Let $X$ be a topological space. Assume that there exists a sequence of simple functions $\phi\_n:X\to X$ (finite range and measurable) with $\lim\phi\_n(x)=x$.
Can we concluded $X$ may be written by a countable union $X=\cup X\_n$ where all $X\_n$'s are all relatively second countable? 



Approximation of the identity by simple functions
The answer is no. The Moore plane is a counterexample.


For $z \in \mathbb{R}^2$ and $r \geqslant 0$ denote by $\bar D(z, r)$ the closed Euclidean disc of center $z$ and radius $r$. Recall that the Moore plane is $M = [0, + \infty) \times \mathbb{R}$, where:


* A basis of neighborhoods of $(x, y) \in (0, +\infty) \times \mathbb{R}$ consists in all discs $\bar D((x, y), r)$ with $0 < r < x$;
* A basis of neighborhoods of $(0, y)$ consists in all discs $\bar D((r, y), r)$, with $r>0$.


This space is a usual example of a separable topological space that is not second-countable. To see that $M$ is separable, remark that the set $D = \{(x, y) \in \mathbb{Q}^2 \mid x> 0\}$ is dense in $M$. However $M$ cannot be written as a countable union of second-countable spaces, otherwise the subspace $\{0\} \times\mathbb{R}$ could also be written as such a union; but the inherited topolgy on $\{0\} \times\mathbb{R}$ is discrete.


To show that an approximation of the identity as required exists in $M$, let $(r\_k)\_{k \in \mathbb{N}}$ be an enumeration of $D$ and define $\phi\_n$ as follows: for $z \in M$, consider $r \geqslant 0$ minimal such that the basic neighborhood of $z$ of radius $r$ contains an element of $\{r\_0, \ldots r\_n\}$, and let $\phi\_n(z) = r\_k$ be such an element with $k$ minimal. Then $\phi\_n(z) \rightarrow z$: indeed, if $(r\_{k\_i})$ is a sequence of elements of $D$ converging to $z$, then for $U$ a basic neighborhood of $z$, there is $i\_0 \in \mathbb{N}$ such that $r\_{k\_{i\_0}} \in U$; and then, for all $n \geqslant k\_{i\_0}$, we have $\phi\_n(z) \in U$. Obviously, $\phi\_n$ has finite range, included in $\{r\_0, \ldots, r\_n\}$. And to show that $\phi\_n$ is Borel, remark that for $k \in \{0, \ldots, n\}$, $\phi\_n^{-1}(\{r\_k\}) = A \sqcup B$, where $A \subseteq \{0\}\times \mathbb{R}$ and $$B = \{z \in (0, \infty) \times \mathbb{R} \mid (\forall 0 \leqslant i < k, \, d(z, r\_k) < d(z, r\_i)) \wedge (\forall k < i \leqslant n, \, d(z, r\_k) \leqslant d(z, r\_i))\},$$ where $d$ is the Euclidean distance in $\mathbb{R}^2$.


$A$ is relatively open in $\{0\} \times \mathbb{R}$ as the inherited topology on this subset is discrete. As $\{0\} \times \mathbb{R}$ is closed in $M$, then $A$ is Borel in $M$. In the same way, $B$ is Borel in $\mathbb{R}^2$, so since the topology inherited from $M$ on $(0, +\infty) \times \mathbb{R}$ is the same as the topology inherited from $\mathbb{R}^2$, and since $(0, +\infty) \times \mathbb{R}$ is open in $M$, we deduce that $B$ is Borel in $M$. So $\phi\_n^{-1}(\{r\_k\})$ is Borel in $M$.


