Find the Mistake to this Problem Let $(a,b)$ be an open interval of real numbers and let $c \in (a,b)$. Describe an open interval $I$ centered at $c$ such that $I \subseteq (a,b)$.
Here is the proposed solution to the problem:
Let $\epsilon >0$ be the radius of an interval I centered at a point $c \in (a,b)$ such that $a < c-\epsilon$ and $c+\epsilon < b$. Then
$$a<c-\epsilon < c < c+\epsilon < b$$
So $I \subseteq (a,b)$, as desired.
 A: His answer is wrong in that he stated "let $\epsilon $ be the radius of an interval centered at c" implying such an interval exists.  Then he claims "such that" giving conditions assuming what is to be proven.
BUT
It could have worked as this:
"Intervals centered a $c $ of every possible radii exist (by simply taking the interval $(c-r,c+r)$ where $r $ is the desired radius).  Simply choose an $r = \epsilon $ such that $a \le c-\epsilon <c < c+\epsilon < b $" would maybe be acceptable to some instructors.  
It would depend on how obvious the statement "for $a < c, \exists \epsilon s.t. a < c- \epsilon <c $" would be considered.
Well... we've exhausted this subject, I think.
A: Easy route: Let $c \in (a,b)$ and take $\epsilon = \frac{1}{2}  \cdot \textrm{min}\{|c-a|,|c-b|\}$. Then just define $I = (c-\epsilon, c+\epsilon)$. 
A: How about $I=\{c\}=[c,c]$? Nowhere do you require $I$ to be open, or to have nonempty interior. You can't say it isn't centered at $c$, either.
This is a dumb answer, but, well, it's a poorly worded question.
