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.

  • $\begingroup$ For your example i would try to show that $f (x) \in ]a,b [$ imply $x $ belongs in open set finite intersection or arbitrary infinite of open set. Otherwise i would exploit simple continuous function properties. $\endgroup$ – user8469759 Sep 16 '15 at 22:31

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 ...

  • $\begingroup$ The open-set definition can handle continuity at single point: for any neighborhood $V$ of $f(x)$, there exists a neighborhood $U$ of $x$ such that $f(U) \subseteq V$. $\endgroup$ – Guillaume F. Jun 17 '18 at 10:29

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