Is there a topology on the full transformation semigroup?

Question

$\mathscr T_X$ 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.

Re Tara B's answer

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.

Answer 1

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.

Answer 2

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 $S_{\mathbb{N}}$.