topology (upper limit and lower limit)

I have to show that upper limit topology and lower limit topology on $\mathbb{R}$ (Real line) are not comparable. But suppose if we take $[a,b)$ and $(a-1,b]$, where $a-1 > a$, then isn't it showing that upper limit topology contains the lower limit topology. Same can be done with lower limit topology also.

• $a-1>a\Rightarrow -1>0\Rightarrow 1<0\cdots$
– Moya
Aug 24, 2015 at 1:45
• It looks like you are trying to show that every open set in one topology contains an open set in another. To show that one topology is finer than the other you need to show that every open set in the latter is an open set of the former. Aug 24, 2015 at 6:36
• @Moya .. Please don't take $a-1$ and $a$ literally. They are just symbols.. Sorry for not mentioning it earlier. Aug 25, 2015 at 13:18

Two topologies $\mathcal{T}_1$ and $\mathcal{T}_2$ on a set $X$ are incomparable iff there exists $A \subset X$ such that $A \in \mathcal{T}_1, A \notin \mathcal{T}_2$ (which shows that $\mathcal{T}_1 \subseteq \mathcal{T_2}$ does not hold) and there also exists some $B \subset X$ such that $B \in \mathcal{T}_2, B \notin \mathcal{T}_1$, which similarly disproves the other inclusion.
For the lower limit topology and the upper limit topology on the reals we can indeed take $A = [0,1)$, which is (basic) open in one, but not open in the other (as there is no set of the form $(a,b]$ that contains $0$ and is contained in $A$), and $B = (0,1]$, with a similar argument regarding $1$ instead of $0$.
• Thanks.. Is there a reason that we are taking $a$ and $b$ to be as endpoints of both topologies? I mean for $[0,1)$ there are infinitely many sets like $(1/4,1/2]$ etc. which are entirely contained in $[0,1)$. Aug 26, 2015 at 14:21
• @user262860 Yes, but none contains $0$. A set $O$ is open iff for every point $p$ in $O$, there is some basic open set of the form $(a,b]$ that contains $p$ and is contained in $O$. Which fails for $0$ only in the case of $[0,1)$. Aug 26, 2015 at 18:36