# Is ${\mathbb R}^n$ with the product topology the same as the metric topology

I have looked at several places into the definition for product spaces. Now all of the definitions I have seen, define the product space topology as generated from the product of sets $U_i$, for which $U_1 \times \dots \times U_n$ is considered open in $X_1\times\dots\times X_n$ when all the $U_i$ are open in $X_i$. The product space is then generated from these sets (infinite unions/finite intersections). So far so good.

Then next as far as I remember, $\emptyset$ is open each $X_i$, which means that $\emptyset\times U_2 \times \dots \times U_n$ is an open set in $X_1 \times \dots \times X_n$.

Now the question arises whether the product topology on ${\mathbb R}^n={\mathbb R}\times\dots\times {\mathbb R}$ is the same as the metric topology. Also shown on e.g. wikipedia to be so, and and another (but clearly equivalent) metric such as discussed in another discussion All of them implicitly assuming in the proof that the first set is not empty.

However, take $U=(0,1)\subset {\mathbb R}$ as an open subset. As far as I can tell there is no open ball in ${\mathbb R}^n$ that will fit in $V=\emptyset\times U\times\dots\times U$, but $V$ is open in ${\mathbb R}\times\dots\times {\mathbb R}$ as far as the definition is concerned, since all of the 'building' sets are open. In fact choosing $\emptyset$ for any of these will just give you ${\mathbb R}^{n-1}$, which is as far as the product space topology definition is concerned an open subset of ${\mathbb R}^n$, but it is not so in the metric sense. From which you could conclude that either the definition is missing something, (i.e. non empty open subsets, rather than open subsets) or the two or not equal.

Or maybe I am missing something altogether.

The product of any set times the empty set is the empty set: $$\emptyset\times U=\emptyset$$
Note that the product is a set of $n$-ples, and the first component must be an element of the first set. But the first set has no elements!
Your $V$ is empty !
Note that $$\emptyset\times U = \emptyset$$ (well, what would be an element in $\emptyset\times U$?) so every set in $\emptyset\times U$ contains an open ball (since there is no such set in the first place).