Metric that induces topology distinct from product topology Find a metric on $\mathbb{R}^n$ that does not induce the same topology as the product topology.My effort:
Consider the metric $$d(\textbf{u},\textbf{w})=\begin{cases}  1   &\text{ if } p \neq q\\
0  &\text{ if } p=q\end{cases}$$
and the open ball $\mathcal{B}_{\frac{1}{2}}(\textbf{u})$ for any $\textbf{u} \in \mathbb{R}^n$. Then $\mathcal{B}_{\frac{1}{2}}(\textbf{u})=\{\textbf{u}\}$ and since each basis element $(w_1,v_1) \times \cdots \times (w_n,v_n)$ in the product topology contains more than one point, there is no basis element in the product topology, contained in $\mathcal{B}_{\frac{1}{2}}(\textbf{u})$ so the topology on $\mathbb{R}^n$ induced by each metric is different.
 A: Yes, your example is fine and induces the discrete topology on $\Bbb R^2$ (where all subsets are open).
Another less trivial example, the post office metric: $$d(x,y) = \begin{cases} 
0 & x=y \\
\|x\| + \|y\| & x \neq y\\
\end{cases}$$
Where $\|x\|$ is the usual Euclidean norm on $\Bbb R^2$.  Here $\underline{0}$ has the usual neighbourhoods (like the Euclidean metric) , but all other points are isolated, as in the discrete metric.
Or the river metric:
$$d(x,y)= \begin{cases}
|x_2 - y_2| & x_1= y_1\\
|x_2| + |y_2| + |x_1- y_1| & x_1 \neq y_1\\
\end{cases}$$
This has the property that all vertical lines are like “almost loose” copies of $\Bbb R$, $0$ has the same neighbourhoods it usually has, but e.g. $(1-\frac1n,1) \not\to (1,1)$ while $(1-\frac1n,0) \to (1,0)$ so the topology is quite different from the usual one (not separable, e.g.) These were early examples in my first year metric topology course back in the days.
A: Yes it is fine. You are indeed describing the discrete topology, the finest topology (notice all singletons are open) you can define on $\mathbb{R}^n$ (I would only clarify what are $p,q$ in your question, for further readers).
