# Continuous functions define a topology?

The WP article on general topology has a section titled "Defining topologies via continuous functions," which says,

given a set S, specifying the set of continuous functions $S \rightarrow X$ into all topological spaces X defines a topology [on S].

The first thing that bothered me about this was that clearly this collection of continuous functions is a proper class, not a set. Is there a way of patching up this statement so that it literally makes sense, and if so, how would one go about proving it?

The same section of the article has this:

for a function f from a set S to a topological space, the initial topology on S has as open subsets A of S those subsets for which f(A) is open in X.

This confuses me, because it seems that this does not necessarily define a topology. For example, let S be a set with two elements, and let f be a function that takes these elements to two different points on the real line. Then f(S) is not open, which means that S is not an open set in S, but that violates one of the axioms of a topological space. Am I just being stupid because I haven't had enough coffee this morning?

• The second thing is plain wrong. The open sets in the initial topology induced by $f$ are the sets $f^{-1}(U)$ where $U$ is open in $X$. Except for $\varnothing$, none of these needs to have open image in $X$. – Daniel Fischer Jan 17 '17 at 16:29
• Two nice answers, wish I could accept both! One thing that still bugs me is that given a collection of functions on S that we label as continuous, it's not obvious whether the implied topology on S exists. Daron's answer only shows that if it does exist, it's unique. Also, the thrust of this section in the WP article seems to be that we should be able to bootstrap all of topology from some universal definition of continuity, which would be a much stronger claim -- or maybe I'm misinterpreting what the author had in mind. – Ben Crowell Jan 17 '17 at 23:26
• I don't think I'm competent to fix up that section in the WP article, especially because I'm not sure of the author's intention, but I've put a note on the article's talk page with a pointer to this question. – Ben Crowell Jan 17 '17 at 23:43
• @BenCrowell Re: your first comment, it always exists. Given a set of maps $f_i: S\rightarrow X_i$ for $i\in I$, with each $X_i$ a topological space, the collection of sets $$\{A\subseteq S:\mbox{ for some i\in I, we have } A=f^{-1}_i(U)\mbox{ for some U open in X_i}\}$$ is a subbase for a topology $\tau$ on $S$. The maps $f_i$ are all continuous with respect to $\tau$, and indeed $\tau$ is the coarsest (= fewest open sets) topology on $S$ with this property. – Noah Schweber Jan 17 '17 at 23:51

Your first question is easily addressed. Yes, on the face of it, there are a proper class of such maps. However, we can restrict attention to topological spaces $X$ which have cardinality no greater than $S$, since we can leave off the part of a space not in the image of $f$. Up to homeomorphism, there are only set-many such $X$ (specifically, $2^{2^{\vert S\vert}}$-many at most). And for each specific space $X$, there are only set-many maps from $S$ to $X$.

Incidentally, there is still one remaining subtlety: actually picking out a set of "enough target spaces". This is potentially an issue, since each homeomorphism type contains a proper class of spaces! This can be handled - without even using choice! - by noting that every such homeomorphism type has a representative of rank (in the cumulative hierarchy sense) $\le\kappa+3$, where $\vert X\vert+\aleph_0\le \kappa$ (this is actually massive overkill but oh well); and the class of all topological spaces of a bounded rank is a set.

EDIT: Daron's answer gives a much slicker way to approach the problem. However, it's worth understanding the brute-force approach above, since that kind of reasoning is useful in other contexts as well where we have to deal with an apparent proper class of objects.

Re: your second question, yep, that's a pretty fundamental mistake. It should go the other way: you take the preimage of open sets in the target space. Specifically, the topology induced by $f$ is $$\{A\subseteq S: A=f^{-1}(U)\mbox{ for some U open in the target space}\}.$$ The discrepancy, of course, is due to the fact that $f\circ f^{-1}(U)\subseteq U$, but these sets are not equal in general.

• Re: my italicized paragraph, this is closely related to Scott's trick. – Noah Schweber Jan 17 '17 at 16:40
• Or just consider topological spaces whose underlying set is $S$ (or a subset thereof if you prefer). – Carsten S Jan 18 '17 at 9:42
• One reason this might be found (psychologically) unsatisfying is that among these spaces of cardinality no greater than $S$ will be a copy of the topological space we want to define in the first place. And we only need to consider what will turn out to be the 'identity map' to define the topology as such. But there's no formal problem here! – Daron Jan 23 '17 at 12:26

The first statement can be patched up using functions into the Sierpinski Space $\{0,1\}$ with topology $\{\varnothing , \{1\}, \{0,1\}\}$. Since a continuous function $X \to \{0,1\}$ can be identified with the open set $f^{-1}(1)$ we see the continuous functions into the Sierpinski space are the same thing as the open subsets of $X$.

• Much slicker than what I did, +1. – Noah Schweber Jan 17 '17 at 16:35
• Do we need non-Hausdorff spaces? – Akiva Weinberger Jan 18 '17 at 11:46
• In the general case yes; since $X$ need not be Hausdorff in the first place there is no reason to believe its topology can be given by functions into a Hausdorff space. Now if $X$ is Tychonoff we can do a similar trick to the above, replacing the Seirpinski space with the closed interval $[0,1]$. In case $X$ is Hausdorff but not Tychonoff I'd wager no suitable 'coordinate' space exists. – Daron Jan 23 '17 at 12:23