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I'm aware that the lower limit topology and k-topology are not comparable: see Show that the Topologies of $\mathbb{R}_l$ and $\mathbb{R}_K$ are not comparable.. However, I'd like to understand why the following line of reasoning is flawed.

This proof depends on the fact that open intervals of the form $(a,b)$ are open in the lower limit topology and are found by taking unions of the half-closed bases.

Take $(-1, 1) - K$ to be an open set in the K-topology that supposedly is not in the lower limit topology.

Then let $$ L = \bigcup_{n \in \mathbb{N}} \bigcup_{ \epsilon \in \mathbb{R}:1/(n+1)<\epsilon<1/n}\left[ \frac{1}{n + 1} + \epsilon, \frac{1}{n} \right) $$ Each interval is of the form $(1/2, 1)$, $(1/3, 1/2)$, etc. which are open in the lower limit topology. Unions of open sets are also open, so $L$ is in the topology. And there are no fractions of the form $\frac{1}{n}$ in $L$; $L$ and $K$ are disjoint.

Moreover, $(L \cap [0,1)) \cup (-1, 0)$ is in the lower limit topology and equals $(-1, 1) - K$.

tmastny
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1 Answers1

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$0 \in (-1,1)- K$, but not in your union $L$, so not in $L \cap [0,1)$ and not in $(L \cap [0,1)) \cup (-1,0)$ either.

Or, as the original argument said: $0$ is not an interior point of $\Bbb R- K$ in the lower-limit topology, while it is interior in the $K$-topology.

Henno Brandsma
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