# Smallest compactification for continuous extension of $\sin(x)$

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).

• 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? – Paul Frost Dec 9 '18 at 9:36

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.
• 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? – Niels J. Diepeveen Jun 29 '16 at 15:54
• So, $\overline{h(X)}$ is homeomorphic to the Closed Topologist's Sine Curve. – yadaddy Jun 29 '16 at 16:03