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I'm learning about extended reals and I was wondering how continuity is defined on it. Consider the function $f:\mathbb{R}\rightarrow\mathbb{R}\cup\{\infty\}$ defined by $f(x)=\frac{1}{|x|}$ for $x\not=0$ and $f(0)=\infty$. It's obvious that $f$ is continuous at any $x\not=0$. But what is the continuity of $x=0$?

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  • $\begingroup$ You could think about the continuity of functions on $\mathbb R\cup\{\infty\}$ by using the topology on that space, instead. You can even extend $f:\mathbb R\cup\{\infty\}\to \mathbb R\cup\{\infty\}$ by letting $f(\infty)=0$. $\endgroup$
    – Ian Coley
    Jan 29, 2014 at 0:45
  • $\begingroup$ Oh you are right! Pre-image of open sets must be open. $\endgroup$
    – user44322
    Jan 29, 2014 at 0:47
  • $\begingroup$ or you take a metric on $\mathbb{R} \cup \{\infty\}$, say $d(x, y) = \rho(|x - y|)$ for $\rho(t) = t / (1 + t)$. $\endgroup$
    – user66081
    Jan 29, 2014 at 0:47
  • $\begingroup$ Map $\mathbb{R}\cup \{\infty\}$ to the circle $S^1$ using stereographic projection and then pull back the topology. Define continuous functions in terms of this topology. $\endgroup$
    – Dan Rust
    Jan 29, 2014 at 1:19

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The term "extended real line" usually refers to the set $[-\infty,\infty]$, i.e., the real line with two infinite points appended. The space you describe has a single infinite point. I can see two natural ways of understanding this space, but it is not clear which of these you intend. Specifically, while the sequence $\langle n \vert n\in\mathbb{N}\rangle$ converges to $\infty$, it is not clear whether the sequence $\langle -n \vert n\in\mathbb{N}\rangle$ diverges or also converges to $\infty$. (In other words does your extended real line look like a circle or a half open interval?) You will need to resolve this before you can fully answer the question.

While continuity can be defined in metric spaces (such as $\mathbb{R}$), it is really a topological concept. I guess that you don't want an overtly topological argument here, though, so we need to think about how to define continuity in this setting where we don't have a proper metric space (since for $\forall x\in \mathbb{R} \, d(x,\infty)=\infty$).

The key (and this is where the topology comes in, even if we aren't using it explicitly), is that we can restate the usual definition of continuity in terms of open intervals rather than the distance function $\vert\cdot,\cdot\vert$:

A function $f$ is continuous at $x$ if for every open interval $J$ containing $f(x)$, there is an open interval $I$ containing $x$ such that $$ \forall x'\in I \, f(x')\in J.$$

When we put it like that, the problem is pretty straightforward. Which are the open intervals that contain infinity? Well, this is where it matters where your $\infty$ sits. They may be sets of the form $(a,\infty]$, or they may be sets of the form $(-\infty,b)\cup (c,\infty)\cup\{\infty\}$. So, work out which sort of open interval is applicable to your situation, pick an arbitrary one (call it $J$), and see if you can find an open interval $I\ni 0$ which gets mapped into $J$.

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  • $\begingroup$ @Daniel Whoops - you're right. I'll fix it now. $\endgroup$
    – Unwisdom
    Jan 29, 2014 at 2:10

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