what they are implying in the proposition, those it have something to do with the distance of f(x) going away from 0?

For the part in blue, why are they picking there epsilon as such?

For the part in red those it simply have to do with the fact that we are taking absolute values? I would think so, but am not 100% on that...