The standard topoplogy $O_{\text {stantard } \mathbb{R}^{d}}$ on $\mathbb{R}^{d}$ is defined as the collection of sets
$$U \in O_{\text {stantard } \mathbb{R}^{d}}$$
such that $$\forall p \in U: \exists r \in \mathbb{R}^{+}:\quad B_{r}(p) \subseteq U$$ where $B_{r}(p)$ is the usual open ball.
My question : Is the $\bf{product}$ $\bf{topology}$ on $\mathbb{R}^{d}$ induced by the standard topology on $\mathbb{R}$ a standard topology on $\mathbb{R}^{d}$? (where obviously $\mathbb{R}^{d}$ is the $d$ times cartesian product of $\mathbb{R}$)
Note : $\bf{Product}$ $\bf{topology}$ on $A \times B$ is the set of all the subsets of $A \times B$ of the form $X \times Y$ where $X,Y$ are open in $A,B$ respectively. (Though I've also seen at some places the product topology is defined a bit different)
The reason I believe it is not a std. topology is that, for example, $(1,2)\times(1,2)\cup(3,4)\times(3,4)$ is open in $\mathbb{R}^{2}$ but I can't find any two open sets in $\mathbb{R}$ whose cartesian product is this set.
Yes .
The proof is simple since the metric in $\mathbb{R}^d$ is equivalent to the $|•|_2$ and clearly this metric is the product metric of $\mathbb{R}$ d times
And moreover all the metric in real vector spaces are equivalent ...
https://en.m.wikipedia.org/wiki/Product_metric#:~:text=From%20Wikipedia%2C%20the%20free%20encyclopedia,which%20metrizes%20the%20product%20topology.