• Path Connected?

    From William Elliot@21:1/5 to All on Sat May 7 06:15:14 2016
    Let K = \/{ {1/n}x[-n,oo) | n in N }
    . . S = R^2 - K

    Is S connected but not path connected?

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  • From David Cullen@21:1/5 to William Elliot on Sun May 8 06:13:33 2016
    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?

    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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  • From William Elliot@21:1/5 to William Elliot on Mon May 9 05:57:12 2016
    On Sun, 8 May 2016, William Elliot wrote:
    On Sun, 8 May 2016, David Cullen wrote:

    Good insights, David.  Here's a simplification of your proof.

    Let K = \/{ [-n,oo)x{1/n} | n in N }.  S = R^2 - K.

    S is connected but not path connected.  Proof ensues.

    For all j in N, let Aj = Rx[1/j,oo) - K, B = Rx(-oo,0].
    A = \/_j Aj and B are disjoint and connected.  Also S = A \/ B.

    If open U,V disconnect S, then wlog A subset U, B subset V.
    Since p = (0,0) in B, there's some r > 0 with B(p,r) subset V.
    For some j in N, 1/j < r.  Find some s with 1/(j+1) < s < 1/j.
    As (0,s) in A /\ B(p,r), (0,s) in U /\ V, contradicting that U and V
    are disjoint.  Thus S is connected.

    Let p be a path from (0,0) to (0,2) within S.
    f = pi_1 o p and g = pi_2 o p are in C([0,1],R).
    Since g(0) = 0, g(1) = 2 for all j in N,
    there's some xj in [0,1] with g(xj) = 1/j.
    Whence for all j, f(xj) < -j;  f is unbounded, a contradiction
    since the domain of f is compact.  Thusly S is not path connected.

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