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  • Disconnecting R^n

    From William Elliot@21:1/5 to All on Sat May 7 06:14:09 2016
    Let E be a finite dimensional Euclidean space. If C is a closed subset
    of E that disconnects E, then is it always true that some component of
    C also disconnects E?

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  • From William Elliot@21:1/5 to David Cullen on Sun May 15 06:09:15 2016
    On Fri, 13 May 2016, David Cullen wrote:

    Is there any set that disconnects R^n that has the property that none
    of its components disconnect R^n?

    Ie is the assumption that such a set be closed known to be essential?

    I doubt it.

    Let q:N -> Q be an enumeration of [0,1] /\ Q,

    K = \/{ {a}x(R - {q^-1(a)} | a in [0,1] /\ Q }.

    Is R^2 - K disconnected?  Is K connected?

    This and similar constructions all seem to fail as
    an example to show the necessity of a closed set.

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  • From David Cullen@21:1/5 to All on Tue May 24 06:40:15 2016
    But R^2 - K constructed this way is connected.  You can see this
    because if you have a disconnection into two open sets U,V then every
    set of the form {i}xR with i irrational is contained in one or the
    other of U,V.  Now either all such sets are in U in which case we are
    done, or one of these sets {j}xR is in V, in which case let s in R be
    the supremum over all points x such that (x,0) is in U with x < j.  s
    must be rational or else (s,0) is in the boundary of U and the boundary
    of V, contrary to U,V being a disconnection.  Still, even if s is
    rational, (s, q^-1(s)) is in the boundary of U and the boundary of V.

    --- SoupGate-Win32 v1.05
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  • From David Cullen@21:1/5 to All on Fri May 13 05:02:12 2016
    Is there any set that disconnects R^n that has the property that none
    of its components disconnect R^n?

    Ie is the assumption that such a set be closed known to be essential?

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From William Elliot@21:1/5 to David Cullen on Wed May 25 05:49:05 2016
    On Tue, 24 May 2016, David Cullen wrote:

    But R^2 - K constructed this way is connected.  You can see this
    because if you have a disconnection into two open sets U,V then every
    set of the form {i}xR with i irrational is contained in one or the
    other of U,V.  Now either all such sets are in U in which case we are
    done, or one of these sets {j}xR is in V, in which case let s in R be
    the supremum over all points x such that (x,0) is in U with x < j.  s
    must be rational or else (s,0) is in the boundary of U and the boundary
    of V, contrary to U,V being a disconnection.  Still, even if s is
    rational, (s, q^-1(s)) is in the boundary of U and the boundary of V.

    What's K?  That set K previous determined in another thread
    to be connected but not path connected.  Please include sufficient
    context in your replies so I can know what you're talking about.

    Note, some usenet readers are less sophisticated than others.

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  • From David Cullen@21:1/5 to All on Wed May 25 18:28:38 2016
    I mean K as defined in the context of the present thread, namely: K =
    \/{ {a}x(R - {q^-1(a)} | a in [0,1] /\ Q }.

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