Question about open sets in the product topology in R^N Hi guys I have a statement that may be true or false. I believe it is false but I cannot prove or even show by example my claim.

If $U$ is open in $\Bbb R^\Bbb N$ then $U= \prod _{n=1} ^\infty U_n$ where each $U_n$ is open in $\Bbb R$.

I think this is false because I know that even for $X \times Y$ not every open set is of the form $U \times V$. Where $U$ and $V$ are open in $X$ and $Y$ respectively.Can someone help me here?
The statement is very false (for the product topology). Suppose the $U_n$ are non-empty open subsets of $\mathbb{R}$. Then $U = \prod_{n+1}^\infty U_n$ is open in the product topology iff for all but finitely many $n$, $U_n = \mathbb{R}$.

Proof: pick $(x_n) \in U$, which can be done (axiom of countable choice, formally) as all $U_n$ are non-empty. Then there is a finite subset $F$ of $\mathbb{N}$ and finitely many open set $O_n \subseteq \mathbb{R}$ for $n \in F$ such that $O = \prod_{n=1}^\infty O_n$ is a basic open product set (so $O_n = \mathbb{R}$ for $n \notin F$) and $x \in O \subseteq U$. Then for $n \notin F$, let $p \in \mathbb{R}$ be arbitrary. Then the point $y_m = x_m, m \neq n$, $y_n = p$ is in $O$ (as $x$ is, essentially, and we modify $x$ at a coordinate where $O_n = \mathbb{R}$), so $y \in U$, so in particular $y_n = p \in U_n$. As $p$ was arbitrary, $U_n = \mathbb{R}$, and this holds for all $n \in F$, so we are done.

There is another topology on $\mathbb{R}^\mathbb{N}$ that has exactly all those $U = \prod_n U_n$ (no restrictions of the all but countable type) as a base, which is, as we have seen now, stricly larger than the product topology, called the box topology. The product topology has much better properties (it's a connected metrisable space in this case, and separable etc.) while in the box topology this product is not even known to be normal in ZFC (IIRC). And it's certainly not connected, first countable, separable etc. in the box topology.

It's true that the statement as it stands already fails for finite products (the open unit circle is not of the form $U_1 \times U_2$ in the plane), but there at least we have that all open sets are (countable even) unions of such product open sets. In the infinite product topology, most of such sets are not even open. But again, if we apply the condition that all but finitely many $U_n$ equal the whole reals, we do get all usual product topology base elements, and we can write all open sets in $\mathbb{R}^\mathbb{N}$ as (countable) unions of those again.
You are correct, the statement is false. Consider some simple examples, such as the unit disc in $\mathbb{R}^2$; there is no way to express this as a product of two open sets in $\mathbb{R}$.

You can prove this explicitly by considering what sets you would have to pick in order to to include all the points on the disc. For example, you know that $(1, 0)$ and $(-1, 0)$ need to be included, so by considering the first entry, you have to have $U_1\subseteq (-1, 1)$, and similarly for $U_2$. However,  you immediately know that this won't work, as you are now including points like $(1, 1)$ in the product $U_1\times U_2$, even though this sits outside the unit circle.

There's a topological fact (sometimes given as a definition, sometimes a theorem, depending on the spaces you're looking at) that every open set in a product topology is a union of products of open sets - though this might be infinite!