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I'm trying to learn a little about topology, and I don't quite understand continuity yet. I use this definition of a continuous map f: f is continuous if the inverse image of every open set is open.

I use the sets $\mathbb{R}$ and $A = \mathbb{R}-\{0\}$, both with standard metric topology (can I do that with $A$? I don't see something going wrong).

Now I look at the map f: $ A \rightarrow \mathbb{R}$

$$ f(x) = \begin{cases} 0 & \text{if }x<0 \\ 1 & \text{if }x>0 \end{cases} $$

Now I ask myself whether this is continuous, so I look at the inverse image $B \subseteq A$ of any open set $C \subseteq \mathbb{R}$ and I notice:

$$ B = \begin{cases} \emptyset & \text{if } 0 \notin C \wedge 1 \notin C \\ (0,+\infty) & \text{if } 0 \notin C \wedge 1 \in C \\ (-\infty,0) & \text{if } 0 \in C \wedge 1 \notin C \\ \mathbb{R} & \text{if } 0 \in C \wedge 1 \in C \\ \end{cases} $$

These sets all seem open to me, which would imply that f is contiuous. But it can't be. What am I missing?

user119879
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    Your map is actually continuous. Maybe you are confused by the fact that f changes value as you cross $0$. But the point is that $0$ doesn't belong to the domain of f, so this is not a problem. – Lor Jan 08 '14 at 16:13
  • Oh. That's unexpected. But thanks! – user119879 Jan 08 '14 at 16:17
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    The domain is not connected because of the 'missing' point at 0. The function $f$ is constant on each connected component, hence continuous. Intuitively, we 'assume' connectivity (at least I do!). – copper.hat Jan 08 '14 at 16:20
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    If you haven't already, you'll probably come across the notion of "connectedness". What you can see happening here is that $A$ consists of two different disjoint "connected" components, $(-\infty,0)$ and $(0,\infty)$. It turns out, which is what you've seen a case of, that a function is continuous if and only if it is so on each of its connected components (as it is in this case). Edit: Okay, that's what copper.hat said as well. – fuglede Jan 08 '14 at 16:20
  • The "non continuous appearance" of f comes from the fact that it admits no continuous extension defined on the whole $\mathbb{R}$. – Lor Jan 08 '14 at 16:22
  • In general, you can only apply the intuitive 'don't take the pencil off the paper' definition of continuity if the domain and range are path-connected subsets of the real line. The domain in your example fails the path-connected condition. – Dan Rust Jan 08 '14 at 16:40
  • Thank you all! It's clear now – user119879 Jan 08 '14 at 16:50

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