You proof is a bit long but it works.
The easy way is to show that B is a subset of C closure.
Would someone please check my work?
Problem:
Let B = {(0,y) : -1 <= y <= 1}
C = {(x,sin(1/x)) : 0<x<=1}
and A = BUC
Prove A is connected.
My proof:
Suppose to the contrary that A is not connected, there exists operators W and V such that: 1) W interset A and V interset A are not empty.
2) [A interset W] U [A interset V] = A.
3) (A interset W) interset (A interset V) = 0.
Now, (0,0) is in either W or V. So suppose WOLOG that (0,0) in W.
Then there exists r>0 such that Nr(0,0) C W.
Now, there exists p in C in Nr(0,0).
B is pathwise connected, so B is connected. Implies W,V cannot separate B, in which implies B is either in W or V.
Since (0,0) is in W, so B is in W.
Similarly, C is connected, because its range is an interval and domain is continuous. In which also implies W and V cannot separate C => C is in W or V.
Now, if C is in W, then A = (BUC) is a subset of W and V interset A would be empty, in which is contradicting our assumption (1). Which implies C is in V.
Now, p is in Nr(0,0), so p is in W; however, p is also in C, that is in V from our original assumption. Then we have p in both W and V, in which contradict our assumption (3) (A interset W) interset (A interset V) is empty.
Thus we conclude that A is connected.
Q.E.D.
You think this is right? Thanks.
C is connected, RIGHT?
The closure of a connected is also connected.
You did that one before.
http://www.mathhelpforum.com/math-he...t-problem.html