The Sorgenfrey plane and the Niemytzki plane are Baire spaces

Question

A space $X$ is called a Baire space if every countable intersection of open dense sets is dense. By the Baire category theorem, every complete metric space is Baire and every locally compact Hausdorff space is Baire.

The Sorgenfrey line is an example of a Baire space (shown here) that is not metrizable and not locally compact.

The Sorgenfrey plane and the Niemytzki/Moore plane are also not metrizable and not locally compact, and are not even normal.

For reference, I'd like a proof that the Sorgenfrey plane and the Niemytzki plane are Baire spaces. Sketch of proof is fine.

Answer 1

For the Moore plane: If a space has an dense subspace which is a Baire space, then the space itself is Baire. The open half-plane is dense in the Moore plane and is Baire.

Answer 2

Both your spaces are examples of pseudocomplete spaces (see my answer here for more links etc.) and are Baire for that reason.

Answer 3

Let $S$ and $T$ be topologies on a set $X$ with $S\subset T$. For $t\in T$ let $t^*=Int_S(t).$ Suppose $Cl_T(t^*)=Cl_T(t)$ for every $t\in T.$ (Equivalently, that $t^*$ is $T$-dense in $t$ for every $t\in T.$) Prove that if $(X,S)$ is Baire then $(X,T)$ is Baire by showing that (1) if $Y\subset X$ is $S$-dense then $Y$ is $T$-dense, and (2) if $\{t_n:n\in \Bbb N\} \subset T$ with each $t_n$ being $T$-dense then consider $Y=\cap \{t_n^*:n\in \Bbb N\}.$

For your Q let $S$ be the "usual" metric topology.

Answer 4

Here is a direct proof for the Sorgenfrey plane $X$. The topology for $X$ admits as a basis the collection of all half-open rectangles $[a,b)\times[c,d)$ with $a<b$ and $c<d$. A simple but useful fact is that any such half-open rectangle contains a closed rectangle $[a',b']\times[c',d']$ with nonempty interior in $X$.

To show that $X$ is Baire, assume we have a sequence of dense open sets $(U_n)_n$. Given an arbitrary nonempty set $O\subseteq X$, we have to show that $O\cap\bigcap_n U_n\ne\varnothing$. The intersection $O\cap U_1$ is open and nonempty by density of $U_1$. So it contains a half-open rectangle, which itself contains a closed rectangle $R_1=[a_1,b_1]\times[c_1,d_1]$ with nonempty interior as explained above. The interior of $R_1$ meets $U_2$ by density of $U_2$, so their intersection contains a half-open rectangle, which itself contains a closed rectangle $R_2$ with nonempty interior, etc. Continuing in this way, we get a nested sequence of closed rectangles $R_n=[a_n,b_n]\times[c_n,d_n]\subseteq U_n$. By compactness with respect to the usual topology of the plane the intersection $\bigcap_n R_n$ is nonempty, and is contained in $O\cap\bigcap_n U_n$.

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The proof above is entirely self contained, not requiring the knowledge that the plane with the Euclidean topology is Baire.

If instead one wants to assume the result that the Euclidean plane is Baire, one can use the following (repeating the comments of David Hartley and the answer by Daniel Wainfleet).

Suppose we have a set $X$ and two topologies $\sigma$ and $\tau$ on it, not necessarily comparable, but satisfying the condition:

(*) Every nonempty member of $\sigma$ contains a nonempty member of $\tau$ and vice versa.

Lemma 1: The dense sets in $X$ are the same for the two topologies. In other words, given $A\subseteq X$, $\operatorname{cl}_\sigma(A)=X$ if and only if $\operatorname{cl}_\tau(A)=X$.

Lemma 2: If $U\subseteq X$ is open and dense in $(X,\sigma)$, then $O=\operatorname{int}_\tau(U)$ is open and dense in $(X,\tau)$.

Proposition: Under assumption (*) above, if $(X,\tau)$ is a Baire space, so is $(X,\sigma)$.

(The proofs are not difficult.)