Let $(X, d)$ be a compact metric space, topological dimension of $X$ denote by $dim_{top}(X)$ and define $dim_{top}(X)\leq n$ if for every $\epsilon>0$ there is an open cover $\mathcal{U}$ of $X$ by open sets with diameter $\leq \epsilon$ such that every point belongs at most $n+1$ sets of $\mathcal{U}$.
Also inductive dimension of $X$ denote by $ind(X)$ and define
1)$ind(X)=-1$ if and only if $X=\emptyset$,
2) $ind(X)\leq n$ where $n=0\ldots, $ if every $x\in X$ and every open set $V$ of $x$ there exists an open set $U\subset V$ such that
$x\in U$ and $ind(FrU)\leq n-1$ where $Fr(U)$ is boundary of $U$.
what is the relation between $dim_{top}(X)$ and $ind(X)$?
please help me to know it.
For separable metric spaces $\operatorname{ind}(X) = \dim(X)$, where $\dim(X)$ is the covering dimension which for compact metric spaces coincides with your $\dim_{\text{top}}$ (Using Lebesgue numbers for covers).
So they're all equal in your case. Read Hurewicz or Engelking on dimension theory.