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The complete question is from Mukres's Topology.

(a) Suppose that $f: \mathbb{R} \to \mathbb{R}$ is "continuous from the right" that is $$\lim_{x \to a^{+}} f(x) = f(a),$$ for each $a \in \mathbb{R}$. Show that $f$ is continuous when considered as a function from $\mathbb{R_\mathcal {l}}$ to $ \mathbb{R}$.

(b) Can you conjecture what functions $f: \mathbb{R} \to \mathbb{R}$ are continuous when considered as maps from $\mathbb{R}$ to $\mathbb{R_\mathcal {l}}$? As maps from $\mathbb{R_\mathcal {l}}$ to $\mathbb{R_\mathcal {l}}$?

NOTE:$\mathbb{R_\mathcal {l}}$ is the topology generated by the basis $\{[a,b)|a,b\in R\}$.

It is easy to prove the first part of the question. But I have no idea about how to figure out the second part of the question. Could you help me?

Thanks.

Srivatsan
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Jichao
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  • Please define $\mathbb R_l$. – Rasmus Sep 22 '11 at 08:29
  • @Rasmus:$\mathbb{R_\mathcal{l}}$ is the topology generated by the basis {$[a, b) |a, b\in \mathbb{R}$}. – Jichao Sep 22 '11 at 08:31
  • First observation: This topology is finer than the usual one. Can you guess a function which is continuous as a function $\mathbb R\to\mathbb R$ but not as a function $\mathbb R\to\mathbb R_l$? In fact, it is quite hard for a function to be continuous from $\mathbb R$ to $\mathbb R_l$. – Rasmus Sep 22 '11 at 08:35
  • You might want to put the definition of $\mathbb R_l$ into the question so that others can find it more easily. – Rasmus Sep 22 '11 at 08:38
  • Definitely not, because any set that is open in $\mathbb{R}$ is definitely open in $\mathbb{R_\mathcal{l}}$. But I could not find out a function which is not continuous as a function $\mathbb{R} \to \mathbb{R}$ but not as a function $\mathbb{R} \to \mathbb{R_\mathcal{l}}$. – Jichao Sep 22 '11 at 08:39
  • Look at the function $x\mapsto x$. – Rasmus Sep 22 '11 at 08:40
  • Yeah, the identity function is a good example, but is there a more precise and general statement? – Jichao Sep 22 '11 at 08:42
  • let us continue this discussion in chat – Jichao Sep 22 '11 at 08:42
  • Does not matter. Thanks for your hint. – Jichao Sep 22 '11 at 08:50

2 Answers2

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Hint (for $f: \mathbb{R} \to \mathbb{R}_\ell$): The continuous image of a connected set is connected. What are the connected components of $\mathbb{R}_\ell$?

kahen
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  • I haven't learnt connectivity before.Anyway, the excercise say we'd return to this question after we have learnt it. So thanks for your hint. – Jichao Sep 22 '11 at 09:14
  • That's a nice idea! – Rasmus Sep 22 '11 at 09:17
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    @Jichao: If you know that the only open and closed subsets of $\mathbb{R}$ are $\emptyset$ and $\mathbb{R}$, then this nice argument can be used without referring to connectedness. Note that for any $x\in\mathbb{R}_l$ and $\epsilon>0$ the sets $[x,x+\epsilon)$ are open and closed in $\mathbb{R}_l$. Then so are the sets $f^{-1}[x,x+\epsilon)$ because $f$ is continuous. – LostInMath Sep 22 '11 at 12:07
  • @LostInMath: Why for any $x \in \mathbb{R_\mathcal{l}}$ and $\epsilon > 0$ the sets $[x, x + \epsilon)$ are open and closed in $\mathbb{R_\mathcal{l}}$. – Jichao Sep 22 '11 at 16:04
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    @Jichao: They are open since they are elements of the generating basis of $\mathbb{R}l$. They are closed since the complements are open: $\mathbb{R}_l\setminus[x,x+\epsilon)=\bigcup{y<x}[y,x)\cup\bigcup_{y>x+\epsilon}[x+\epsilon,y)$ is a union of open sets, hence open. – LostInMath Sep 22 '11 at 16:41
  • @LostInMath: Thanks for your detailed explaination. – Jichao Sep 22 '11 at 23:35
5

To get you started on $f:\mathbb{R}_\ell \to \mathbb{R}_\ell$:

If $f$ is continuous as a function from $\mathbb{R}_\ell$ to $\mathbb{R}_\ell$, then it must be continuous as a function from $\mathbb{R}_\ell$ to $\mathbb{R}$, since every open set in $\mathbb{R}$ is also open in $\mathbb{R}_\ell$. Thus, as a function from $\mathbb{R}$ to $\mathbb{R}$ it must be continuous from the right. However, this isn’t enough: $f(x)=-x$ is continuous as a function from $\mathbb{R}$ to $\mathbb{R}$, but it’s not continuous as a function from $\mathbb{R}_\ell$ to $\mathbb{R}_\ell$. Why? If you can see what keeps this function from being continuous from $\mathbb{R}_\ell$ to $\mathbb{R}_\ell$, you’ve a good chance of working out exactly which functions from $\mathbb{R}_\ell$ to $\mathbb{R}_\ell$ are continuous.

Brian M. Scott
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