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mtayab1994
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Homework Statement
Let A be a non empty part of ℝ, such that A and A complement are two open subsets of ℝ.
Homework Equations
1- Prove that A is not bounded above.
2- Assuming that A complement is non-empty and let x in A complement.
and let B={xεA such that x≤t}. Prove that B is non empty and has a lower bound m such that m≥x.
3- Prove that: [tex]m\notin A[/tex] and [tex]m\notin A^{c}[/tex] . Conclude that A=ℝ.
4- Let F be non-empty set in ℝ such that F and F complementary are closed in R. What can we say about the set F?
The Attempt at a Solution
1- A is an open subset of R so we're going to use a proof by contradiction.
Suppose that a=sup(A) so for all x in A : x≤a . But since A is open then there exists an ε>0 such that: [tex]]a-\epsilon,a+\varepsilon[\cap A\neq\varnothing[/tex] so that means there is another number say a+ε/2 that's also in A so therefore A is unbounded above.
2- B is a subset of A so there for B is nonempty. Second part I know I have to conduct a proof by contradiction but I don't know how to start . Well I can suppose that B doesn't have a lower bound.
Any help will be appreciated. Thank you very much.