Timeline for Spaces $X$ with every compactification $0$-dimensional with $\beta X\setminus X$ not locally compact
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Nov 26 at 21:10 | comment | added | Anonymous | Excuse the typo in the last comment; the $K \setminus A$ should be $K \setminus f(A)$. | |
| Nov 26 at 18:20 | comment | added | Anonymous | The claim is not that the closure of $D$ in $K$ is homeomorphic to $\beta \omega$ but rather that there is a point $y$ in $K \setminus A$ that has two pre-images. Maybe it would be clearer to say that $y$ has at least two pre-images, but two is all that is needed for a contradiction. Does that clarify things? I have edited the answer to include an ``at least''. | |
| Nov 26 at 18:15 | history | edited | Anonymous | CC BY-SA 4.0 |
added 11 characters in body
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| Nov 26 at 16:41 | comment | added | Jakobian | The closure of $N$ in $\beta\omega$ is homeomorphic to $\beta\omega$, but why would the same be true for closure of $D$ in $K$? | |
| Nov 25 at 12:00 | comment | added | Jakobian | When you say that $f$ is two-to-one on the closure of $N$, which fact do you use? If $g:\mathbb{N}\to\mathbb{N}$ is a two-to-one map for example, then it extends to $\hat g :\beta\mathbb{N}\to \mathbb{N}\cup\{\infty\}$ which cannot be two-to-one. | |
| Nov 25 at 11:23 | history | edited | Anonymous | CC BY-SA 4.0 |
A better proof is given.
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| Nov 25 at 11:06 | history | undeleted | Anonymous | ||
| Nov 24 at 19:59 | history | deleted | Anonymous | via Vote | |
| Nov 24 at 16:23 | history | edited | Anonymous | CC BY-SA 4.0 |
added 434 characters in body
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| Nov 24 at 0:09 | comment | added | Jakobian | Oh sorry, that $\beta\mathbb{N}\setminus \mathbb{N}$ is not separable is a corollary of this answer! | |
| Nov 23 at 23:58 | comment | added | Jakobian | I've never seen this fact. Is it because if you take a maximal almost disjoint family $\mathcal{A}$ on $\mathbb{N}$, then $\{\overline{A}\setminus\mathbb{N} : A\in\mathcal{A}\}$ is a pairwise disjoint family of non-empty clopen sets of size $\mathfrak{c}$? | |
| Nov 23 at 21:32 | comment | added | Anonymous | $\beta {\mathbb{N}} \setminus {\mathbb{N}}$ itself is not separable. It also contains an uncountable discrete subset. | |
| Nov 23 at 15:45 | comment | added | Jakobian | Do there exist subspaces $A\subseteq \beta\mathbb{N}$ (equivalently $A\subseteq \beta\mathbb{N}\setminus \mathbb{N}$) such that $A$ is not separable? | |
| Nov 11 at 11:34 | history | answered | Anonymous | CC BY-SA 4.0 |