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Michael line can be seen here. It said this space has point-countable base by using a countable base for the usual topology on $R$ with all singletons from $R\setminus Q$. I try to prove it is true. It is easy for any $x\in R\setminus Q$, because it exists the open set ${x}$ which intersects the given base at most countable. However, if $x\in Q$, can we get an open set such that intersects the base at most countable? I'm afraid maybe it will include uncountable points from $R\setminus Q$. Somebody help me!

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Note that point countable means that each point belongs to only finitely many members of the family. (You seem to be using locally countable in your question.) –  Arthur Fischer Feb 20 '13 at 11:08
    
@Arthur: ‘... to only countably many ...’. –  Brian M. Scott Feb 20 '13 at 11:10
    
@Brian: Doesn't finite = countable? (oops...) –  Arthur Fischer Feb 20 '13 at 11:14
    
@Arthur: Depends on how good the beer is. hic –  Brian M. Scott Feb 20 '13 at 11:14
    
O, My heaven...! –  Paul Feb 20 '13 at 11:15
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1 Answer

up vote 3 down vote accepted

Let $$\mathscr{B}=\{(p,q):p,q\in\Bbb Q\text{ and }p<q\}\cup\big\{\{x\}:x\in\Bbb R\setminus\Bbb Q\big\}\;.$$ Then $\mathscr{B}$ is a point countable base for the Michael line because $\mathscr{B}_0=\{(p,q):p,q\in\Bbb Q\text{ and }p<q\}$ is countable: any $x\in\Bbb R$ belongs to at most countably many members of $\mathscr{B}_0$ and to at most one of the singletons in $\mathscr{B}$.

Note that any base for the Michael line must include $\big\{\{x\}:x\in\Bbb R\setminus\Bbb Q\big\}$, so no base can be locally countable: any open set containing a point $q\in\Bbb Q$ will necessarily contain uncountably many of the irrational singletons.

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