Continuity of abs(1/x)

Question

I know that when $x \in \mathbb R$ then $\frac {1}{x}$ is continuous on $\mathbb {R^{\ast-}}$ and on $\mathbb {R^{\ast+}}$

Considering $f(x) = abs(\frac{1}{x})$ where $x \in \mathbb {R}$ and $f(x) \in \mathbb {R^+} \cup \infty^+ $

Is this function continuous ? I feel like adding $\infty^+$ to $\mathbb {R}$ might be sufficient.

Answer 1

The function is a continuous function on $\mathbb R\setminus \{0\}$.

The function is not defined on $\mathbb R$, because it is not defined on $0$.

If you add $\infty$ to $\mathbb R$, then you need to also redefine what continuity means, since continuity is defined on $\mathbb R$ (or, more generally, in topological spaces, but you still have to define the new topology...)

Answer 2

I'm not clear what definitions you're using - and that matter does need attention, as others have already explained - but whatever definitions you eventually adopt, if the truth of your claim that the function $g(x) = 1/x$ is continuous is preserved (by whatever changes you will have had to make), then it will follow that the function $f$ is also continuous.

That is because $f$ is the composite function $\operatorname{abs} \mathop{\circ} g$, and the composite of two continuous functions is continuous (regardless of whatever else is going on).

To make sure of this, you will have to make sure that your function $\operatorname{abs}$ is continuous, of course!

The usual function $\mathbb{R} \to \mathbb{R}$, $x \mapsto |x|$ is continuous (prove it!), but again, it is not clear exactly how your function $\operatorname{abs}$ relates to this usual 'abs' function.

If you get stuck, there is a Wikipedia article that should help: Projectively extended real line.

If, as seems likely, you want to distinguish between $+\infty$ and $-\infty$ (these are the standard notations, to be preferred to your $\infty^+$), then it is not clear (to me, at least) how you would decide which of them $1/0$ is to be!

Still, this other Wikipedia article might be worth a look: Extended real number line.

It looks as if you might be content with continuity on an extended set of non-negative real numbers. The only unambiguous standard notation for the usual, unextended set of non-negative real numbers (there are several Stack Exchange questions on this and related topics - this, for example) is $\mathbb{R}_{\geqslant0}$ (or if you prefer, $\mathbb{R}_{\geq0}$), so you might want to define something like this: $$ \overline{\mathbb{R}}_{\geqslant0} = \mathbb{R}_{\geqslant0} \cup \{+\infty\} = \{x \in \overline{\mathbb{R}} : x \geqslant 0\}, $$ (I've no idea if that's standard, but probably not) and then examine the continuity of functions defined on that subset of the extended real line $\overline{\mathbb{R}}$.

The trouble then is: what is the point of having a function $\operatorname{abs}$?!

I just don't see how you're going to have a function $x \mapsto 1/x$, defined at $0$, continuous on an extended version of $\mathbb{R}$, and taking both strictly positive and strictly negative values.

Plenty to think about!