Is it possible to make a Polish space Polisher? Suppose that $τ⊆σ$ are two topologies on a set $M$. Is it possible for $(M,τ)$ and $(M,σ)$ to be both Polish spaces?
In other words, if $(M,d_1)$ and $(M,d_2)$ are complete separable metric spaces, is it possible that the topology induced by $d_1$ is strictly finer or strictly coarser than the one induced by $d_2$?
A special case. Can we find a topologically stronger or weaker metric $d$ than the usual one on $\mathbb{R}$ such that $(\mathbb{R}, d)$ is a complete separable metric space?
 A: Yes, this is definitely possible.
Whenever $(X,\tau)$ is a Polish space and $B\subseteq X$ is a Borel set we can find a topology $\sigma$ on $X$ such that:

*

*$\tau\subseteq\sigma$,

*$(X,\sigma)$ is a Polish space

*$B$ is clopen in $\sigma$

*$(X,\tau)$ and $(X,\sigma)$ have the same Borel $\sigma$-algebra

This is easy to do in specific cases, for example when $B$ is closed. In that case you can check that if $\mathcal B$ is a basis for $\tau$, then taking as $\sigma$ the topology whose basis is given by $\mathcal B\cup\{A\cap F\mid A\in\mathcal B\}$ works (indeed this is the same as taking $(X\setminus F)\sqcup F$ with the disjoint union topology).
In particular putting on $\Bbb R$ the disjoint union topology obtained by thinking about $\Bbb R$ as $(\Bbb R\setminus[0,1])\sqcup[0,1]$ gives a Polish topology on $\Bbb R$ finer than the usual one. For an explicit complete metric on the latter example let $d_1$ be the usual Euclidean distance on $[0,1]$ and let $d_2$ be any bounded complete metric on $\Bbb R$ inducing the Euclidean topology (such as $d_2(x,y)=\min\{1,|x-y|\}$). Then $$d(x,y)=\begin{cases}
d_1(x,y) & x,y\in[0,1] \\
d_2(x,y) & x,y\in\Bbb R\setminus[0,1] \\
2 & \text{otherwise}
\end{cases}$$
is a complete metric inducing on $\Bbb R$ the disjoint union topology of $(\Bbb R\setminus[0,1])\sqcup[0,1]$
