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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.

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Please define $\mathbb R_l$. –  Rasmus Sep 22 '11 at 8:29
    
@Rasmus:$\mathbb{R_\mathcal{l}}$ is the topology generated by the basis {$[a, b) |a, b\in \mathbb{R}$}. –  Jichao Sep 22 '11 at 8:31
    
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 8:35
    
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 8:38
    
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 8:39

2 Answers 2

up vote 5 down vote accepted

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

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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 9:14
    
That's a nice idea! –  Rasmus Sep 22 '11 at 9:17
    
@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
    
@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

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.

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