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Anonymous
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Notice that if a set $S$ is a regular-closed subset of the space $X$ (that is, if $S$ is the closure of an open subset of $X$), then $S$ is b-open. Furthermore, any open subset is b-open. Therefore, if $X$ has a non-empty proper regular-closed set $S$, then $S$ and $X \setminus S$ shows that $X$ is b-disconnected. In particular, every Hausdorff space having at least two pointpoints, say $p$ and $q$, is b-disconnected. (Take $S$ to be the closure of a neighborhood of $p$ such that $q \not\in S$.)

To get an example of a b-connected space, let $X$ be the set $\omega$ of natural numbers with the topology consisting of sets of the form $A_n = \{0, 1, \cdots n\}$ along with the empty set and $X$. (Here $\omega$ can be replaced with any ordinal $\lambda \geq 1$; if $\lambda = 2$, this is Sierpinski's two-point space.) Notice that any non-empty open subset of $X$ contains $0$ and that the closure in $X$ of the non-empty subset $S$ is $\{m, m+1, \cdots\}$ where $m$ is the smallest element of $S$. It follows that a non-empty subset $S$ of $X$ is b-open if and only $0 \in S$, and, in particular, there are no two disjoint non-empty b-open subsets of $X$.

Notice that if a set $S$ is a regular-closed subset of the space $X$ (that is, if $S$ is the closure of an open subset of $X$), then $S$ is b-open. Furthermore, any open subset is b-open. Therefore, if $X$ has a non-empty proper regular-closed set $S$, then $S$ and $X \setminus S$ shows that $X$ is b-disconnected. In particular, every Hausdorff space having at least two point, say $p$ and $q$ is b-disconnected. (Take $S$ to be the closure of a neighborhood of $p$ such that $q \not\in S$.)

To get an example of a b-connected space, let $X$ be the set $\omega$ of natural numbers with the topology consisting of sets of the form $A_n = \{0, 1, \cdots n\}$ along with the empty set and $X$. (Here $\omega$ can be replaced with any ordinal $\lambda \geq 1$; if $\lambda = 2$, this is Sierpinski's two-point space.) Notice that any non-empty open subset of $X$ contains $0$ and that the closure in $X$ of the non-empty subset $S$ is $\{m, m+1, \cdots\}$ where $m$ is the smallest element of $S$. It follows that a non-empty subset $S$ of $X$ is b-open if and only $0 \in S$, and, in particular, there are no two disjoint non-empty subsets of $X$.

Notice that if a set $S$ is a regular-closed subset of the space $X$ (that is, if $S$ is the closure of an open subset of $X$), then $S$ is b-open. Furthermore, any open subset is b-open. Therefore, if $X$ has a non-empty proper regular-closed set $S$, then $S$ and $X \setminus S$ shows that $X$ is b-disconnected. In particular, every Hausdorff space having at least two points, say $p$ and $q$, is b-disconnected. (Take $S$ to be the closure of a neighborhood of $p$ such that $q \not\in S$.)

To get an example of a b-connected space, let $X$ be the set $\omega$ of natural numbers with the topology consisting of sets of the form $A_n = \{0, 1, \cdots n\}$ along with the empty set and $X$. (Here $\omega$ can be replaced with any ordinal $\lambda \geq 1$; if $\lambda = 2$, this is Sierpinski's two-point space.) Notice that any non-empty open subset of $X$ contains $0$ and that the closure in $X$ of the non-empty subset $S$ is $\{m, m+1, \cdots\}$ where $m$ is the smallest element of $S$. It follows that a non-empty subset $S$ of $X$ is b-open if and only $0 \in S$, and, in particular, there are no two disjoint non-empty b-open subsets of $X$.

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Anonymous
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Notice that if a set $S$ is a regular-closed subset of the space $X$ (that is, if $S$ is the closure of an open subset of $X$), then $S$ is b-open. Furthermore, any open subset is b-open. Therefore, if $X$ has a non-empty proper regular-closed set $S$, then $S$ and $X \setminus S$ shows that $X$ is b-disconnected. In particular, every Hausdorff space having at least two point, say $p$ and $q$ is b-disconnected. (Take $S$ to be the closure of a neighborhood of $p$ such that $q \not\in S$.)

To get an example of a b-connected space, let $X$ be the set $\omega$ of natural numbers with the topology consisting of sets of the form $A_n = \{0, 1, \cdots n\}$ along with the empty set and $X$. (Here $\omega$ can be replaced with any ordinal $\lambda \geq 1$; if $\lambda = 2$, this is Sierpinski's two-point space.) Notice that any non-empty open subset of $X$ contains $0$ and that the closure in $X$ of the non-empty subset $S$ is $\{m, m+1, \cdots\}$ where $m$ is the smallest element of $S$. It follows that a non-empty subset $S$ of $X$ is b-open if and only $0 \in S$, and, in particular, there are no two disjoint non-empty subsets of $X$.