Let $X$ be a second countable space and $X$ is scattered.
A scattered space is a space for which every not empty subset has an isolated point.
How to show $X$ is countable?
Thanks ahead.
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11Can those voting to close please explain to the OP why you are doing so? – user1729 Apr 29 '13 at 13:51
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The question as it is written fails to address a central point: why would anyone suspect a scattered space is countable in the first place? – Carl Mummert May 01 '13 at 13:05
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Suppose that $X$ is uncountable. Let $\mathscr{B}$ be a countable base for $X$. Let $\mathscr{B}_0=\{B\in\mathscr{B}:|B|\le\omega\}$, let $A=\bigcup\mathscr{B}_0$, and let $Y=X\setminus A$. Then $|A|\le\omega$, so $Y$ is uncountable. Let $y\in Y$ be arbitrary. Suppose that $B\in\mathscr{B}$ and $y\in B$; then $B\nsubseteq A$, so $B\notin\mathscr{B}_0$, and therefore $|B|>\omega$. Finally, $|A|\le\omega$, so $|B\cap Y|=|B\setminus A|>\omega$. Thus, not only is $y$ not an isolated point of $Y$, but in fact every open nbhd of $y$ contains uncountably many points of $Y$.
Brian M. Scott
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I'm looking for a classical reference for this argument. Do you have any? Thanks – Lorenzo Jan 29 '24 at 14:56
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1@Lorenzo: I didn’t, since I’ve always thought of it as folklore, but I’ve done a bit of digging and found that although the result is not stated explicitly there, essentially this argument appears in Kap. VIII, §3 of Felix Hausdorff, Grundzüge der Mengenlehre, Leipzig, 1914. In my notation above he does explicitly show that $A$ is countable, from which this result follows immediately. (Mind you, it’s in German, and his notation differs significantly from current notation.) – Brian M. Scott Jan 30 '24 at 00:53
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Thanks a lot! In the meantime I've found a more recent reference that explicitly proves this result (through Cantor-Bendixson' argument): Zbigniew Semadeni, Banach spaces of continuous functions. Vol. I, Section 8.5. – Lorenzo Jan 30 '24 at 09:26
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