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In this note by G. Eric Moorhouse, which appears to be some course notes handout, it is stated on page 3:

The [two-point] above is the Stone-Čech compactification of $(0,1)≃\mathbb{R}$;that is, $\beta\mathbb{R} ≃ [0, 1]$. This is the most general compactification of $\mathbb{R}$ in a sense that we proceed to describe.

Then he argues, complete with diagrams, that for every embedding of $(0,1)$, there is a unique extension

On the other hand, it was my understand, as similar to how it is set out by Henno Brandsma in this answer, that the Stone-Čech compactification does not generally admit an explicit description, only existence proofs using the axiom of choice. And moreover we can say that it has cardinality $2^{2^\mathfrak{C}}$ (or at least $2^\mathfrak{c}$ I'm not sure) so there's no way that $[0,1]$ could be homeomorphic to $\beta\mathbb{R}.$ Also, as stated in this question, no sequence in $X$ will have its limit in $\beta X\setminus X$, whereas the endpoints of $[0,1]$ are sequential limit points of $(0,1)$.

So what's going on here? Probably just a mistake by Moorhouse, right?

bof
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John Daniels
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    Looks like a big mistake. – bof Mar 31 '18 at 23:56
  • For instance ${(t,\sin\frac1t):0\lt t\lt1}$ is a subset of $\mathbb R\times\mathbb R$ which is homeomorphic to $(0,1),$ and its closure in $\mathbb R\times\mathbb R$ is a compactification of $(0,1).$ – bof Apr 01 '18 at 00:00

2 Answers2

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Yes, Moorhouse is just wrong. Not every map $(0,1)\to Y$ extends continuously to $[0,1]$ as his pictures suggest. For instance, as you approach the ends of the curve, the curve could oscillate infinitely so that the limit of $f(t)$ as $t$ approaches $0$ (or $1$) does not exist.

Eric Wofsey
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    @bof See the disclaimer here: http://ericmoorhouse.org/handouts/ I've alerted Dr. Moorhouse of the error. His website contains great resources...please inform him of any other errors you may find so that these resources can improve. Thanks! – Atsina May 08 '18 at 03:56
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By definition, any map $(0,1)\to X$ where $X$ is compact Hausdorff should factor through $\beta (0,1)$. If you consider the topologist's sine curve $(0,1)\to [-1,1]^2, x\mapsto {(x,\sin\frac1x)}$, it clearly fails to factor through $[0,1]$.

Arnaud Mortier
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