Suppose f,g: D->R are both continuous on D. Define h: D->R by h(x) =max{f(x),g(x)}. Show that h is continuous on D.
Let $\displaystyle x\in D$ and $\displaystyle \varepsilon>0$ be given. Find $\displaystyle \delta_f,\delta_g$ according to the definition of continuity at $\displaystyle x$ for the functions $\displaystyle f$ and $\displaystyle g$. Show that $\displaystyle \min(\delta_f,\delta_g)$ works for $\displaystyle h$.
Wait, Im looking at an example in my book and i'm getting confused. So I know I have to show that i guess |f(x)-f(y)|<delta as well as |g(x)-g(y)|<delta..So then I can let delta be the min{f(x),g(x)}. I'm not sure im confusing myself because this example is the closest one and I'm not sure espicially since i'm looking at a straight example with one fuction f(x). So im not really sure how to even go about this problem.
Crap, I keep forgetting that this form lumps analysis in with topology. Is D a metric space or just a topological space? Which definition of continuity are we working with here? (Of course they're equivalent in any context where they're both valid, but it'll be easier if we're on the same page.)
Yikes!..Well my book is called intro to analysis, and unfortunately it doesn't specifiy and we haven't even discussed that in class so I'm not sure. However, looking at both I feel like i'm torn between them like it could possibly be either or but again my book doesn't specify.
Analysis, okay. So let's say D is a metric space, and d is the metric. Find $\displaystyle \delta_f,\delta_g$ such that $\displaystyle d(x,y)<\delta_f\Rightarrow|f(x)-f(y)|<\varepsilon$, and $\displaystyle d(x,y)<\delta_g\Rightarrow|g(x)-g(y)|<\varepsilon$. This is just using the fact that both f and g are continuous at x. Now let $\displaystyle \delta=\min(\delta_f,\delta_g)$. So both of the above relations hold if we replace $\displaystyle \delta_f$ and $\displaystyle \delta_g$ with $\displaystyle \delta$, right? Now by definition, for every x in D, we have h(x)=f(x) or h(x)=g(x). Can you finish from here?