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Let $f: [0, \infty) \to \mathbb{R}$ be continuous and bounded and let $K$ be a Hausdorff compactification of $[0, \infty)$ such that $f$ has an extension to a continuous function $F$ on $K$. As an example, if $f(x) = \arctan(x)$ we can take the one-point-compactification $[0, \infty]$ and set $F(\infty) := \frac{\pi}{2}$.

If $f(x) = \sin(x)$ then what is the minimal Hausdorff compactification $K$ of $[0, \infty)$ on which $f$ has a continuous extension? (Is it the Closed Topologist's Sine Curve?) The Stone-Cech compactification is the largest one (up to equivalence).

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  • $\begingroup$ You asked your question more than two years ago, so perhaps you are no longer interested in that subject area. But could you define what the "minimal compactification" should be? $\endgroup$ – Paul Frost Dec 9 '18 at 9:36
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Take $f(x) = \frac{x}{1+x}: X = [0,\infty) \rightarrow [0,1]$, which is an embedding and $g(x) = \sin(x): [0,\infty) \rightarrow [-1,1]$.

Then $h= (f,g): X \rightarrow [0,1] \times [-1,1]$ is continuous (defined by $h(x) = (f(x),g(x)))$ and as it separates points and closed sets, $h$ is a homeomorphism from $X$ onto $h[X]$ and $g$ can be extended from $\overline{h[X]}$ by the second projection onto $[-1,1]$. So $(\overline{h[X]},h)$ is a compactification that extends the $\sin(x)$ function.

I'd wager this is the minimal way to do it (we need the extra function as $\sin(x)$ is not injective), though I cannot offer a proof for now. The technique is quite generally applicable.

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  • $\begingroup$ Doesn't minimality follow from the fact that the extension is injective on the remainder $\overline{h[X]}\setminus h[X]$, because the mapping onto a smaller compactification is a quotient map? $\endgroup$ – Niels J. Diepeveen Jun 29 '16 at 15:54
  • $\begingroup$ So, $\overline{h(X)}$ is homeomorphic to the Closed Topologist's Sine Curve. $\endgroup$ – yadaddy Jun 29 '16 at 16:03
  • $\begingroup$ @yadaddy I think that's correct. $\endgroup$ – Henno Brandsma Jun 29 '16 at 16:23

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