Talk:Closure of Real Interval is Closed Real Interval

Nitpicking/comments:

The proof currently assumes $a\neq b$, but this is not stated as a hypothesis (it doesn't need to be).

The $N_\epsilon$ that is said to exist by definition in the first half of the proof uses an $\epsilon$ that might be different from the $\epsilon$ said to exist when $x=a$. This seems like it should be something like
 * If $x=a$, then $\exists \epsilon'$ such that $0 < \epsilon' < \epsilon$

and make use of the fact that $N_\epsilon$ is an interval. Alternatively, the statement that $N_\epsilon$ exists should be quantified. Right now it looks like one $N_\epsilon$ is being chosen. Mostly, care needs to be taken to ensure that $(c,d)$ and these epsilons all mesh.

In the last part of the proof, when $x<a$, a particular value of $\epsilon$ can be easily selected if you want (say, $(x+a)/2$. I don't know approach is pedagogically superior, or if anyone even has a preference.

Qedetc 01:24, 6 June 2011 (CDT)


 * Good call. Feel free to follow this up. --prime mover 01:35, 6 June 2011 (CDT)

I got carried away and changed it substantially. Let me know if it's too cluttered because of the cases or anything. Qedetc 19:54, 6 June 2011 (CDT)