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) is a limit point of B, it follows
that B U {(0,0)} = A is connected.
To show that A is not path connected, suppose to the contrary that there
is a path f : [0,1]->A with f(0) = ((1/pi),0)and f(1)=(0,0) Clearly (0,1)
not in f([0,1])but (0,1)in cl(f([0,1])) Hence f([0,1]) is not closed and
therefore f([0,1])is not compact.
On the other hand, f([0,1])is supposed to be compact since it is the
continuous image of a compact set.
This is a contradiction and we are forced to conclude that A is not path
connected.
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