I am trying to answer a question related to topologies in the real numbers.
Say I have two different topologies in $\Bbb R$, for example the standard and lower limit topologies. I know that a set $A$ can be open in $\Bbb R$ with both of these topologies, but I've been told that the conditions for openness are different.
For example, say $X = (0,1)$. People tell me that $A$ is open in $\Bbb R$ with the standard topology simply by looking at the definition of a topology in the real numbers. I feel like that "just look at it" approach only work for the standard topology.
Looking at an example:
Say I have $A = [-1, 1)$ with the subspace topology inherited from $\Bbb R$ with the upper limit topology.
If $B$ is a subset of $A$, how do I figure out if $B$ is open in $A$? How do I figure out if $B$ is even open in $\Bbb R$ in the first place? What does openness look like in the upper limit topology?
- $B = [0,1)$ -> I have been told that $B$ is open in $\Bbb R$ because $B = [-1, 1) \cap [0, \infty)$ and the latter set is open in $\Bbb R$. This doesn't resolve the underlying issue of why a set $[a,b)$ is open in $\Bbb R$, let alone in $A$.
- $B = (-1,0]$ -> Same issue as for 1.
- $B = (-1,0)$ -> My thoughts here are if $X$ from above is open in $\Bbb R$ for the reasons discussed, then $B$ must also be open in $\Bbb R$. The complement of $B$ in $A$ is $[-1, -1]\cup [0,1)$. This brings me to the exact same problem as 1.
- $B = [-1,0]$ -> If the idea that a set like $X$ is open in $\Bbb R$ using the definition of a topology holds here, then $B$ is similarly closed. We can also say that by inspecting the complement of $B$, we have no open intervals $O$ in which $B = A \cap O$. Hence it must not be open in $A$.
For 1. and 2., no matter how I tackle them, I run into the problem of needing to know that some set - $A$, $B$, a complement, etc. - is open in $\Bbb R$ with the upper limit topology.
Any suggestions would be excellent.