In Theorem 3.1(3) on page 10 of Hutchinson's paper 'Fractals and Self similarity' (https://maths-people.anu.edu.au/~john/Assets/Research%20Papers/fractals_self-similarity.pdf) they state in point (iii) that an invariant set K of a finite collection of contraction maps is the union of singletons $k_{i_1,\dots i_p, \dots}$, which are the intersection $\bigcap_{p=1}^{\infty} K_{i_1\ldots i_p} $. I understand why this union of singletons is a subset of $K$, but why are these two sets equal? This is not clearly explained in the paper.
We know that $K$ can be written as the finite union of $K_{i_1,\dots,i_p}$, which is the invariant set after $p$ arbitrary contractions from the collection are applied, but I am not sure if this also holds for infinite unions and intersections.
I was thinking that we can write $K = \bigcup_{i_1, \dots, i_p} K_{i_1, \dots, i_p}$ for finite $p$. The sets are nested in each other like $K \supset K_{i_1} \supset \dots \supset K_{i_1\dots i_p} \supset \dots$.
We can now take the limit $p \to \infty$ to get $K = \lim_{p \to \infty} \bigcup_{i_1, \dots, i_p} K_{i_1, \dots, i_p} = \bigcup_{i_1=1}^N \bigcup_{i_2=1}^N \dots \bigcup_{i_p=1}^N \dots K_{i_1\dots i_p \dots} = \bigcup_{i_1=1}^N \bigcup_{i_2=1}^N \dots (\bigcap_{p=1}^{\infty}K_{i_1, \dots,i_p}) = \bigcup_{i=1}^N \bigcup_{i_2=1}^N \dots \{k_{i_1\dots i_p\dots}\}$
So by letting $p$ go to infinity we get an infinite number of finite unions of singletons, by also making use of the fact that since the sets are nested in each other, $K_{i_1\dots i_p \dots} = \bigcap_{p=1}^{\infty}K_{i_1,\dots,i_p}$. Are all steps taken well-defined/allowed? If not, what would a correct proof look like?
It essentially stems from the fact that $K = \bigcup_{i=1}^{N} K_i$. To see what I mean, fix $x \in K$. Then there exists $1 \le i_1 \le N$ and $x_1 \in K$ such that $x = S_{i_1}(x_1)$. Thus, $x \in K_{i_1}$. Applying the same process to $x_2$ gives $1 \le i_2 \le N$ and $x_2 \in K$ so that $x = S_{i_1}(x_1) = S_{i_1 i_2} (x_2)$ so that $x \in K_{i_1i_2}$. Proceeding inductively we get a sequence $(i_1,i_2,\ldots)$ with the property that $x \in \bigcap_{p=1}^\infty K_{i_1 \ldots i_p}$. So every $x \in K$ appears in one of the singletons.