Is "the smallest open set containing zero" a well defined concept? A function is said to be continuous at zero iff:
$\lim_{x \rightarrow 0}{f(x)} = f(0)$
Could this be the same as saying:


*

*Let $\Delta$ = the smallest open set containing zero

*$f(x) = f(0), \forall x \in \Delta$
Am I misunderstanding what a limit is, or are these two definitions equivalent?
Edit:  I've got a few responses saying that there is no such set as $\Delta$.  I agree.  However, it seems to me that the expression:
$\lim_{x \rightarrow 0}{f(x)} = f(0)$
is definitively claiming to evaluate $f$ at some unspecified value or values $x \neq 0$ but not claiming any particular values.  To be specific, for any particular value you could name, the claim that the limit exists does not claim it needs to evaluate f at that point.  So, what exactly is it claiming? 
 A: As others have commented, no smaller set exist (for a standard usage of smaller, at least). In terms of open sets, what you can say is that $f$ is continuous at $p$ if for every open set $V$ containing $f(0)$, there is an open set $U$ containing $0$ such that $f(U) \subseteq V$. This allows you to take arbitrarily small intervals, in particular, which approximates your intuitive idea of taking the 'smallest' such set. 
Many concepts in analysis, in particular, use this same thought process: defining notions by making sense of approximations up to arbitrary precision.
A: There is no smallest open set containing zero. 
Any open set containing zero by definition contains an open interval containing zero and that interval contains a smaller open interval containing zero.
A: There is no such thing as “the smallest open set containing $0$”. That's so because if $A$ is an open subset of $\mathbb R$ (I'm assuming that you're working in $\mathbb R$) and $0\in A$, then $A\supset(-\varepsilon,\varepsilon)$, for some $\varepsilon>0$. If $A\varsupsetneq(-\varepsilon,\varepsilon)$, take $A^\star=(-\varepsilon,\varepsilon)$; otherwise, take $A^\star=\left(-\frac\varepsilon2,\frac\varepsilon2\right)$. In each case, $A^\star$ is an open set containing $0$ and $A^\star\varsubsetneq A$.
