Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.