He closed unit interval [0,1] is a compact subspace of the real line E. In the following proof of this theorem the order of the various statements has been jumbled, and a couple of extra inappropriate statements have been added. Please remove the inappropriate statements and then order the remaining appropriate statements. Enter your answer as a string of capital letters corresponding to the correct order of the appropriate statements. Don't leave any spaces between the capital letters. For instance, enter DCBA... if you think the first appropriate statement is D, the second is C, the third is B, the fourth is A, and so on. Proof.
A Suppose that s=≠1.
B Let U∈F be chosen with s∈U.
C We need to show that s=1 as this will mean that some finite subcover of F has union equal to [0,1]
D Let U∈F be chosen with 1∈U.
E By the Least Upper Bound property of the real line the set X has a least upper bound s.
F Choose ϵ>0 with (s−ϵ,s+ϵ)⊆U.
G By the Intermediate Value Theorem the set X has a least upper bound s.
H Let F be a finite subfamily of F whose union contains [0,s−ϵ].
I Note that: - X is non-empty because 0∈X. - if 0≤y≤x and if x∈X then y∈X. - X is bounded above by 1 .
J Then F ∪{U} has union containing [0,s+ϵ].
K Consider the set X={x∈[0,1]:[0,x] is contained in the union of a finite su of F}.
L But then s is not the least upper bound of X. Hence we must have s=1.
M Let F be an open cover of [0,1].
N Let U∈F be chosen with s=1∈U. Choose ϵ>0 with the open set (1−ϵ,1]=[0,1]∩(1−ϵ,1+ϵ) contained in U.
P Then the union of the finite family F ∪{U} contains [0,1].
Q Therefore s∈X.
R Let F be a finite subfamily of F whose union contains [0,1−ϵ/2].