Continuity on the one-point compactification of $R^n$

Question

We have seen that the one-point compactification of $R^n$, given as $R^n\cup \{\infty\}$ for a point not in the set, and denoted $\hat{R^n}$, is homeomorphic to the sphere $S^n$. I equip $\hat{R^n}$ with the chordal metric, given by $$d(x,y)=|\hat{\pi}(x)-\hat{\pi}(y)| $$ where $\hat{\pi}$ is the extended stereographic projection, mapping $\infty$ to $e_{n+1}$.

It is straight-forward to show that $$d(x,\infty)=\frac{2}{(1+|x|^2)^{1/2}} $$ In other words; the open balls $B_d(\infty,r)$ is of the form \begin{equation} N(\infty,s)=\hat{R^n}-\overline{B(0,s)} \end{equation} for some suitable $s$. And so, a basis for a topology on $\hat{R^n}$ consists of all the usual open euclidean balls $B(x,r)$ and balls around $\infty$ of the form above.

Now, in my textbook, the author claims that this in particular implies that a function $f:\hat{R^n}\rightarrow \hat{R^n}$ is continuous at $a$ if and only if $\lim_{x\rightarrow a}f(x)=f(a)$ in the usual euclidean sense. However, I have trouble seeing why this is. How do you tie this limit to the open balls above?

Thanks

Answer 1

A function $f:\hat{R^n}\rightarrow \hat{R^n}$ is continuous at $a$ if and only if $\lim_{x\rightarrow a}f(x)=f(a)$ in the usual euclidean sense.

I think this description is misleading so it is better to forget it. :-)

The exact and correct description of continuity of $f$ in the “usual Euclidean sense” that is in terms of $\Bbb R^n$ is complicated and not very convenient. Namely, let $f:\hat{\Bbb R^n}\rightarrow \hat{\Bbb R^n}$ and $a\in \hat{\Bbb R^n}$. The function $f$ is continuous at $a$ iff

$a, f(a)\in \Bbb R^n$ and $\lim_{x\in\Bbb R^n, x\rightarrow a }f(x)=f(a)$. The latter also requires that $a$ has a neighborhood $U\subset\Bbb R^n$ such that $f(x)\in\Bbb R^n$ for each $x\in U$.

$a\in \Bbb R^n$, $f(a)=\infty$ and $\lim_{x\in\Bbb R^n, x\rightarrow a}f(x)=\infty$. The latter means that for each $t>0$ the point $a$ has a neighborhood $U\subset\Bbb R^n$ such that $f(x)\in(\Bbb R^n\setminus \overline{B(0,t)})\cup\{\infty\}$ for each $x\in U$.

$a=\infty$, $f(a)\in\Bbb R^n$ and $\lim_{x\in\Bbb R^n, x\rightarrow\infty}f(x)=f(a)$. The latter means that for each neighborhood $V\subset\Bbb R^n$ of the point $f(a)$ there exists $s>0$ such that $f(x)\in V$ for each $x\in \Bbb R^n\setminus \overline{B(0,s)}$.

$a, f(a)=\infty$ and $\lim_{x\in\Bbb R^n, x\rightarrow\infty}f(x)=\infty$. The latter means that for each $t>0$ there exists $s>0$ such that $f(x)\in(\Bbb R^n\setminus \overline{B(0,t)})\cup\{\infty\}$ for each $x\in \Bbb R^n\setminus \overline{B(0,s)}$.