1. ## Continuity, Topology

I'm having difficulty understanding how you're supposed to do inverse mapping of a function. Here's the question I think if I see how this is supposed to be done, I'll be able to tackle the rest

The functionn f:(R,T)->(R.S), is given by f(x)=1 if x is an element of [2,3) and f(x)=0 otherwise,

Is f continuous if T=S= Sorgenfrey Topology.

2. Originally Posted by blimp
I'm having difficulty understanding how you're supposed to do inverse mapping of a function. Here's the question I think if I see how this is supposed to be done, I'll be able to tackle the rest

The functionn fR,T)->(R.S), is given by f(x)=1 if x is an element of [2,3) and f(x)=0 otherwise,

Is f continuous if T=S= Sorgenfrey Topology.
You can't expect to just tell you the answer, right? We need to see some work? What is the Sorgenfrey topology? Of course, it is what usually referred to as the Sorgenfrey line or "lower limit topology", the topology which has $\mathfrak{B}=\left\{[a,b):a,b\in\mathbb{R}\right\}$ as a base?

So, what are your ideas? You should know that if it's continuous it is sufficient to show that the preimage of every basic open set is open. So, what are the cases for $f^{-1}\left([a,b]\right)$?

3. Originally Posted by Drexel28
You can't expect to just tell you the answer, right? We need to see some work? What is the Sorgenfrey topology? Of course, it is what usually referred to as the Sorgenfrey line or "lower limit topology", the topology which has $\mathfrak{B}=\left\{[a,b):a,b\in\mathbb{R}\right\}$ as a base?

So, what are your ideas? You should know that if it's continuous it is sufficient to show that the preimage of every basic open set is open. So, what are the cases for $f^{-1}\left([a,b]\right)$?
well, I was thinking that the mapping defines the second sorgenfrey line, giving [0,1) and that to see if it's continuous do an inverse mapping of these elements and see if this result is still open in the Sorgenfrey line. This is where my issue lies $f^{-1}\left(1\right)$ gives me an element in [2,3)? which I guess is open, and $f^{-1}\left(0\right)$ gives me something like (-infinity,0)U[1,infinity)? I don't know if that is right or open

4. Originally Posted by blimp
well, I was thinking that the mapping defines the second sorgenfrey line, giving [0,1) and that to see if it's continuous do an inverse mapping of these elements and see if this result is still open in the Sorgenfrey line. This is where my issue lies $f^{-1}\left(1\right)$ gives me an element in [2,3)? which I guess is open, and $f^{-1}\left(0\right)$ gives me something like (-infinity,0)U[1,infinity)? I don't know if that is right or open
Notice that $[2,3)$ is both open and closed and so is $\mathbb{R}-[2,3)$. So, if $1\in[a,b)$ then $f^{-1}([a,b))=[2,3)$ which is open. If $0\in[a,b)$ then $f^{-1}([a,b))=\mathbb{R}-[2,3)$ which is open. Now, what if both or none are in there?

5. Originally Posted by Drexel28
Notice that $[2,3)$ is both open and closed and so is $\mathbb{R}-[2,3)$. So, if $1\in[a,b)$ then $f^{-1}([a,b))=[2,3)$ which is open. If $0\in[a,b)$ then $f^{-1}([a,b))=\mathbb{R}-[2,3)$ which is open. Now, what if both or none are in there?
I'm not sure, will it give the union of the two spaces: R? If neither you're not going to get anything in [2,3)?

6. Originally Posted by blimp
I'm not sure, will it give the union of the two spaces: R? If neither you're not going to get anything in [2,3)?
I left you literally the easiest part.

If $1,0\notin [a,b)$ then what in $\mathbb{R}$ maps to it? What if both are in?