Timeline for Mistake on article about Bohr compactification?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| 2 days ago | comment | added | stgo | Right, I almost forgot my own argument! I wanted $b\mathbb{R}\setminus\mathbb{R}$ to be path-connected. Guess it's not going to be that simple, but this is a good lead. Thank you very much! | |
| 2 days ago | comment | added | KP Hart | Alas, $b\mathbb{R}\setminus\mathbb{R}$ is not compact, so not homeomorphic to that torus. | |
| 2 days ago | vote | accept | stgo | ||
| 2 days ago | comment | added | stgo | Great! Thanks for clarifying. What you say about my second question makes a lot of sense, although I can't properly see it myself as I don't have access to [4]. Despite this, if it is as you say, then that solves all of my problems, as I only needed the homeomorphism of $b\mathbb{R}\setminus\mathbb{R}$ with that torus from this paper. I'll see if I can prove it. | |
| 2 days ago | comment | added | KP Hart | First question: the map in the first half of the proof is continuous and injective on the remainder $b\mathbb{R}\setminus\mathbb{R}$, it maps that set onto $\{-1\}\times(b\mathbb{R}\setminus\mathbb{R})$. On the other hand: the closure of $f[\mathbb{R}]$ in the product is obtained by taking its union with the set $\{-1\}\times b\mathbb{R}$, so that map is not onto the remainder $B\mathbb{R}\setminus\mathbb{R}$. | |
| 2 days ago | comment | added | KP Hart | Second question first: It appears that Theorem 4 is based (implicitly) on Theorem 2: the space $B\mathbb{R}$ has weight $\mathfrak{c}$ and hence is indeed embeddable into that torus, but the references to [4] indicate that the embedding Ivanov had in mind derives from the natural embedding of $b\mathbb{R}$ into that torus plus the homeomorphism from Theorem 2. So that proof is invalid as well, and your reference supports this. | |
| 2 days ago | comment | added | stgo | $b\mathbb{R}$ is not path-connected: mathoverflow.net/questions/483682/… | |
| 2 days ago | comment | added | stgo | Thank you for your answer! So, the map it mentions is continuous but fails to have a continuous inverse (or even an inverse at all)? And in regards to the Addendum, are you saying that the result proves ${\rm B}\mathbb{R}\setminus\mathbb{R}\simeq b\mathbb{R}$, or that it would prove that if it were correct? Because if this homeomorphism held true, there would still be a contradiction I believe: Theorem 4 implies that ${\rm B}\mathbb{R}\setminus\mathbb{R}$ is path-connected, as a product of path-connected spaces, but $b\mathbb{R}$ cannot be path-connected (see link in comment below). | |
| 2 days ago | history | edited | KP Hart | CC BY-SA 4.0 |
Addendum
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| 2 days ago | history | edited | KP Hart | CC BY-SA 4.0 |
$e$ became $f$
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| 2 days ago | history | answered | KP Hart | CC BY-SA 4.0 |