- #1
jojo12345
- 43
- 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 [tex] [0,1]=A\cup B [/tex] 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, [tex]z=\text{inf}B\in B [/tex]. Note that [tex]z\not =0[/tex] because [tex]A\cap B=\emptyset[/tex].
Now, because [tex]B[/tex] is open in the interval, there is some [tex]\epsilon >0[/tex] such that [tex]b=(z-\epsilon,z+\epsilon)\cap [0,1]\subseteq B[/tex]. [tex]b[/tex] cannot contain [tex]0[/tex], again, because this would violate the disjointness of [tex]A[/tex] and [tex]B[/tex]. Also, [tex]b[/tex] cannot contain any numbers less than [tex]z[/tex] because [tex]z[/tex] is a lower bound on [tex]B[/tex]. However, the only way this last sentence can be true is if [tex]z=0[/tex], 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 [tex] [0,1]=A\cup B [/tex] 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, [tex]z=\text{inf}B\in B [/tex]. Note that [tex]z\not =0[/tex] because [tex]A\cap B=\emptyset[/tex].
Now, because [tex]B[/tex] is open in the interval, there is some [tex]\epsilon >0[/tex] such that [tex]b=(z-\epsilon,z+\epsilon)\cap [0,1]\subseteq B[/tex]. [tex]b[/tex] cannot contain [tex]0[/tex], again, because this would violate the disjointness of [tex]A[/tex] and [tex]B[/tex]. Also, [tex]b[/tex] cannot contain any numbers less than [tex]z[/tex] because [tex]z[/tex] is a lower bound on [tex]B[/tex]. However, the only way this last sentence can be true is if [tex]z=0[/tex], 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?