If a continuous function is bigger than another at a point, then it is so in an entire neighbourhood...proof? Let $f,g: \mathbb{R}\to \mathbb{R}$, both be continuous at zero. If $f(0)>g(0)$, show that there exists positive $\delta$ such that for all $y,z \in (-\delta , \delta)$, $f(y)>g(z)$.
Attempt at a proof:
I used the obvious choice for $\epsilon >0$, namely $\epsilon:=f(0)-g(0)>0$, and as a $\delta$ I went for the minimum of $\delta_1$ and $\delta_2$ in the definitions of continuity (at zero) of $f$ and $g$ respectively. Then I have both inequalities $|f(x)-f(0)|<f(0) - g(0)$ and $|g(\tilde{x}) - g(0)|<f(0)-g(0)$, hold for $x\in (-\delta , +\delta)$.
Here is where I am stuck, maybe the choice of $\epsilon$ was bad? Any suggestions would be great.
 A: Hint- its easier to start with $F=f-g$. The assumption is that $F(0)>0$. Need to prove: $F(x)>0$ on a neighbourhood of $0$. Try $\epsilon = F(0)/2$.
A: For every $\epsilon>0$, there exists a $\delta>0$ such that whenever $|x|<\delta$ we have
$$f(0)-\epsilon<f(x)\tag 1$$
and
$$g(0)+\epsilon >g(y)\tag 2$$
From $(1)$ and $(2)$, we have
$$f(x)-g(y)>f(0)-\epsilon -(g(0)+\epsilon)=f(0)-g(0)-2\epsilon$$
Now take any $\epsilon \le \frac12(f(0)-g(0))$ and conclude.
A: Let's go step by step.
$h(x) =f(x) -g(x), $
$h$ is continuos and $h(0)=:a >0.$
Continuos at $x=0:$
Let $\epsilon >0$ be given, there is $\delta >0$ such that 
$|x| \lt \delta$ implies 
$|h(x) -a| \lt \epsilon$, or
$-\epsilon \lt h(x) -a \lt \epsilon $, or
$a -\epsilon \lt h(x) \lt a+ \epsilon.$
Choose $\epsilon =\epsilon_0 \lt a$, I.e.
$0 < a- \epsilon_0.$
Then there exists a  $\delta_0 >0$ such that
$|x| < \delta_0$ implies 
$0 \lt h(x).$
Now the tricky part:
$g$ is continuous at $x=0:$
For a given $\epsilon_1/2 >0$ there  is a $\delta_1>0$ such that 
$|x|<\delta_1$ implies $|g(x)-g(0)| < \epsilon_1/2.$
For $ |x|\lt \delta_1$, $|y |\lt \delta_1:$
$|g(x)-g(y)| \lt $
$|g(x)-g(0)| + |g(y)-g(0)| \lt \epsilon_1,$ 
or
$-\epsilon_1 \lt g(x) - g(y)\lt \epsilon_1.$
We have :
$f(x)-g(y) =$
$(f(x)-g(x))+(g(x) -g(y)) >$
$a-\epsilon_0 + (g(x)-g(y)) > $
$a -\epsilon_0 -\epsilon_1.$
Choose for example:
$ \epsilon_0 =a/4,$  $\epsilon_1=a/4,$ and $\delta :=\min(\delta_0,\delta_1).$
Finally:
For $x,y \in (-\delta,\delta);$
$f(x)-g(y) \gt 0.$
A: Hint: Don't choose that $\varepsilon$. Use its half.
A: Choose any number $k$ such that $f(0)>k>g(0) $. Such a number $k$ exists because reals are dense. Now $f(0)>k$ and $f$ is continuous at $0$ so if we take a small neighborhood of $0$ values of $f$ in this neighborhood can be ensured to be closer to $f(0)$ than to $k$ and thus all these values will be greater than $k$. Similarly another neighborhood of $0$ exists where values of $g$ are less than $k$. The intersection of these two neighborhoods is our desired neighborhood where any value $f$ is greater than any value of $g$. 
A: You are almost there. 
You need to create two disjoint neighborhood, centered at $f(o)$ and $g(0)$ and show that each function has its values in the corresponding neighborhood for a small enough interval centered at $0$.
With $\epsilon = (f(0)-g(0))/2$ your goal will be achieved.    
