On Sun, 4 May 2008, Virgil wrote:
> In article <Pine.BSI.4.58.0805040058470.2687@[EMAIL PROTECTED]
>,
> William Elliot <marsh@[EMAIL PROTECTED]
> wrote:
>
> > On Sat, 3 May 2008, Virgil wrote:
> > > amu <amu786la@[EMAIL PROTECTED]
> wrote:
> > >
> > > > Let B = {(x, sin(1/x)): x in R-{0}}
> >
> > > > Theorem: Putting a limit point of a connected set into a connected
set
> > > > does not disconnect the set (i.e., the set remains connected).
Thus,
> > > > since B is connected and (0,0)
> > >
> > The complete and full theorem:
> > connected K, K subset A subset cl K ==> A connected.
> >
> > > How is B connected? One can separate it into the disjoint open sets
> > > {(x,y): x,y in R and x > 0} and {(x,y): x,y in R and x < 0}
> > >
> > B \/ {(0,0)} is connected.
>
> But B, by itself, is not.
>
Correct. The usual topologist sin curve is
cl { (x,sin 1/x) | 0 < x }
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