Slight curiosity. I've learned not to question too much in topology and basically, acquiesce. In a sense, thinking hard or trying to be smart in this area of study is a suicide mission for newbies. Unless for uber futuristic geniuses of course.
Anyway, so a continuous function $f:X \to Y$, in topology is defined as
$f^{-1}:Y \to X$ maps open sets to open sets.
Sure. I don't see how this relates to previously learned continuity of "the graph can be drawn without lifting the pen" or the more rigorous definition using $\lim f(x)$. So this isn't my question.
My question is, well, if we are equipping $X$ with the indescrete topology $\tau=\{X,\phi\}$ then… does any $f:X \to Y$ become continuous?
Well, in this topology, the only open sets are the entire set and the null set. In other words, I take any element $x \in X$ and that element is… wait, as I am writing this, I have another question; is this $x$ open in $X$? My confusion is, when we say open in $X$ we often talk about "sets" or "subsets." Here, Iv've hand picked an element and asked if it is open or not. $X$ itself is open, (as a set) but then, are each elements of it also open too? Or do I have to take all $x$'s in $X$ collectively?
Okay, that's one question, and so back to my original query, say if $x$ is open in $X$, then, whatever I take from $Y$, say $y \in Y$, the inverse $f^{-1}$ will map to some element in $X$…which is an element of an open set $X$. Whatever way I define $f$, as long as there is an inverse, in this case, is $f$ is continuous regardless?
My actual question is dependent on the answer to my spontaneous question but can someone clear these up for me?
Best Answer
First, usually there is no map $f^{-1}:Y\to X$, what we have is a map $f^{-1} : P(Y) \to P(X)$ defined by $f^{-1}(E) = \{ x\in X | f(x) \in E \}$. And this is always well defined.
About your first question :
If you are equiping $X$ with the indiscrete topology, then you don't have continuity in most cases : if your topology on $Y$ is separated, the only continuous functions are constant. Indeed, if $f$ take two values $a$ and $b$, as $Y$ is separated, there is an open set $U$ that contain $a$ but not $b$, that means that $f^{-1}(U) \neq \emptyset$ (because there is an x that verify $f(x) = a$) and $f^{-1}(U) \neq X$ (because there an x that verify $f(x) = b \not\in U$) Hence $f^{-1}(U)$ is not open