On Saturday, May 7, 2016 at 6:15:16 AM UTC-6, William Elliot wrote:
Let K = \/{ {1/n}x[-n,oo) | n in N }
. . S = R^2 - K
Is S connected but not path connected?
Yes.
In the below though I have reversed the x and y so for me K is the
union over natural n of all [-n, infinity)x{1/n}.
To see that S is connected, consider the sets:
X_0 := (-infinity, infinity)x(1, infinity)
X_n := ((-infinity, -n)x(1/(n+1), infinity)) union ((-infinity, infinity)x(1/(n+1), 1/n)) for natural n
X := the union over all of the X_i
Y := (-infinity, infinity)x(-infinity,0)
Z := (-infinity, infinity)x{0}
Now, S is the disjoint union of X, Y, and Z
X and Y are open and connected
(to see X is connected, use the well known theorem that the union over
any set of pairwise intersecting connected sets is connected)
Suppose disjoint open sets U, V form a disconnection of S, then without
loss of generality
X is contained in U and Y is contained in V (otherwise you could
obtain a disconnection of the connected sets X, Y;
noting that X and Y cannot both be contained in U because no open set
is contained in Z)
now the point (0, 0) in Z cannot be a member of U or V because every
open neighborhood of (0, 0) contains points from both X and Y.
This contradicts U, V being a disconnection, and so S is connected.
To see that S is not path connected:
Suppose there is a path from (0, 0) to (0, 2)
Then there are continuous functions f,g:[0, 1]->R with (f(0), g(0)) =
(0, 0), (f(1), g(1)) = (0, 2), and for all a in [0, 1], (f(a), g(a)) is
in S.
by the extreme value theorem, f attains a minimum value m. Let M be a
positive integer so that -M < m.
by the intermediate value theorem, there exists a c in [0, 1] so that
g(c) = 1/M.
now, f(c) is in [-M, infinity) so (f(c), g(c)) is not in S. This is a contradiction,
so S is not path connected.
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