Is the radial plane radial?

Question

A space is radial provided that for each $A\subseteq X$ and each $p\in \overline A$ there is a transfinite sequence $(x_\alpha)_{\alpha<\lambda}$ of points of $A$ converging to $p$. (Here $\lambda$ is a limit ordinal and can always be taken to be a regular cardinal.)

In the Radial plane, sets $U$ are open provided for every point $x\in U$ and line $L$ passing through $x$, there exists an open subinterval $S\subseteq L$ with $x\in S\subseteq U$.

I hadn't been aware of the use of "radial" to describe topological spaces with this transfinite sequence property. Possibly of interest are my comments to Intuition behind nets. — Dave L. Renfro, Oct 10 '23 at 19:33
@Dave see e.g. https://www.sciencedirect.com/science/article/abs/pii/B9780444503558500453 for the use of this language — Steven Clontz, Oct 10 '23 at 19:42

score 2 · Accepted Answer · answered Oct 10 '23 at 19:37

No, because names are hard and topology was a mistake. /s

Here's a better reason: in this answer, Eric Wofsey points out that $\mathbb Q^2$ is dense in the radial plane, but the radial plane limit of any sequence in $\mathbb Q^2$ must be on a line segment between two points of $\mathbb Q^2$. However, there are only countably-many such line segments, and PatrickR clarifies that these cannot cover $\mathbb R^2$ (in fact, they miss a Euclidean-dense subset of $\mathbb R^2$) due to the Baire category theorem. Therefore there is a point of the radial plane $\mathbb R^2=cl(\mathbb Q^2)$ that is not the limit of any sequence from $\mathbb Q^2$. Finally, since $\mathbb Q^2$ is countable, this point cannot be the limit of any transfinite sequence either.

https://math.stackexchange.com/questions/4785088 gives more details for the last sentence. — PatrickR, Oct 14 '23 at 05:57

Is the radial plane radial?

1 Answers1