Dtjytyk

I have started to read Engelking's book "Theory of Dimesions: Finite and Infinite" and the author states (pp. $3$) that:

The small inductive dimension (also called Menger-Urysohn dimension) is a topological invariant, i.e. $hcolon Xto Y$ homeo implies $mathrm{ind}X=mathrm{ind}Y$.

Here I post my tentative proof because I want to be sure it is correct.
Any suggestion is appreciated.

As I understand it, the proof relies on the fact that any homeomorphism preserves boundaries.

Notation: Here $U(x)$, $V(x)$ denote an open nhbd of point $xin X$; $Fr[.]$ is the usual boundary operator.

For the convenience of the reader, this is the definition given in the text:

Definition:

(MU$1$) $mathrm{ind} X=-1$ iff $X=emptyset$.

(MU$2$) $mathrm{ind} Xle n$ iff $forall xin X$ $forall U(x)$ $exists V(x)$ s.t. $V(x)subseteq U(x)$ and $mathrm{ind} Fr[V(x)]le n-1$.

(MU$3$) $mathrm{ind}X=n$ iff $n-1<mathrm{ind}Xle n$.

(MU$4$) $mathrm{ind} X=infty$ iff $mathrm{ind}X>n $ for every $n=-1,0,dots$

Attempt:

(MU$1$) It is obvious.

(MU$2$) Induction on $n$. The case $n=0$ follows by the fact that $Fr[V(x)]=emptyset$; if $nge 1$, we want to prove it for $n+1$. Assume $mathrm{ind}Xle n+1$, then $mathrm{ind}Fr[V(x)]le n$ and by the inductive hypothesis we get
begin{equation}tag{*}
label{eqn:*}
mathrm{ind}Fr[V(x)]=mathrm{ind},h(Fr[V(x)])=mathrm{ind}Fr(h[V(x)])
end{equation}
and we are done.

The remaining cases (MU$3$) and (MU$4$) are obtained in a similar way using eqref{eqn:*}.

Does it work?
Thank you in advance for your help.

If $h:X to Y$ is a homeomorphism then $operatorname{ind}(X) = operatorname{ind}(Y)$.

Maybe first show: $$phi_n:text{for all homeomorphisms } h text { between all spaces } X text{ and } Y: operatorname{ind}(X) le n implies operatorname{ind}(Y) le n$$ by induction on $n in {-1}cup mathbb{N}$.

If $n=-1$, we have $operatorname{ind}(X)le -1 iff X =emptyset$ and as $h$ is a homeomorphism (so a bijection that preserves cardinality), $Y=emptyset$ and so also $operatorname{ind}(Y) =-1 le -1$. This covers the base case.

If $phi_n$ holds, then so does $phi_{n+1}$: Assume $operatorname{ind}(X) le n+1$ and let $y in Y$, and $U_y$ be an open neighbourhood of $y$. As $h$ is a homeomorphism, there is some $x in X$ and an open $U_x$ containing $x$, such that $h(x)=y$ and $h[U_x]=U_y$.

Because of the assumption $operatorname{ind}(X) le n+1$ there exists (by clause MU2 of the definition) an open neighbourhood $V_x$ of $x$ such that $V_x subseteq U_x$ and $operatorname{ind}(operatorname{Fr}(V_x)) le n$. The restriction of $h$ to $operatorname{Fr}(V_x)$ is a homeomorphism with this subspace to $operatorname{Fr}(h[V_x])$, by standard topological facts on homeomorphisms. So by the induction assumption $phi_n$ we know that

$$operatorname{ind}(operatorname{Fr}(h[V_x]) le n$$

and this shows that $h[V_x]$ is the required open neighbourhood that allows us, again from MU2, to conclude that $operatorname{ind}(Y) le n$, as $y$ and $U_y$ were arbitrary.

So $phi_n$ holds by induction.

So if $X$ and $Y$ are homeomorphic spaces via $h$, $Y$ and $X$ are homeomorphic via $h^{-1}$ so for all $n in {-1} cup mathbb{N}$ we have $operatorname{ind}(X)=operatorname{ind}(Y)$ from both inequalities. The infinite case is an immediate corollary: if $X$ is infinite dimensional, we couldn't have $operatorname{Y} le n$ for some $n$ as $phi_n$ applied to $h^{-1}$ gives a direct contradiction.