Let $X = (-1,1)^{\Bbb N}$ have the product topology. Is the subset $(0,1)^{\Bbb N}$ open?

Question

Let $X = (-1,1)^{\Bbb N}$ have the product topology. Is the subset $(0,1)^{\Bbb N}$ open?

To consider whether $(0,1)^{\Bbb N}$ is open I know that it is if I can find a basic open set that is a subset of this set, but I'm tangled up with the definitions. Is it that

$(0,1)^{\Bbb N}$ is open if one can find a basic open set $\prod_{n \in \Bbb N} V_n \subset (0,1)^{\Bbb N}$ such that $\color{red}{V_n = (0,1)^\Bbb N}$ for all but finitely many $n$.

or is it that

$(0,1)^{\Bbb N}$ is open if one can find a basic open set $\prod_{n \in \Bbb N} V_n \subset (0,1)^{\Bbb N}$ such that $\color{red}{V_n = (-1,1)^\Bbb N}$ for all but finitely many $n$.

I have some confusion with this since we're dealing with a subspace of $(-1,1)^\Bbb N$.

Answer 1

Your space $X$ is the product of infinitely many copies of $(-1, 1)$, so for the basic open in $X$, that is the interval that almost all of the $V_n$ must be equal.

(In general, a subbasis of a topological product space $\prod X_n$ is $\prod U_n$ where all but one $U_n$ is $X_n$, and thus a basis will have only finitely many $U_n \neq X_n$. Everyone thinks this is strange the first time they encounter infinite products, and would intuitively go for the so-called box topology instead. But there are deep theoretical reasons why this is objectively the best choice, mostly linked to it being the coarsest topology that makes each coordinate projection continuous.)

And clearly, if some $V_n$ are equal to $(-1, 1)$ (and none of the $V_n$ are empty), then $\prod V_n$ cannot be a subset of $(0, 1)^{\Bbb N}$, as $\prod V_n$ will contain points that have at least some negative coordinates.

Answer 2

Neither, but the second is closest.

Simply finding a basic open subset is insufficient to show that a set is open (e.g. on the real line, $(0, 1]$ contains the basic open set $(0, 1)$, but $(0, 1]$ is not open). In order to establish that a set $A$ is open with respect to the topology generated by a basis $\mathcal{B}$, you need to show that, for all $x \in A$, there exists a $U \in \mathcal{B}$ such that $x \in U \subseteq A$. That is, every point in the set is contained in a basic open set (again, which is true of most points of $(0, 1]$, but not for $x = 1$).

On the other hand, this means that any non-empty open set must contain at least one basic open set! While this is isn't sufficient to be non-empty and open, it is definitely necessary. This necessary condition matches your second statement. If you can show that this condition is false, then $(0, 1)^\Bbb{N}$ is definitely not open (but the converse is not true in general).

The (usual) basis of the product topology consists of products of open subsets of $(-1, 1)$, taking the form $\prod_{n \in \Bbb{N}} U_n$, such that all but finitely many $\mathcal{U}_n$ are equal to the full space $(-1, 1)$. Does $(0, 1)^\Bbb{N}$ contain any such basic open sets? No it doesn't; any such $\prod_{n \in \Bbb{N}} U_n$ contains points with $0$ coordinates (in $n$th position, where $n$ is such that $\mathcal{U}_n = (-1, 1)$), and $(0, 1)^\Bbb{N}$ contains no such points. Thus, the necessary condition for openness has failed, and $(0, 1)^\Bbb{N}$ is not open in the product topology.

Answer 3

For any product of topological spaces $X_i$, if $U\subseteq \prod_{i\in I} X_i$ is open and non-empty, and $\pi_j:\prod_{i\in I} X_i\to X_j$ is the projection onto the $j$th factor, then necessarily $\pi_j(U) = X_j$ for all but finitely many $j$.

This is because if we take $x\in U$, then there is some basic open set $V$ with $x\in V\subseteq U$ (that is, $V = \bigcap_{k=1}^m \pi_{j_k}^{-1}(U_k)$ for some $j_1, ..., j_m\in I$ and $U_k\subseteq X_{j_k}$ open), and then we have the inclusions $\pi_j(U)\supseteq \pi_j(V)$, and since $\pi_j(V) = X_j$ for all but finitely many $j$ (it holds for all $j\in I\setminus\{j_1, ..., j_m\}$), the same must hold for $\pi_j(U)$.

In particular, $\pi_n((0, 1)^\mathbb{N}) = (0, 1)\neq (-1, 1)$ for all $n\in\mathbb{N}$, and the above condition doesn't hold. The set cannot be open.