Is R_d x R Locally Euclidean of Dimension 1?

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SUMMARY

The topological space R_d x R, where R_d represents the set of real numbers with the discrete topology and R denotes the Euclidean topology, is established as locally Euclidean of dimension 1. Each point in R_d has a neighborhood homeomorphic to {0} in R^0, while points in R have neighborhoods homeomorphic to R. Consequently, the space R_d x R exhibits local properties akin to R, confirming its local Euclidean nature.

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PsychonautQQ
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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.
 
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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.
 
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