Suppose each U_alpha is connected. Form a graph with one vertex called v_\alpha for each U_\alpha, and with vertex v_\alpha connected to v_\beta iff U_\alpha intersect U_beta is nonempty. Prove or disprove: X is connected if and only if the graph is connected.
Could I get a hint for this? Not homework, just self study.
Martin: Oh, I took uniform measure over [0,1] and a step function. Your function is simpler (constant or identity?) but the measure is a combination of the Dirac measures. // Summarizing to myself basically.. :)
The uncountability is making this tricky. I want to say assume we have a separation U,V of X, and show this leads to a separation of the graph.
Suppose a collection subsets U_\alpha of a set X is given. Define a topology on each U_\alpha. Define a topology on X by declaring a subset V open if and only if V intersect U_\alpha is open in U_\alpha for every \alpha. I have verified this is a topology. Question:
Suppose each U_alpha is connected. Form a graph with one vertex called v_\alpha for each U_\alpha, and with vertex v_\alpha connected to v_\beta iff U_\alpha intersect U_beta is nonempty. Prove or disprove: X is connected if and only if the graph is connected.
I want to say take two of the U_\alpha, one in U, one in V, and show this leads to a contradiction, but I have a problem because the "path" of the graph joining of them my be uncountable.
Sorry for the interruption: can anyone tell me if this fails to be a counterexample and why? X = [1,2] union [3,4] with the usual topology. U1 = [1,2], U2 = [3,4], U3=X all with the usual topology. (Should verify that this is indeed ok.) (v1, v3) is an edge and so is (v2, v3). So the graph is connected. But, clearly the original space is not.
Ok so we take a connected component and wish to show the complement is open. The complement is open if and only if its intersection with every U_alpha is open. But this is now a trivial statement
Let our two vectors be w_1, w_2. Pick the one with smaller modulus, and let epsilon be half than that modulus. It suffices to show the ball around 0 with radius epsilon contains no point of the lattice.
