Suppose $A=\bigcup_i A_i$ is f-invariant. Then $f^n(A_i)\subseteq A$ for all $n\in\mathbb{N}$?

Question

Suppose $A=\bigcup_{i\in I}A_i, A_i\subseteq X,$ for some index set $I$ and $A$ is $f$-invariant for some function $f\colon X\to X$,

Does this imply that, for each $i\in I$, $$ f^n(A_i)\subseteq A~\forall n\in\mathbb{N}? $$

My answer is yes:

$A$ f-invariant means $$ f(A)=\bigcup_{i\in I}f(A_i)\subseteq A. $$

Moreover, $A_i\subseteq A$.

Both together should give $$ f^n(A_i)\subseteq f^n(A)=f^{n-1}(f(A))\subseteq f^{n-1}(A)=f^{n-2}(f(A))\subseteq f^{n-2}(A) $$

and repeating this again and again, I get $$ f^n(A_i)\subseteq f(A)\subseteq A. $$

Am I right?