1. ## Right continuous

Notation:
$\displaystyle (\mathbb{R},\tau_s)$ Standard topology on R
$\displaystyle (\mathbb{R}.\tau_l)$ Lower limit topology on R

Prove $\displaystyle f:\mathbb{R}\to\mathbb{R}$ is right continuous on $\displaystyle \mathbb{R}$ iff. f is continuous on $\displaystyle \mathbb{R}_l$ when considered as a function from $\displaystyle \mathbb{R}_l\to\mathbb{R}$.

$\displaystyle (\Rightarrow)$
Since f is right continuous on R, we know $\displaystyle \lim_{x\to x^+_0}f(x)=f(x_0)$
$\displaystyle f^{-1}((a_i,b_i))=\cup_{i\in I}[a_i,b_i)$
Where $\displaystyle (a_i,b_i)$ is open in X. I am not sure about the saying the invers equals the union of half closed and open sets. I am sort of lost.

2. ## Re: Right continuous

Originally Posted by dwsmith
Notation:
$\displaystyle (\mathbb{R},\tau_s)$ Standard topology on R
$\displaystyle (\mathbb{R}.\tau_l)$ Lower limit topology on R

Prove $\displaystyle f:\mathbb{R}\to\mathbb{R}$ is right continuous on $\displaystyle \mathbb{R}$ iff. f is continuous on $\displaystyle \mathbb{R}_l$ when considered as a function from $\displaystyle \mathbb{R}_l\to\mathbb{R}$.

$\displaystyle (\Rightarrow)$
Since f is right continuous on R, we know $\displaystyle \lim_{x\to x^+_0}f(x)=f(x_0)$
$\displaystyle f^{-1}((a_i,b_i))=\cup_{i\in I}[a_i,b_i)$
Where $\displaystyle (a_i,b_i)$ is open in X. I am not sure about the saying the invers equals the union of half closed and open sets. I am sort of lost.
So, you've got the right idea. Right continuity on $\displaystyle \mathbb{R}$ is equivalent to saying that $\displaystyle \displaystyle f^{-1}(a,b)=\bigcup [c,d)$, but this right hand union is a union of sets open in $\displaystyle \mathbb{R}_\ell$...so.

3. ## Re: Right continuous

Originally Posted by Drexel28
So, you've got the right idea. Right continuity on $\displaystyle \mathbb{R}$ is equivalent to saying that $\displaystyle \displaystyle f^{-1}(a,b)=\bigcup [c,d)$, but this right hand union is a union of sets open in $\displaystyle \mathbb{R}_\ell$...so.
I don't know.

4. ## Re: Right continuous

Originally Posted by dwsmith
I don't know.
Yes you do. The right hand side of that last equality is a union of open sets in $\displaystyle \mathbb{R}_\ell$. What's true about a union of open sets?

5. ## Re: Right continuous

Originally Posted by Drexel28
Yes you do. The right hand side of that last equality is a union of open sets in $\displaystyle \mathbb{R}_\ell$. What's true about a union of open sets?
It just seems to easy. What about the other direction? I have no idea what to do for that one.

6. ## Re: Right continuous

Originally Posted by dwsmith
It just seems to easy. What about the other direction? I have no idea what to do for that one.
Well, do you have an idea? You use the same formula. Just show that the preimage of an interval looking like that implies that the right limit is the actual value.

7. ## Re: Right continuous

Originally Posted by Drexel28
Well, do you have an idea? You use the same formula. Just show that the preimage of an interval looking like that implies that the right limit is the actual value.
$\displaystyle f^{-1}((a_i,b_i))=[a_i,b_i)$ which is open since f is continuous. By the well ordering principle, $\displaystyle \bigcup [a_i,bi)$ has a least element say y.

Therefore, $\displaystyle \lim_{x\to y^+}f(x)=f(y)$

Does this work?