Hi,
Could anyone help me prove that " an open interval A is connected".
I tried proving it by contradiction and had 3 cases for A and taking two open sets U=(a,b) and V=(c,d)
case 1: b<c and showed it was impossible for A to be in U union V
case 2: b=c similarly
case 3: b>c similarly
Thank you
?? You are simply asserting that an interval is connected which is what was supposed to be proved!
Chaoticmath, if the interval (a, b) were NOT connected, there would exist two open sets, A, B, such that and . Take x in A, y= B and show that there must exist points in (x, y) that are not in either A or B.
Thank you that helped!
Here is what I have, please let me know if it's correct:
Take x in U, y in V such that x/=y (/= is for different), x<y and x,y belong to (a,b).
By a theorem we saw, we know (x,y)/=0. So let z belong to (x,y).
Now consider, the set A1={m in V : m>n for all n in U}
A1 included in A and A1 /= 0.
Let z be the greatest lower bound of A1. Since V is open then z is not in V so z is in U.
By another theorem we saw, there exists an open interval A2 containing z so there exists z2 in A2 such that z2>z but this contradicts the fact that z is the greatest lower bound.
Someone else read it and told me I may still have to show that my A1 was in A because U could be the union of many open sets, but I'm not sure I understand that.
Thank you for your help
Definition: A set is connected if and only if is not the union of two separated sets (R.L. Moore)
Definition: Two sets are separated if each contains neither a point nor a limit point of the other.
(Remember that Prof. Moore did not think that a point set could be empty).
Lemma: If then is either in or is a limit point of .
For the purpose of contradiction suppose that where are separated sets.
Without loss of generality we can say that .
Now consider these: .
Can you find a contradiction in all of that?