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 $A$?



Lebesgue dimension of closures satisfying the Z-set condition
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.


