# Lebesgue dimension of closures satisfying the Z-set condition

Given any subspace $A\subset X$ of a topological space with Lebesgue dimension $\le N$.

Let $\bar{A}$ denote the closure of $A$. Assume, that the pair $(\bar{A},A)$ satisfies the Z-set condition, i.e. there is a homotopy $H:\bar{A}\times [0;1]\rightarrow \bar{A}$, such that $H_1=id$ , Image$(H_t)\subset A$ for all $t<1$.

Does this imply, that the Lebesgue dimension of $\bar{A}$ is at most the Lebesgue dimension of $A$?

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It's Lebesgue, not Lebesque. –  TonyK Sep 7 '10 at 14:18

## 1 Answer

No for general topological spaces, yes for metrizable ones (and I believe the argument can be generalized to all normal spaces).

Bad example: $X=\{a,b,c\}$ with open sets $\emptyset$, $X$, $\{a\}$, $\{a,b\}$, $\{a,c\}$. Let $A=\{a\}$, then $\bar A=X$. The homotopy is given by $H_1=id$, $H_t\equiv a$ for $t<1$. The dimension of $A$ is 0 but the dimension of $\bar A$ is 1.

On the positive side, let me begin with a quick and dirty proof in the case when $\bar A$ is a compact metric space. Let $\{U_i\}$ be an open covering of $\bar A$. We need to find a refined covering of multiplicity at most $N+1$ where $N=\dim A$. It suffices to find a continuous map $f:\bar A\to A$ and an open covering $\{V_j\}$ of $A$ such that $\{f^{-1}(V_j)\}$ is a refinement of $\{U_i\}$. Indeed, in this case we can find a refinement of $\{V_j\}$ of multiplicity at most $N+1$ and its $f$-preimage is the desired refinement of $\{U_i\}$.

In the compact case, let $\{V_j\}$ be the covering by $(\rho/3)$-balls where $\rho$ is the Lebesgue number of the covering $\{U_i\}$. Then, for some $t$ sufficiently close to 1, the map $f=H_t$ satisfies the desired property: the preimage of every $V_j$ has diameter less than $\rho$ and hence is contained in some of the sets $U_i$. Indeed, suppose the contrary. Then there is a sequence $t_k\to 1$ and sequences $x_k,y_k\in \bar A$ such that $|x_ky_k|\ge\rho$ but $|H_{t_k}(x_k)H_{t_k}(y_k)|<2\rho/3$. Due to compactness we may assume that $x_k$ and $y_k$ converge to some $x,y\in\bar A$. Then $|xy|\ge\rho$ but $|H_1(x)H_1(y)|\le 2\rho/3$, a contradiction.

In the general metric space case, let $\rho(x)$ denote the local Lebesgue number of $\{U_i\}$ at $x$, that is the supremum of $\rho$ such that the ball $B_\rho(x)$ is contained in one of the set $U_i$. Note that $x\mapsto\rho(x)$ is a positive 1-Lipschitz function on $\bar A$. It is easy to construct a continuous function $u:\bar A\to[0,1)$ such that the distance from $x$ to $H_t(x)$ is less that $\rho(x)/10$ for all $x\in\bar A$ and all $t>u(x)$. Then the map $f:\bar A\to A$ given by $f(x)=H_{u(x)}(x)$ and the covering of $A$ by the balls of the form $B_{\rho(x)/10}(x)$, $x\in A$, will do the job.

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