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.
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