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.
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