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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?

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This sounds like a HW exercise from Munkrees. Is this for HW? What have you tried so far? – wckronholm May 31 '11 at 14:45
No this is not homework. I just would like to know the general known fact. – student May 31 '11 at 14:52
In my experience, the $K$-topology on $\mathbb{R}$ is useful mainly for providing counterexamples to various statements and to help one navigate the plethora of definitions in point-set topology. In other words, the $K$-topology is a learning tool. So, the best thing to do is to start working through the definitions yourself. If you've already started this process, describe what you've tried and where it has taken you. – wckronholm May 31 '11 at 15:54
Actually I think that the topology is the one generated by $B$ and not by $K$. – Giovanni De Gaetano May 31 '11 at 17:53

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:

  1. $\cal{T} \subset \cal{U}$.
  2. 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.

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