Suppose f: [a,b] -> R and g: [a,b]-> R are both continuous. Let T={x: f(x)=g(x)}. Prove that T is closed.
Suppose $\displaystyle x_0\in [0,1]\setminus T\Longrightarrow f(x_0)-g(x_0)=c\neq 0 \Longrightarrow $ since [/tex]f(x)=g(x)[/tex] is continuous on $\displaystyle [0,1]$ we get that there exists $\displaystyle \epsilon > 0$ s.t.
$\displaystyle f(x)-g(x)\neq 0\,\,\,\forall\,x\in I_\epsilon:=(x_0-\epsilon,x_0+\epsilon)\Longrightarrow I_\epsilon\subset [0,1]\setminus T$ $\displaystyle \Longrightarrow [0,1]\setminus T$ is open ...
Tonio
I think the second way is a bit easier to do...
Since both f and g are continuous, you know that for a sequence $\displaystyle \{ x_n \} $ such that $\displaystyle x_n \rightarrow x_0 $, we have $\displaystyle f(x_n) \rightarrow f(x_0) , g(x_n) \rightarrow g(x_0) $.
Now, think about this, if the sequence $\displaystyle x_n $ is from T, then we should have $\displaystyle f(x_n) = g(x_n) $, right? Well, you just need to prove that $\displaystyle x_0 $ is also in T. In other words, you need something like $\displaystyle f(x_0)=g(x_0)$, which is true, but why?
More generally the follow works whenever the codomain is Hausdorff.
Consider that $\displaystyle A(f,g)=\left\{x\in[a,b]:f(x)=g(x)\right\}=\varphi^{-1}(\Delta)$ where $\displaystyle \Delta$ is the diagonal and $\displaystyle \varphi:[a,b]\mapsto\mathbb{R}\times\mathbb{R}$ given by $\displaystyle x\mapsto(f(x),g(x))$