Q:

Could there be a finite self-similar set?

While trying to explore the idea of self-similar sets, I came across two (supposedly equivalent) definitions, one based on homeomorphism over topological space, and the other based on contraction maps.
Since a topological space can be finite, my intuition is that there could possibly exist self-similar sets that are finite. Is this correct? And if so, how do we define a contraction map over a finite set? Could anyone help clear my mind with some concrete examples?

A:

In order to define a contraction map, we need a metric, but finite metric spaces are necessarily discrete. In particular, then, if $X$ is a finite set with more than one point, and if $d$ is a metric on $X,$ then there exist distinct $x_0,y_0\in X$ such that $$d(x_0,y_0)=\min\{d(x,y)\mid x,y\in X,x\ne y\}.$$ In order for $f:X\to X$ to be a contraction map, then, we would have $$d\bigl(f(x_0),f(y_0)\bigr)<d(x_0,y_0),$$ so that $d\bigl(f(x_0),f(y_0)\bigr)=0,$ and so $f(x_0)=f(y_0)$.
If (as I suspect) your contraction map definition of self-similarity requires an injective contraction map, it then follows that finite metric spaces are not self-similar, unless they are empty or singletons.
Added:  If you don't require your contraction maps to be injective, then it's actually quite simple to define an appropriate metric and non-trivial contraction map on a finite set of any size. I'll leave the empty and singleton cases to you. Suppose $X$ is an $n$-element set for some finite $n>1,$ say with elements $x_1,\ldots,x_n.$ Then define a metric $\rho$ on $X$ by $$\rho(x_k,x_m)=\begin{cases}0 & \text{if }k=m\\\min\left(\frac1k,\frac1m\right) & \text{otherwise,}\end{cases}$$ and define $f:X\to X$ by $f(x_k)=x_{k+1}$ for $1\le k\le n-1$ and $f(x_n)=x_n.$
Not all metrics will allow this, however. The standard discrete metric is given by $$d(x,y)=\begin{cases}0 & \text{if }x=y\\1 & \text{otherwise,}\end{cases}$$ and the only contraction maps on $X$ (as far as $d$ is concerned) are the constant maps $X\to X.$ Such constant maps will, of course, be contraction maps regardless of the metric, but they will be injective precisely when the set in question has one point at most.

Added: The non-surmorphism" definition given by Wikipedia is somewhat problematic, and depends on what is meant by homomorphism in this case.
My first thought upon reading the definition is that it is a typo, and is referring non-surjective maps $f_s:X\to X$ such that $f_s$ is a homeomorphism $X\to f_s(X)$ for each $s\in S.$ If this is the case, then among topological spaces with finite underlying sets, only the empty space is self-similar in this fashion, since we can simply allow $S$ to be empty in that case, while if $X\ne\emptyset,$ any injective function $f:X\to X$ is necessarily surjective, and we can't allow $S$ to be empty.
If, on the other hand, it is intended to indicate that each $f_s:X\to X$ is to be a non-surjective continuous function, then (as pointed out by Henning Makholm) the non-definition works for all finite topological spaces except for singletons, as any finite set can be "covered" by finitely-many images of constant self-maps (the empty set again covered by an empty family of such), which are necessarily continuous
(regardless of topology), and will be non-surjective precisely when the set has more than one point.

