Let $f_1, \ldots$, $f_N:X\to X$ be contractions in the complete metric space $X$, and $K$ the self-similar set with respect to the $f_i$. If $A\subset X$ is compact and
$$A\subset f_1(A)\cup\cdots\cup f_N(A)$$
what I am trying to prove is that then $A\subset K$.
So I was doing is the following, I was trying to prove that
$$ f_1(A)\cup\cdots\cup f_N(A) \subset f_1(K)\cup\cdots\cup f_N(K)=K$$
so we pick a $x \in f_1(A)\cup\cdots\cup f_N(A)$ that means that we can write $x=f_i(a)$ for some $i$ and $a \in A$, but I don't know how to continue because I actually don't have anything that relates $A$ and $K$, another thing I was trying to do is to define $D=f_1(A)\cup\cdots\cup f_N(A) $ and $K=f_1(K)\cup\cdots\cup f_N(K)$, so $D,K$ are compact, and then use a result of compact spaces, but I really don't know which one could help me, because I don't know anyone that involves subsets of images of continous functions.
So, Can someone help me to prove the above theorem?
And other thing, Under the same hypothesis of the previous problem, if $B\subset X$ is nonempty and
$B\supset f_1(B)\cup\ldots\cup f_N(B)$,
then how do you prove that $\bar B \supset K$.
so Does @JonhMa's answer is the same in this case?
Thanks a lot in advance.
One might assume that $A$ is nonempty, or the claim is trivial. I am following the notations in Kigami's book. Recall that one can define a metric $\delta$ on $\mathcal C(X)$, the space of nonempty compact subset of $X$, given by
$$\delta(A, B) = \inf\{ r > 0: A \subset U_r(B),\ \ B\subset U_r(A)\}.$$
This is a metric and the mapping $F : \mathcal C(X) \to \mathcal C(X)$ defined by $$F(A) = f_1(A) \cup \cdots f_n(A)$$
turns out to be a contraction in $(\mathcal C(X), \delta)$. Now let $A\in \mathcal C(X)$ such that $A \subset F(A)$. Then we have
$$A \subset F(A) \subset F^2(A) \subset \cdots \subset F^n(A) \cdots $$
and note that this sequence in $\mathcal C(X)$ converges to $K$ as $n\to \infty$, as $F$ is a contraction. Thus for all $r>0$, there is $N$ so that $\delta (F^N(A), K)<r$. Thus $$A \subset F^N(A) \subset U_rK.$$ As $r$ is arbitrary, $$A \subset \bigcap_{r>0} U_r(K)\Rightarrow A\subset \overline K = K. $$
Remark If we have instead $F(A) \subset A$, then we can conclude $K \subset A$. To see this, note that we have
$$\cdots F^n(A) \subset F^{n-1}(A) \subset \cdots \subset F(A) \subset A$$
for all $n$. Note again that $F_n(A) \to K$ in $\delta$. Thus for all $r>0$ there is $N$ so that $\delta (F^N(A), K) < r$. This implies
$$K \subset U_r F^N(A) \subset U_r(A)$$
Then the same argument implies that $K \subset \overline A = A$.
Further remark The above two results does not hold if $A$ is not compact. For example, let $X = [0,1]$ and $f_1 (x) = \frac x3$, $f_2(x) = \frac x3 + \frac 23$. Then the fixed point of $F$ is the Cantor set $\mathcal C$. If $A = (0,1)$, then $F(A) \subset A$, but $\mathcal C$ is not a subset of $A$ as $\{0,1\} \in \mathcal C$.