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Lemma

Let be $(X,\le)$ an ordered set and let be $\infty\notin X$. So we define a new order $\preccurlyeq$ in $X^\infty:=X\cup\{\infty\}$ such that correspond to $\le$ in $X\subseteq X^\infty$: so for any $x_1,x_2\in X^\infty$ we define $x_1\preccurlyeq x_2$ iff $x_1,x_2\in X$ and $x_1\le x_2$ or if $x_2=\infty$ (or else $x_1=\infty$: so it is easy to show that $(X^\infty,\preccurlyeq)$ is an order relation in $X^\infty$ such that correspond to $X$ for any $x_1,x_2\in X\subseteq X^\infty$.

So using the above lemma we can add to $\Bbb{R}$ the points $\infty$ and $-\infty$ and so we can order $\overline{\Bbb{R}}:=\Bbb{R}\cup\{\infty,-\infty\}$ so that for any $x\in\overline{\Bbb{R}}$ it results that $-\infty\le x\le\infty$.

Then if $+$ and $\cdot$ are the usual sum and product in $\Bbb{R}$, for any $x\in \Bbb R$ we define $$ \begin{align} \infty+x &= \infty \\ -\infty+x &= -\infty \\ \infty+\infty &= \infty \\ -\infty-\infty &= -\infty \\ \infty-\infty=0\\ \frac{x}\infty &= 0 =\frac{x}{-\infty}\\ \frac{\infty}\infty=1\\ \frac{\infty}{-\infty}=-1=\frac{-\infty}\infty \end{align} $$ and for $\infty\ge x>0> y\ge -\infty$, $$\begin{align} x\cdot\infty &= \infty \\ x\cdot(-\infty) &= -\infty \\ y\cdot\infty &= -\infty \\ y\cdot(-\infty) &= \infty \\ 0\cdot\infty &=0 = 0\cdot(-\infty).\\ \end{align} $$ then we impose that the arithmetic of $\infty$ and $-\infty$ has all properties of arithmetic of usual real arithmetic.

So I asked to me if using two times the one-point compactification I can define on $\overline{\Bbb{R}}$ the usual topology that we all know: so could someone show to me how formally to do this, please?

As reference here is defined the one-point compactification.

Antonio Maria Di Mauro
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  • You can't get all properties of arithmetic for $\pm\infty$ with those definitions, as that leads to contradictions. For example: $\infty = \infty + 0 = \infty + (\infty-\infty) = (\infty + \infty) - \infty = \infty - \infty = 0$. – celtschk Apr 05 '20 at 18:16

2 Answers2

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Arithmetic has little to do with this. You could consider $[0,\infty)\subset\mathbb{R}$ and do its Aleksandrov compactification, which is obviously homeomorphic to $[0,\infty]\subset\bar{\mathbb{R}}$; similarly, $[-\infty,0]$ is the Aleksandrov compactification of $(-\infty,0]$. Now glue the two at $0$.

What's this “glueing”? The full space is $\bar{\mathbb{R}}=[-\infty,0]\cup[0,\infty]$. Points different from $0$ have as a basis of neighborhoods the neighborhoods they have in the subspace they sit in. As neighborhoods of $0$, take the sets $U_-\cup U_+$, where $U_-$ is a neighborhood of $0$ in $[-\infty,0]$ and $U_+$ is a neighborhood of $0$ in $[0,\infty]$.

This coincides with the usual (order) topology on $\bar{\mathbb{R}}$.

egreg
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This is a specific construction for ordered spaces. There is a maximal one, called the Nachbin compactification (mentioned in this paper and its references, aka the Wallman ordered compactification). For $\Bbb R$ this is $\overline{\Bbb R}$.

This Nachbin compactification is only equal to the Aleksandrov compactification for specific ordered spaces, like $[0,\infty)$ e.g.

Henno Brandsma
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