David Cullen
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But R^2 - K constructed this way is connected. You can see this
because if you have a disconnection into two open sets U,V then every
set of the form {i}xR with i irrational is contained in one or the
other of U,V. Now either all such sets are in U in which case we are
done, or one of these sets {j}xR is in V, in which case let s in R be
the supremum over all points x such that (x,0) is in U with x < j. s
must be rational or else (s,0) is in the boundary of U and the boundary
of V, contrary to U,V being a disconnection. Still, even if s is
rational, (s, q^-1(s)) is in the boundary of U and the boundary of V.