Path Connected?

[William Elliot]

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

Is S connected but not path connected?

[David Cullen]

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.

[William Elliot]

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.