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NihilTico
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Homework Statement
Let [itex]A=\mathbf{C}-[/itex]{[itex]z:Re(z)[/itex] and [itex]Im(z)[/itex] are rational}. Show that [itex]A[/itex] is a connected set.
Homework Equations
My book gives the definition of a disconnected set as a set that satisfies three conditions. A set [itex]A[/itex] is disconnected if there exist two open sets [itex]U[/itex] and [itex]V[/itex] in [itex]\mathbf{C}[/itex], that satisfy (i) [itex]A\cap U\cap V=\emptyset[/itex]; (ii) [itex]A\cap U\ne\emptyset[/itex] and [itex]A\cap V\ne\emptyset[/itex]; (iii) [itex]A\subset U\cup V[/itex].
The Attempt at a Solution
I feel that I should assume that the set is disconnected and derive a contradiction from this, but I haven't an idea where to start when I do that. Another approach I entertained was to show that there exists a polygonal arc between any two points such that it is entirely contained within the set [itex]A[/itex], of course, I couldn't figure out how that would work either.
So, going off of the former attack mentioned, assuming that the set [itex]A[/itex] is disconnected only gives me that there exist two sets that satisfy the above three arguments. My intuition in this case tells me that a contradiction would most easily arise by considering (i) and (iii). But isn't it the case that I could construct two sets that satisfy all three of these conditions by requiring [itex]U[/itex] to be the set of all pairs (x,y) such that at least one of x and y are rational and by letting [itex]V[/itex] be the set of all pairs (x,y) such that both of x and y are irrational? I see no contradiction in that, and it seems to say to me that it is disconnected.
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