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Lower limit topology ($\R_l$)
- Basis: $\{[a,b) \mid a,b \in \R\text{ s.t. } a < b\}$
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K-topology ($\R_K$)
- Basis: $\{(a,b), (a,b) - K \mid a,b \in \R\text{ s.t. } a < b \}$, where $K = \{ 1/n \mid n \in \Z_+ \}$.
$\R_l \nsubseteq \R_K$
Consider a base element $[a,b)$. At the point $a$, no open interval $(c,d)$ containing $a$ is a subset of $[a,b)$.
$\R_K \nsubseteq \R_l$
Let $B_2 = (-1, 1) - K$. At $B_2 \notin \R_l$ because any base element $[0,b)$ containing 0 must hit $1/n$ for some $n \in \Z_+$ by Archimedean Property of $\Z_+$. Thus, $\forall b > 0, [0,b) \nsubseteq B_2$.