Is R_d x R Locally Euclidean of Dimension 1?

[PsychonautQQ]

Let us look at the topological space R_d x R where R_d is the set of real numbers with the discrete toplogy and R the euclidean topology. This set is not second countable, because R_d has no countable basis.

I am wondering if this space is locally euclidean, and if so, of what dimension? Given a point (x,y) of R_d x R, is there a neighborhood that is homeomorphic to a neighborhood of R^n for some n?

Well every point in R_d has a neighborhood homeomorphic to {0} in R^0, and every point in R has a neighborhood homeomorphic to a neighborhood in, well, R. So I think R_d x R is locally euclidean of dimension 1.

 

[fresh_42]

That's my understanding, too. $(x,y) \in U_{(x,y)} = \{x\} \times (y-\varepsilon,y+\varepsilon) \sim (0,1)$ is an open neighborhood which is locally $\mathbb{R}$. The trouble starts the moment we try to take advantage of this fact.

 

[math771]

To show this, we can take a point (x,y) in R_d x R and consider the neighborhood U = {x} x (y-1,y+1). This neighborhood is homeomorphic to (y-1,y+1) in R, which is a 1-dimensional open interval. Therefore, R_d x R is locally euclidean of dimension 1 at every point.

Additionally, we can see that R_d x R is not locally euclidean of any higher dimension. This is because for any n > 1, there is no way to find a neighborhood of a point (x,y) that is homeomorphic to a neighborhood in R^n. This is because R_d has no countable basis, so we cannot construct a basis for R_d x R using countable products of open sets in R^n.

In summary, R_d x R is locally euclidean of dimension 1, but not locally euclidean of any higher dimension. This is due to the lack of a countable basis in R_d.