I'm trying to find a set of $\mathbb R^2$ that is a a subbasis, but not a basis for a topology on $\mathbb R^2$. I think I have a set, but am having trouble going about proving this.
My set is $S=\bigl\{(a,b)×(-∞,∞)│a,b∈\mathbb R\bigr\}∪\bigl\{(-∞,∞)×(c,d)│c,d∈\mathbb R\bigr\}$. I represented this pictorially, which is how I figured it out.
Any help on getting started proving would be helpful. Thank you!
On
What you have written down is just a standard subbase for the product topology on $\mathbb{R}^2$. The "official" subbase for a product $\prod_{i \in I} X_i$, is just the set $\{\pi_j^{-1}[O]: j \in I, O \subseteq X_j \text{ open } \}$, but we can restrict these $O$ to a fixed base or subbase in the space $X_j$, for each $j$, to get a smaller equivalent subbase. In your subbase you use the base of open intervals in each copy of $\mathbb{R}$ and the sets $\pi_1^{-1}[(a,b)] = (a,b) \times \mathbb{R}$ and $\pi_2^{-1}[(c,d)]= \mathbb{R} \times (c,d)$. This subbase generates the product topology on the plane which equals the topology induced by the standard metric, e.g.
It's not a base as a set like $(a,b) \times (c,d)$ cannot be written as a union of elements from it, but is open in $\mathbb{R}^2$.
So your example is the canonical one, one could say; a similar example can be found in any product space.
Your example $S$ is certainly not a basis, because every open set would be expressible as a union of elements of $S$. Now, is it a subbasis? If so, the smallest topology containing $S$ is the usual topology on $\mathbb R^2$ Since every one of the sets in $S$ is open, you're certainly not going to get more than the usual topology. So it would be enough to show that you get every open set in some basis for the usual topology, say the open disks. I'm sure you can take it from here.