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May 7, 2016, 8:15:16 AM5/7/16

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Let K = \/{ {1/n}x[-n,oo) | n in N }

. . S = R^2 - K

Is S connected but not path connected?

May 8, 2016, 8:13:35 AM5/8/16

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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.

May 9, 2016, 7:57:15 AM5/9/16

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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.

> 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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