What is inductive dimension? I cannot understand the inductive dimension properly. I read something on Google but mostly there only are conditions or properties. Not a definition. I got to know about it from the book “ The fractal geometry of nature”. ( I am a 12 grader.)
 A: It's a recursive definition. We define all the spaces of dimension $-1$: we define there to be only one, the empty set with its one and only topology). We then define the spaces of dimension $0$ to be the spaces which have the following properties:


*

*It's not of dimension $-1$ (i.e. non-empty), and

*Every open subset of the space contains a (typically smaller) open subset, whose boundary is contained in the first open set, but whose boundary has dimension $-1$ (i.e. has empty boundary).


Dimension $1$ spaces are defined similarly, with $-1$ replaced with $0$, etc, defining spaces of all integer dimensions. Not every space is covered by this recursive definition, so we define their inductive dimension to be $\infty$.
As an example, take the real line. To show it has dimension $1$, we should first show it does not have dimension $-1$ or $0$.
Showing it doesn't have dimension $-1$ is easy, since it's non-empty. Suppose it had dimension $0$. Then if I take an open set, say $(0, 1)$ for example, I should be able to find an open subset of $(0, 1)$ with empty boundary. Suppose $U \subseteq (0, 1)$ is such an open set. Then $U$ is bounded, so we must have a supremum $\alpha$ of $U$. But then $\alpha$ lies in the boundary of $U$, so the boundary isn't empty after all. Thus $\mathbb{R}$ is not $0$-dimensional.
Let's verify the second property. Take an arbitrary open subset $V \subseteq R$. Then $V$ must contain an open interval $I = (\alpha, \beta)$, and by shrinking the open interval as necessary, I can assume without loss of generality that $\alpha, \beta \in V$. Note that the boundary of $I$ is $\lbrace \alpha, \beta \rbrace$. I claim that this is a $0$-dimensional subspace of $\mathbb{R}$.
Note that $X := \lbrace \alpha, \beta \rbrace$ is not empty, so it's not $-1$-dimensional. The subspace topology on $X$ is discrete, so the full list of open sets are $\emptyset, \lbrace \alpha \rbrace, \lbrace \beta \rbrace, \lbrace \alpha, \beta \rbrace$, all of which have empty boundaries. So, $X$ has dimension $0$.
Thus, I have shown, by definition, that $\mathbb{R}$ has dimension $1$.
A: The usual definition of the small inductive dimension $\operatorname{ind}(X)$ is as follows and defines $\operatorname{ind}(X) \le n$ by recursion: $\operatorname{ind}(X)$ is a function from topological spaces $X$ to $\{-1,0,1, 2 \ldots, \infty\}$ defined as follows
(i) $\operatorname{ind}(\emptyset) =-1$
Suppose we know already what $\operatorname{ind}(Y) \le n$ means for all spaces $Y$ and some $n \in \{-1, 0, \ldots\}$, next we define this for $n+1$:
(ii) $\operatorname{ind}(X) \le n+1$ iff for every $x \in X$, and for all open sets $U$ that contain $x$, there is an open set $V$ that contains $x$ and such that $\overline{V} \subseteq U$ and $\operatorname{ind}(\partial V) \le n$.
A space $X$ has $\operatorname{ind}(X) = n$ iff $\operatorname{ind}(X) \le n$ and not $\operatorname{ind}(X) \le n-1$.
A space has $\operatorname{ind}(X) = \infty$ iff for no $n \in \{-1, 0, 1,2, \ldots\}$ we have $\operatorname{ind}(X) \le n$.
Sometimes we see the $n=0$ defined as the base case ($X$ has a clopen base, which means a base of sets with empty boundary, as $\partial V = \emptyset$ iff $\operatorname{int}(V) = \overline{V} = V$. The $\operatorname{ind}(\emptyset) = -1$ is a sort of trick to make this original base case the first inductive stage. The intuition behind this definition is that the boundary of a set "should" have a dimension one lower than the set we start with. This seems to hold for all normal subsets of the plane, e.g.
Sets $X$ with $\operatorname{ind}(X) = 0$ include all discrete spaces (all sets are clopen), spaces like $\mathbb{Q}$ (as sets of the form $(a,b) \cap \mathbb{Q}$ form a base for it, and these are clopen iff $a,b$ are both irrational.
We have $\operatorname{ind}(\mathbb{R}) \le 1$ as open intervals have discrete boundary sets. It does not have $\operatorname{ind}(\mathbb{R}) \le 0$ as this would imply that $\mathbb{R}$ is disconnected. So $\operatorname{ind}(\mathbb{R}) = 1$, as it should be.
Next, Brouwer was the first to show (using his fixed point theorem) that $\operatorname{ind}(\mathbb{R}^n) = n$ for all $n$. This is a deep theorem (to get the exact value, $\le n$ is easier. As homeomorphic spaces have the same dimension (everything only depends on the topology), this was the first proof that different $n$ the spaces $\mathbb{R}^n$ were not homeomorphic. This was an open problem for a while (and space filling curves has been found, so people did not trust their intuitions about dimension very much).
A: The small inductive dimension can be defined inductively by  


*

*$\text{ind}(\emptyset)=-1$  

*$\text{ind}(\{x\})=0$  

*$\text{ind}(X)$ is the smallest number $n$ such that for all $x\in X$ and every open set $U\ni x$ there is an open set $V\ni x$ with $\bar{V}\subseteq U$ such that  $\text{ind}(\partial V)\leq n-1$
https://en.wikipedia.org/wiki/Inductive_dimension#Formal_definition
