# Continuous functions in the indiscrete topology?

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?

• No, if $\color{red}{Y}$ has the indiscrete topology, then every map $f\colon X\to Y$, where $X$ is any topological space, is continuous. – egreg May 3 '16 at 22:35
• I don't think one would have much luck connecting the indiscrete topology with the notion of drawing something without lifting one's pen; maybe continuity of functions $\mathbb R\rightarrow\mathbb R$ under the usual topology is supposed to correspond to that, but the indiscrete topology definitely isn't trying to model that situation. – Milo Brandt May 4 '16 at 1:03
• The first paragraph says "I gave up on this concept a while ago, but there's something I'm not clear about". One should always ask more questions. And that is the only way to learn mathematics. Asking more questions, then through the answers you slowly understand, and develop intuition until you don't need to verbally ask the questions, you just answer them yourself. Mathematics is about your tenacity, and your endurance in the face of ignorance. And it has nothing to do with geniuses. – Asaf Karagila May 4 '16 at 4:35
• The definition of continuous function is stated wrongly; in general there is no $f^{-1}:Y\to X$ at all. What does always exist is a map $\def\P{\mathcal P}\P(Y)\to\P(X)$ where $\P(X)$ denotes the power set of$~X$, the set of all subsets of$~X$. Somewhat confusingly this function is usually written $f^{-1}$; it is used here. (It is confusing because if $f$ happens to be invertible, there are two different functions with the same name $f^{-1}$. However whenever applied to an argument, you can tell which one is meant; in computer science this is called overloading and overload resolution.) – Marc van Leeuwen May 4 '16 at 5:30
• If people only spent some of the time that it takes to write a question on trying to think through the definitions... – Carsten S May 4 '16 at 6:59

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.

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

First, the open sets in $X$ are by definition subsets of $X$; if $x\in X$, it makes no sense to ask whether $x$ is open. It is meaningful to ask whether the set $\{x\}$ is open, however.

Now suppose that $X$ is given the indiscrete topology, so that the only open subsets of $X$ are $\varnothing$ and $X$ itself, and let $f:X\to Y$. Then $f$ is continuous if and only if the following statement is true:

whenever $U$ is an open subset of $Y$, $f^{-1}[U]$ is an open subset of $X$.

Since the only open subsets of $X$ are $\varnothing$ and $X$, in this special case we can rewrite the condition:

whenever $U$ is an open subset of $Y$, $f^{-1}[U]$ is either empty or $X$.

Suppose that $U\subseteq Y$, and $f^{-1}[U]=\varnothing$; then $U\cap f[X]=\varnothing$. In other words, $U$ is disjoint from the range of $f$. Now suppose that $f^{-1}[U]=X$; then $U\supseteq f[X]$. In other words, $f$ is continuous if every open subset of $Y$ either contains or is disjoint from $f[X]$.

In general this will not be the case. Suppose, for example, that $Y$ is $T_0$, and $f[X]$ contains at least two points. Let $y_0$ and $y_1$ be distinct points of $f[X]$. Since $Y$ is $T_0$, either there is an open set $U$ in $Y$ such that $y_0\in U$ and $y_1\notin U$, or there is an open set $U$ in $Y$ such that $y_1\in U$ and $y_0\notin U$. In either case $U$ is not disjoint from $f[X]$, since it contains one of the points $y_0$ and $y_1$, but it also does not contain $f[X]$, since it fails to contain one of the points $y_0$ and $y_1$. Thus, $f^{-1}[U]$ is neither $\varnothing$ nor $X$, and $f$ is not continuous.

Matters are very different if $X$ has the discrete topology. Then every subset of $X$ is open, so $f^{-1}[U]$ is open in $X$ for every subset $U$ of $Y$, open or not, and $f$ is therefore automatically continuous.

$f^{-1}$ is not necessarily a function so $f^{-1}:Y \rightarrow X$ doesn't always make sense, but you can talk about pre-images of open sets under $f$. Also, points of $X$ are not technically subsets of $X$, so you shouldn't talk about $x \in X$ being open. You can ask if $\{x\} \subset X$ is open, which it isn't in the indiscrete topology unless $\{x\} = X$; as you stated the only opens are $X$ and $\varnothing$.

It's not true that all $f:X \rightarrow Y$ are continuous when X has the indiscrete topology. Take $X=Y=\{1,2,3\}$ with the indiscrete topology on X, and the discrete topology on Y (all subsets are open). Then the identity function $f(x) = x$ is not continuous since $f^{-1}(\{2\}) = \{2\}$ which is not open in $X$.