Question

Exercise 3.2.12. Let A be an uncountable set and let B be the set of real numbers that divides A into two uncountable sets; that is, s E B if both {{ : 2 € A and r < s} and {x : x € A and x > s} are uncountable. Show B is nonempty and open. T

Answer 1

1 First we will show that B +0. We will show B is not empty for bounded set. Proposition : Every uncountable set has bounded uncountable subset. Proof: Let A be uncountable subset. Define A = (-n, n) na Then there exists some N such that for all n > N, , is uncountable. If it were not true then would be countable for all n N. But ỦA =Ů ((-n,n) 4) = A n=1 n-1 But countable union of countable set is countable which shows A is also countable which contradicts our given assumption. Therefore there must exists some N such that for all n > N, A', is uncountable. 2 B is not empty. Without loss of generality, we assume A is bounded. Define X. = {a € A: a <r} and P= {r ER: X, is countable }. Sincet A is bounded so, there exists r such that X, is uncountable. Since P is bounded above, by AoC, there exists a = sup P and by definition of supremum, for all e > 0, Xate is uncountable. Similarly, define Yo = {a e A: a > s} and Q = {8 € R:Y, is countable }. Again by boundedness of A, we define B = inf Q. Then 8 >a otherwise a > 3 implies A is countable set. For any e > 0, by definition Xate and Y3-e are uncountable subsets of A. We may choose e > 0 such that Xaten Y3-e = 0 then there always exist some reR such that r > a for all a e Xote and 2 <a for all a € X3-c. Hence r e B and B +0.

3 B is open. Letre B, we wish to show that there exists € > 0 such that X, is uncountable for all r-e <m<r. If X, where countable for any e then choose en = then Xr-, is countable for every n. So UX,-- = X, n=1 is also countable which is a contradiction.