1. ## Continuous Function Problem.

If $\displaystyle f$ & $\displaystyle g$ are continuous on $\displaystyle R$, let

$\displaystyle S=\{x \in R: f(x) \geq g(x) \}$. Also let $\displaystyle (s_n) \subseteq S$. Suppose lim$\displaystyle (s_n)=s$, show that $\displaystyle s \in S$.

Well since f and g are continuous, we know

$\displaystyle \lim(s_n) = s$ then
$\displaystyle \lim(g(s_n)) = g(s)$ and $\displaystyle \lim(f(s_n)) = f(s)$

And $\displaystyle g(s)$ and $\displaystyle f(s)$ exists. I just don't know how to show $\displaystyle f(s) \geq g(s)$.

2. Originally Posted by hockey777
If $\displaystyle f$ & $\displaystyle g$ are continuous on $\displaystyle R$, let

$\displaystyle S=\{x \in R: f(x) \geq g(x) \}$. Also let $\displaystyle s \subseteq S$. Suppose lim$\displaystyle (s_n)=s$, show that $\displaystyle s \in S$.
Look at that notation! There is sonething wrong with the symbols.
I guess I should also add I know the $\displaystyle \lim(f(s_n)) \geq \lim(g(s_n))$
4. Suppose that $\displaystyle g(s) > f(s)$ then let $\displaystyle \varepsilon = \frac{{g(s) - f(s)}}{2} > 0$.
$\displaystyle \left( {\exists N} \right)\left[ {n \geqslant N \Rightarrow \left| {g(s_n ) - g(s)} \right| < \varepsilon \,\& \,\left| {f(s_n ) - f(s)} \right| < \varepsilon } \right]$
Now you also know that $\displaystyle g(s_n ) \leqslant f(s_n )$.