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Def.: A topological space $X$ is self-similar if there is a proper subspace $Y \subsetneq X$ such that $Y \simeq X$ (homeo.).

Def.: A topological space $X$ is infinite self-similar if there is a sequence of subspaces $(X_k)_{k \geq 0}$ such that:

$$X=X_0 \supsetneq X_1 \supsetneq X_2 \supsetneq \cdots$$ and $X_i \simeq X$ $ \forall \ i \geq 0$.

Indeed, "infinite self-similar $\Rightarrow$ self-similar" is true.

But, what about to the other implication?

I mean, "self-similar $\Rightarrow$ infinite self-similar" is true? I suspect no.

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Your suspicion is incorrect. Assume that $X$ is a "self-similar'' topological space.. So there is a proper subspace $Y$ of $X$ homeomorphic to $X$ say via a homeomorphism $f : Y \to X$. The sequence of subspaces $X_i$ of $X$ defined by:

$$ \begin{align*} X_0 &= X \\ X_{i+1} &= f^{-1}[X_i] \end{align*} $$

then shows that $X$ is "infinite self-similar".