I have a question about Lusin space.
Definition A Hausdorff topological space, $(X,\tau)$ is said to be a Lusin space if, there exists a topology $\tau'$ (on $X$) stronger than $\tau$ such that $(X,\tau')$ is a Polish space. Eqivalently, $(X,\tau)$ is a Lusin space, if there exists a Polish space $(Y,\sigma)$ and a continuous bijective mapping from $Y$ to $X$.
My question
Let $X$ be a Lusin space. we adjoin an isolated point $\Delta$ to $X$. Then $X_{\Delta}:=X\cup\{\Delta\}$ is a Lusin space?
More generally, if $(Y,\sigma)$ and $(Z,\rho)$ are Polish spaces, then the topological sum $(Y \amalg Z, \sigma \amalg \rho)$ is a Polish space. For simplicity of notation, assume that $Y\cap Z = \varnothing$, then we can describe the topological sum such that the underlying set is $Y\cup Z$, and the topology is
$$\sigma \amalg \rho = \{ U \subset (Y\cup Z) : (U\cap Y) \in \sigma, (U\cap Z) \in \rho\}.$$
If $\mathscr{B}_\sigma$ is a basis of $\sigma$ and $\mathscr{B}_\rho$ a basis of $\rho$, then
$$\mathscr{B}_{\sigma\amalg \rho} = \{ U\cup V : U\in \mathscr{B}_\sigma, V\in \mathscr{B}_\rho\}$$
is a basis of $\sigma\amalg \rho$. Since $\mathscr{B}_{\sigma\amalg\rho}$ is countable when $\mathscr{B}_\sigma$ and $\mathscr{B}_\rho$ are both countable, it follows that the topological sum of two second countable spaces is second countable.
If $d_Y$ is a metric on $Y$ inducing $\sigma$ such that $(Y,d_Y)$ is a complete metric space, and $d_Z$ a metric on $Z$ inducing $\rho$ and making $(Z,d_Z)$ complete, then
$$d_{Y\amalg Z}(p,q) = \begin{cases} \frac{d_Y(p,q)}{1+d_Y(p,q)} &, p,q \in Y\\ \frac{d_Z(p,q)}{1+d_Z(p,q)} &, p,q\in Z\\ \quad 1 &, \text{otherwise}\end{cases}$$
is a metric on $Y\amalg Z$ inducing $\sigma\amalg \rho$ in which $Y\amalg Z$ is complete.
Hence: The topological sum of two Polish spaces is Polish.
It follows that the topological sum of two Lusin spaces is a Lusin space:
If $f_1 \colon (Y_1,\sigma_1) \to (X_1,\tau_1)$ and $f_2\colon (Y_2,\sigma_2) \to (X_2,\tau_2)$ are continuous bijections from the Polish space $(Y_k,\sigma_k)$ to the Lusin space $(X_k,\tau_k)$, $k\in \{1,2\}$, then
$$\begin{gathered} f_1\amalg f_2 \colon (Y_1\amalg Y_2, \sigma_1 \amalg \sigma_2) \to (X_1\amalg X_2, \tau_1 \amalg \tau_2)\\ (f_1\amalg f_2)(p) = \begin{cases} f_1(p) &, p \in Y_1 \\ f_2(p) &, p\in Y_2\end{cases} \end{gathered}$$
is a continuous bijection, showing that $(X_1\amalg X_2, \tau_1\amalg \tau_2)$ is a Lusin space.
Adjoining an isolated point $\Delta\notin X$ to $X$ means forming the topological sum of $X$ and the one-point space $\{\Delta\}$. Since singleton spaces are Polish, hence Lusin, it follows that the space obtained from adjoining an isolated point to a Lusin space is again a Lusin space.