# Is there a topology on the full transformation semigroup?

$\mathscr T_X$ will denote the set of all functions from a non-empty set $X$ into itself, with the binary operation of composition $\circ$ making it a semigroup, called the full transformation semigroup on $X$.

Is there a topology on the set $\mathscr T_X$ such that $\circ:\mathscr T_X\times \mathscr T_X\longrightarrow \mathscr T_X$ is a continuous function with respect to the product topology on $\mathscr T_X\times \mathscr T_X?$

(i.e., is there a topology on $\mathscr T_X$ making it a topological semigroup?)

Clearly, two (one if $X$ is a one-element set) topologies always work: the discrete and the indiscrete topology on $\mathscr T_X$ make the composition continuous as any function into an indiscrete space is continuous and any function from a discrete space is. (And the product of two discrete spaces is discrete.) I will call those two topologies trivial.

These topologies don't seem useful at all, so I will re-write the question. Let $\operatorname{card}(X)>1.$

Is there a non-trivial topology on $\mathscr T_X$ making it a topological semigroup?

(or at least, is there a construction of such a topology depending on $\operatorname{card}(X)$ which yields non-trivial topologies at least in some cases?)

I cannot think of any general approach to this question and I think I may not have the tools -- I know virtually nothing about topological semigroups. I would be grateful for any help, be it in the form of a hint, a reference, or a full or partial answer to the question. Also, please don't hesitate to comment on anything even remotely related to this.

I may be mistaken but I think your example works only for finite sets. In general, I think, when we have a semigroup $S$ and a non-empty proper subset $A\subset S,$ then $\{\emptyset, A,S\}$ forms a good topology iff the following two conditions are satisfied:

$(1)$ $A$ is a subsemigroup of $S;$

$(2)$ $S\setminus A$ is an ideal in $S.$

Let's say that in this situation, we call $A$ saturated and $S\setminus A$ prime. (I'm not sure if this is standard nomenclature, but I can imagine it might be.)

Suppose a non-empty proper subset $A\subset \mathscr T_X$ is saturated. Then there is $\phi\cdot\operatorname{id}=\phi\in A$ and so $\operatorname{id}\in A.$ Let $\psi\in S_X.$ Then $\psi\psi^{-1}=\operatorname{id}\in A,$ and so $\psi\in A.$ Therefore $S_X\subseteq A.$

But also, let $\mathscr T_X\ni\chi\mathscr J\operatorname{id}.$ Then for some $\alpha,\beta\in\mathscr T_X,$ we have $\alpha\chi\beta=\operatorname{id}.$ Hence $\alpha\chi\in A,$ and so $\chi\in A.$ Therefore the $\mathscr J$-class $J_{\operatorname{id}}$ is contained in $A.$

But for an infinite set $X,$ we have the proper containment $S_X\subsetneq J_{\operatorname{id}},$ because there are functions from $X$ to $X$ whose rank is equal to $\operatorname{card}(X)$ but which aren't permutations. So $S_X$ cannot be saturated.

I think for an infinite set $X$ there will be no such $A$ at all. I'm unable to prove this but I think we can obtain any function in $\mathscr T_X$ by composing functions in $J_{\operatorname{id}}.$ If that's true, then if $A$ were saturated, then it would be a subsemigroup containing a set generating the whole $\mathscr T_X$ and so $A=\mathscr T_X.$

$S_X$ clearly works for finite sets $X$ though. It's a subsemigroup of $\mathscr T_X$ and its complement is an ideal because the composition of functions of which at least one doesn't have the maximal rank cannot have the maximal rank either. And for finite $X,$ a function $\phi: X\longrightarrow X$ is a permutation iff it has the maximal rank. I believe $S_X$ is the only saturated subsemigroup of $\mathscr T_X$ for $X$ finite.

EDIT The statement

"when we have a semigroup $S$ and a non-empty proper subset $A\subset S,$ then $\{\emptyset, A,S\}$ forms a good topology iff the following two conditions are satisfied: $(1)$ $A$ is a subsemigroup of $S;$ $(2)$ $S\setminus A$ is an ideal in $S.$"

is false. When a semigroup isn't a monoid it may not be true. For example, Let $\mathbb N=\{1,2,\ldots\}$ be the additive semigroup of natural numbers. Let $A=\{1\}.$ Then $\{\emptyset,A,\mathbb N\}$ is a good topology on $\mathbb N,$ because the inverse image of $A$ under addition is empty. This is impossible in a monoid. It is also impossible in a monoid for the inverse image of $A$ to be equal $S\times S$ because the image of $S\times S$ under the semigroup operation is equal to $S.$ I have to think about it some more.

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This is no mere semigroup, but a monoid. There is a natural topology on any set of continuous maps between two spaces – called the compact–open topology – but this does not have any good properties if your spaces are not nice. If, however, $X$ is a compactly-generated Hausdorff space, then everything is good and your $\mathscr{T}_X$ is a topological monoid. –  Zhen Lin Mar 24 '12 at 2:35
@ZhenLin But here $X$ is just a set, not a topological space, and I take all functions from $X$ to $X.$ According to what you said, if I take the discrete topology on $X$ and $X$ is a finite set, then I will obtain a good topology on $\mathscr T_X,$ right? (Since then every function from $X$ to $X$ is continuous.) But it won't work for infinite $X.$ Am I correct? –  user23211 Mar 24 '12 at 2:46
A discrete space is a compactly-generated Hausdorff space, and if you want just any topology on $\mathscr{T}_X$ making it into a topological monoid then you can use either the discrete or indiscrete topologies. –  Zhen Lin Mar 24 '12 at 5:47
@ZhenLin I understand, thank you very much. I have read some Wikipedia articles about these things and if I'm not mistaken these two constructions lead to the discrete and the indiscrete topology on $\mathscr T_X$ respectively. But these are trivial in the following sense: any semigroup equipped with one of those topologies becomes a topological semigroup. I should have excluded them in my question, especially given that I'm making a false statement because of not excluding them. I have edited the question. –  user23211 Mar 24 '12 at 9:22
This is another of your questions that I think might be more appropriate for mathoverflow (once you've changed it to not allow my answer). –  Tara B Mar 26 '12 at 22:06

What about the product topology on $X^X$ with $X$ taken discrete?

A subbasis is given by all sets of the form $$U(x,y) = \{f\in X^X\mid f(x) = y\}$$ for $x,y\in X$. Now if $g\circ h \in U(x,y)$, then for all $$(g',h')\in U(h(x),y)\times U(x,h(x))$$ we have $$(g'\circ h')(x) = g'(h'(x)) = g'(h(x)) = y$$ And $U(h(x),y)\times U(x,h(x))$ is an open neighbourhood of $(g,h)$. Therefore $\circ$ is continuous and if $X$ is infinite, the product topology is neither discrete nor indiscrete.

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Thank you very much! –  user23211 Mar 28 '12 at 22:02
@ymar: I'm still wondering though, whether these three topologies might be the only ones. Or whether one could find a topology which is "truly" nontrivial. –  Sam Mar 29 '12 at 18:54
I will be very happy to hear more from you if you come up with anything. I doubt these are the only ones though. Note Tara B's example, which does work for finite sets. However, I suspect there may be trouble finding Hausdorff ones. (And I would love to see if I'm right!) –  user23211 Mar 29 '12 at 19:01
This is nice! $\hspace{1cm}$ –  Tara B Apr 1 '12 at 11:35

This is a somewhat silly answer, but if you really only want to exclude the discrete and indiscrete topologies as 'trivial', then the answer is yes. However, my example is nearly as trivial. Let the open sets be $\emptyset$, $S_X$ (the symmetric group on $X$) and ${\cal T}_X$. Then the inverse image of $S_X$ under composition is $S_X \times S_X$, and so composition is continuous with respect to the product topology.

I expect what you really want is a topology that has infinitely many open sets when $X$ is infinite (and is not the discrete topology)?

EDIT As ymar pointed out, this only works when $X$ is finite. For example in ${\mathcal T}_{\mathbb{N}}$ the preimage of $S_{\mathbb{N}}$ under composition contains the pair $(\alpha,\beta)$, where $n\alpha = n+1$ for all $n\in \mathbb{N}$ and $n\beta = n-1$ for $n\geq 2$ and $1\beta = 1$. Neither $\alpha$ nor $\beta$ is in $S_{\mathbb{N}}$.

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Thanks for this answer! Well, yes, that wouldn't be bad. I asked this question because I was wondering how we could give more structure to the semigroups $\mathscr T_X,$ $\mathscr B_X,$ $\mathscr P_X,$ $\mathscr I_X$ (all subsemigroups of the semigroup of binary relations $\mathscr B_X$). I did realize that perhaps I would have better chances for an answer if I asked about $S_X$ first after a while since posting the question but I decided to wait to see what happens here. :) –  user23211 Mar 26 '12 at 21:44
Sorry, it's a pretty useless answer really. –  Tara B Mar 26 '12 at 22:05
It's not! It answers the question, and I wouldn't have come up with this myself. I will accept this answer if no all-revealing answer comes along. –  user23211 Mar 26 '12 at 22:08
Could you please see my edit? If I'm to ask this question on MO, I'd like to make sure I have the simple stuff figured out, and I feel I don't... –  user23211 Mar 27 '12 at 12:45
I haven't had time to look at it in detail yet, but I will later. You are quite right that my example only works for $S$ finite. I didn't think that through enough. –  Tara B Mar 27 '12 at 15:37