Let A be a collection of sets, then We defined A’ = Union of all elements...

Let A be a collection of sets, then We defined A’ = Union of all elements of A.

Definition: If A = NOT the union of C and D , where C and D are non empty sub-collection,

such that C’ intersect D’ = empty

Then A is coherent.

Prove: if A is coherent then A’ is connected (i.e A’ is not the union of two separated sets in standard topology)

Expert Solution

Consider the contrapositive statement " if A' is disconnected then A is not coherent." It is enough to prove this contrapositive statement for our proof.

For let A' be disconnected, then there exist nonempty subsets B' and C' of A' such that A' = C' U B' and C' B' = . Correspondingly there exists nonempty sub collections C and B of A such that C' and B' are the union of all elements of C and B respectively. Clearly C U B A. Now let a A then a A' either a C' or a B' but not in both.

either a C or a B

a C U B

A C U B

A = C U B.

Hence there exists nonempty sub collections C and B of A such that A = C U B and C' B' = . Therefore A is not coherent.

Therefore if A' is disconnected then A is not coherent, which implies if A is coherent then A' is connected.

Colby Messinger answered 2 years ago

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