Let $K$ denote the set of all numbers of the form $1/n$ where $n$ is a positive integer. Let $B$ be the collection of all open intervals $(a,b)$, along with all sets of the form $(a,b)-K$. The topology generated by $B$ is called the $K$-topology on $\mathbb{R}$. When $\mathbb{R}$ is given this topology we denote it by $\mathbb{R}_{K}$. Is $\mathbb{R}_{K}$ a Baire space?
Is $\mathbb{R}_{K}$ a Baire space?
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0Actually I think that the topology is the one generated by $B$ and not by $K$. – 2011-05-31
1 Answers
Yes, $\mathbb{R}_{K}$ is Baire. We can show this by comparing with the normal Euclidean topology $\cal{T}$ on $X = \mathbb{R}$. Call the topology on $\mathbb{R}_{K}$ $\cal{U}$. Then we have the 2 following facts:
- $\cal{T} \subset \cal{U}$.
- Every member of $\cal{U}$ has non-empty interior in $\cal{T}$.
Fact 2. follows from the fact that $K$ is nowhere dense in the usual topology. In that case we have that $(X,\cal{T})$ Baire implies $(X, \cal{U})$ Baire, using the simple lemmas (in above notation, satisfying 1 and 2)
Lemma 1
If $O$ is open and dense in $(X,\cal{U})$, then its interior in $(X,\cal{T})$ is (open and) dense in $(X,\cal{T})$.
Lemma 2 A set $D$ that is dense in $(X,\cal{T})$ is also dense in $(X, \cal{U})$.
Its simple proofs can be found in my posting, where I use it to show that the Sorgenfrey line (lower limit topology on $\mathbb{R}$) is also Baire, using this technique.