# Thread: clearification on Continuous Lemma

1. ## clearification on Continuous Lemma

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...

2. Originally Posted by 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...
The reason for the choice of epsilon is it is half of the height of the function. Note that it is nothing special and fraction of the hight of the function would have also worked. ex: $\displaystyle \frac{f(c)}{10}$

If you break it up into two cases it may be clearer

case I.

let $\displaystyle f(c) > 0$ then it states that there is a neighborhood centered at c $\displaystyle (c-\delta,c+\delta)$ such that for every $\displaystyle y \in (c-\delta,c+\delta) \implies f(y) >0$

What the lemma states is that if a function is positive there are points "close" to c that also have positive values.

Case II. would be the simialr but with $\displaystyle f(c) <0$ and so forth.