# Stone–Čech compactification with topologist's sine curve

The Stone–Čech compactification $\beta((0, 1))$ of $(0, 1)$ can be described as the closure of $(0,1)$ in $\prod_{j\in J} \mathbb{R}$ where $J$ is the set of bounded continuous functions $f : (0,1) \rightarrow \mathbb{R}$. Also recall that for any compactification $Y$ of $(0,1)$ there is a unique map $\beta((0,1)) \rightarrow Y$ which is the identity on $(0,1)$. Let $Y$ be the topologist's sine curve and give a formula for the image in $Y$ of a point $(x_f)_{f \in J}$ in $\beta((0,1))$.

The compactification takes points, say $0.5$, to $(f(0.5),\dots, f_j(0.5))$, but I'm not really sure how to proceed from there.

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For each $f\in C^*(Y)$ let $\mathbb{R}_f$ be a copy of $\mathbb{R}$, and define $$e_Y:Y\to\prod_{f\in C^*(Y)}\mathbb{R}_f:y\mapsto \left\langle f(y):y\in C^*(Y)\right\rangle\;;$$ $e$ is the evaluation map and is a homeomorphism of $Y$ into $\prod_f\mathbb{R}_f$.

Similarly, for each $f\in C^*((0,1))$ let $\mathbb{R}_f$ be a copy of $\mathbb{R}$, and define $$e:(0,1)\to\prod_{f\in C^*((0,1))}\mathbb{R}_f:x\mapsto \left\langle x,f(x):f\in C^*((0,1))\right\rangle\;;$$ $\beta((0,1))$ is the closure of $e[(0,1)]$ in $\prod_f\mathbb{R}_f$.

Let $\varphi:(0,1)\to Y:x\mapsto \langle x,\sin\frac1x\rangle$, and let $\Phi:\beta((0,1))\to Y$ be the Čech-Stone extension of $\varphi$. Given a point $$x = \left\langle x_f:f\in C^*((0,1))\right\rangle$$ in $\beta((0,1))$, we want to determine $\Phi(x)$. The first step is to define the function $$\hat\varphi:\prod_{f\in C^*((0,1))}\mathbb{R}_f\to\prod_{f\in C^*(Y)}\mathbb{R}_f$$ that takes $x=\left\langle x_f:f\in C^*((0,1))\right\rangle$ to the point $y=\left\langle y_f:f\in C^*(Y)\right\rangle$ such that $y_f = x_{f\circ\varphi}$ for each $f\in C^*(Y)$. (Clearly $f\circ\varphi\in C^*((0,1))$ whenever $f\in C^*(Y)$, so this makes sense.) I’ll leave it to you to verify that $\hat\varphi$ is continuous. Now let $\Phi = e_Y^{-1}\circ \big(\hat\varphi\upharpoonright\beta((0,1))\big)$; $\Phi$ is clearly a continuous map from $\beta((0,1))$ to $Y$, and it’s not hard to see that $\Phi\circ e=\varphi$, i.e., that $\Phi$ is the Čech-Stone extension of $\varphi$. This gives us a concrete description of $\Phi$.

In particular, start with $x = \left\langle x_f:f\in C^*((0,1))\right\rangle\in\beta((0,1))$. Then $\hat\varphi(x) =$ $\left\langle y_f:f\in C^*(Y)\right\rangle$, where $y_f=x_{f\circ\varphi}$ for each $f\in C^*(Y)$, and $\Phi(x)$ is the unique point $p\in Y$ such that $e_Y(p) = \left\langle y_f:f\in C^*(Y)\right\rangle$, i.e., such that $f(p)=x_{f\circ\varphi}$ for each $f\in C^*(Y)$.

Now two of the functions in $C^*(Y)$ are the projection maps to the axes. For each $\langle x,y\rangle \in Y$ let $\pi_0(\langle x,y\rangle)=x$ and $\pi_1(\langle x,y\rangle)=y$. Let $x = \left\langle x_f:f\in C^*((0,1))\right\rangle\in\beta((0,1))$, and suppose that $\Phi(x) = \langle a,b\rangle \in Y$. Let $y = \left\langle y_f:f\in C^*(Y)\right\rangle=e_Y(\langle a,b\rangle)=\hat\varphi(x)$. From the preceding paragraph we know that $a = \pi_0(\langle a,b\rangle) = x_{\pi_0\circ\varphi}$ and $b=\pi_1(\langle a,b\rangle)=x_{\pi_1\circ\varphi}$. In other words, $\Phi(x)$ is completely determined by two coordinates of $x$, $x_{\pi_0\circ\varphi}$ and $x_{\pi_1\circ\varphi}$. Specifically, $$\Phi(x) = \langle x_{\pi_0\circ\varphi},x_{\pi_1\circ\varphi}\rangle\in Y\;.$$

As a quick sanity check, suppose that $a\in (0,1)$. The corresponding point of $\beta((0,1))$ is $\left\langle f(a):f\in C^*((0,1))\right\rangle$, which is sent by $\Phi$ to \begin{align*} \left\langle\pi_0(\varphi(a)),\pi_1(\varphi(a))\right\rangle &= \left\langle \pi_0\left(\left\langle a,\sin\frac1a\right\rangle\right), \pi_1\left(\left\langle a,\sin\frac1a\right\rangle\right)\right\rangle \\ &= \left\langle a,\sin\frac1a\right\rangle \\ &= \varphi(a)\;, \end{align*} just as it should be.

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