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Thread: Is the Excluded Point Topology a scattered space?

  1. #1
    MHF Contributor Matt Westwood's Avatar
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    Is the Excluded Point Topology a scattered space?

    Let $\displaystyle S$ be a set.

    Let $\displaystyle p \in S$.

    Then the excluded point topology $\displaystyle T = (S, \tau)$ is given by:

    A subset of $\displaystyle H \in \tau$ is open iff $\displaystyle p \notin H .$


    The argument given in Counterexamples in Topology by Steen and Seebach (counterexamples 13-15: 5) goes:

    Let $\displaystyle x \in S$. Then $\displaystyle \{x\} \subseteq S$ is open in $\displaystyle T$ as $\displaystyle \notin \{x\}$ .

    So by definition $\displaystyle x$ can not be a limit point and so is an isolated point.

    So any subset $\displaystyle H \subseteq S$ is not dense-in-itself, because it contains isolated points.

    So by definition $\displaystyle T$ is scattered.


    There's a flaw in the above, in that $\displaystyle p$ is easily shown to be a limit point (the only one possible in $\displaystyle T$, in fact) and so $\displaystyle \{p\}$ has no isolated points in it, so is dense-in-itself.

    So it appears that $\displaystyle T$ is not scattered after all, as it contains (exactly) one dense-in-itself subset.


    Is Steen and Seebach wrong? Or have I missed something, e.g. "$\displaystyle H$ is required to be open to be described as dense-in-itself" or something? But I can't find any such words.
    Last edited by Plato; Jun 8th 2011 at 02:21 PM. Reason: de-latexed as the commands don't work
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    Please use the [tex][/tex] tags.
    [tex]\{x\}\subseteq S[/tex] gives $\displaystyle \{x\}\subseteq S$.
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