- #1
jojo12345
- 42
- 0
Hi,
I'm studying for the final exam in my first course in topology. I'm currently recalling as many theorems as I can and trying to prove them without referring to a text or notes. I think I have a proof that the closed interval [0,1] is connected, but it's different than what I have in my notes. I was hoping I might just verify that my proof is sound.
Proof:
Assume [0,1]=A\cup B where A and B are clopen, disjoint, and not empty. Further, assume that A contains 0. Because B is closed in the interval and the interval is closed in the reals, B is closed in the reals and contains its infimum, z=\text{inf}B\in B. Note that z\not =0 because A\cap B=\emptyset.
Now, because B is open in the interval, there is some \epsilon >0 such that b=(z-\epsilon,z+\epsilon)\cap [0,1]\subseteq B. b cannot contain 0, again, because this would violate the disjointness of A and B. Also, b cannot contain any numbers less than z because z is a lower bound on B. However, the only way this last sentence can be true is if z=0, which is absurd. Thus the initial separation of the interval into clopen, disjoint, not empty sets is impossible.
Does this work out, or did I overlook something?
I'm studying for the final exam in my first course in topology. I'm currently recalling as many theorems as I can and trying to prove them without referring to a text or notes. I think I have a proof that the closed interval [0,1] is connected, but it's different than what I have in my notes. I was hoping I might just verify that my proof is sound.
Proof:
Assume [0,1]=A\cup B where A and B are clopen, disjoint, and not empty. Further, assume that A contains 0. Because B is closed in the interval and the interval is closed in the reals, B is closed in the reals and contains its infimum, z=\text{inf}B\in B. Note that z\not =0 because A\cap B=\emptyset.
Now, because B is open in the interval, there is some \epsilon >0 such that b=(z-\epsilon,z+\epsilon)\cap [0,1]\subseteq B. b cannot contain 0, again, because this would violate the disjointness of A and B. Also, b cannot contain any numbers less than z because z is a lower bound on B. However, the only way this last sentence can be true is if z=0, which is absurd. Thus the initial separation of the interval into clopen, disjoint, not empty sets is impossible.
Does this work out, or did I overlook something?