Is $[0,1]$ the Stone-Čech compactification of $(0,1)$?

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.


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 $[0,1]$.

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General Topology

Compactification

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