I have a doubt about how to prove continuity using the definition in terms of open sets. The $\epsilon$-$\delta$ definition of continuity is not very pleasant to work with, however, I know what must be done: we must show that for any given $\epsilon$ there will be a $\delta$ satisfying those inequalities.
But what about the other definition? Let $(M_1, d_1)$ and $(M_2, d_2)$ be metric spaces. A function $f: (M_1, d_1) \to (M_2, d_2)$ is called continuous if for every open set $U$ in $M_2$, the preimage $f^{-1}(U)$ is open in $M_1$. I understand this definition, but how to use it in practice? For instance, let $\mathbb{R}$ denote the real line as a metric space with the usual metric $d(x,y) = |y-x|$, then how can I show using this definition that the function $f:\mathbb{R} \to \mathbb{R}$ given by $f(x)=a_0+a_1x+a_2x^2$ is continuous at every point of $\mathbb{R}$?
My problem is that I don't know even where to start. What procedure should we take?
Thanks very much in advance.
Best Answer
Note that the open-set definition does not handle continuity at single points, but rather as a whole (i.e. at all points).
Let $U\subset\mathbb R$ be an open set. We have to show that $f^{-1}(U)$ is open. So if $a\in f^{-1}(U)$, we must prove that some open ball around $a$ is also in $f^{-1}(U)$. That is, we need to exhibit some $\delta$ such that $|a-x|<\delta$ implies $f(x)\in U$. In order to find such $\delta$, we make use of the fact that $U$ is open, that is that there exists $\epsilon>0$ such that $|y-f(a)|<\epsilon$ implies $y\in U$. As you'll notice, we've ended up in doing simply the the $\epsilon\delta$ proof ...