Small inductive dimension of subspace: is regularity needed Use the definition of small inductive dimension $\operatorname{ind}$ that, for integer $n \geq 0$, space $X$ has $\operatorname{ind} X \leq n$ if at each $x \in X$, for each neighborhood $U$ of $x$ there is a neighborhood $V$ of $X$ with $V \subset U$ and $\operatorname{ind} \operatorname{bdy} V \leq n - 1$.
(In other words, do not require such $V$ to have the additional property that $\operatorname{cls} V \subset U$.)
A standard result is that if $Y$ is a subspace of a topological space $X$, then $\operatorname{ind} Y \leq \operatorname{ind} X$.
This appears, for example, as 1.1.2 in Engelking,Dimension Theory (1978), but with the hypothesis that $X$ be regular.
Is regularity actually needed? If so, where in the proof is it used?
 A: $\DeclareMathOperator{\ind}{ind}\DeclareMathOperator{\bd}{bd}$No, regularity is not needed for the small inductive dimension, however if you look at the large inductive dimension instead, regularity is not even enough to get this inequality. I'm not sure how the inequality is proved in Engelking's book, but here is a proof that doesn't use regularity.
Lemma: Let $X$ be a topological space and $A\subseteq X$ be any subspace. Then $\ind(A)\leq\ind(X)$.
Proof: If $\ind X=-1$ there is nothing to prove, since both $X$ and $A$ are empty. We proceed now by induction, so suppose that for all spaces $Y$ with $\ind Y\leq n-1$ and all $B\subseteq Y$ we know that $\ind B\leq n-1$. We have a space $X$ with $\ind X\leq n$, a subspace $A\subseteq X$, and we want to prove $\ind A\leq n$. Let $a\in A$ and let $G$ be an open nbhd of $a$ in $A$, so that $G=H\cap A$ for some $H\subseteq X$ open. Since $\ind X\leq n$ we can find an open $a\in V\subseteq H$ in $X$ such that $\ind\bd(V)\leq n-1$. Consider now $U=V\cap A$ which is an open neighbourhood of $a$ in $A$. Since we have $\bd^A(U)\subseteq\bd(V)\cap A$, where $\bd^A(U)$ denotes the boundary of $U$ as computed in $A$. By inductive hypothesis $\ind(\bd(V)\cap A)\leq n-1$, since this is a subspace of the $n-1$ dimensional space $\bd(V)$, and so also by inductive hypothesis $\ind(\bd^A(U))\leq n-1$, showing that $\ind A\leq n$.
A: It is not used in that particular proof but it is used in some characterizations, for example $\operatorname{ind} X\le n$ iff for every $x\in X$ and every closed set $F$ with $x\notin F$ there is a partition $L$ between $x$ and $F$ such that $\operatorname{ind} L\le n-1$.
A partition, $L$, between $x$ and $F$ is a closed set with the property that $X\setminus L=U\cup V$ with $U$ and $V$ open and disjoint, and such that $x\in U$ and $F\subseteq V$.
This characterization contains a characterization of regularity.
