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.
2026-02-23 06:36:13.1771828573
topology (upper limit and lower limit)
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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 $0$.