[proofplan] We prove both directions directly from the domain-relative $\varepsilon$-$\delta$ definition of continuity at $a$. In the forward direction, an open interval containing $f(a)$ contains a symmetric interval around $f(a)$, and continuity sends sufficiently nearby points of $E$ into that smaller interval. In the reverse direction, we test the assumed neighbourhood condition on the particular interval $(f(a)-\varepsilon,f(a)+\varepsilon)$ to recover the $\varepsilon$-$\delta$ condition. [/proofplan] [step:Send a sufficiently small relative neighbourhood of $a$ into a given interval around $f(a)$] Assume that $f$ is continuous at $a$. Let $V \subset \mathbb{R}$ be an open interval with $f(a) \in V$. Since $V$ is an open interval containing $f(a)$, there exists $\varepsilon > 0$ such that \begin{align*} (f(a)-\varepsilon,f(a)+\varepsilon) \subset V. \end{align*} By continuity of $f$ at $a$, applied to this $\varepsilon$, there exists $r > 0$ such that for every $x \in E$, \begin{align*} |x-a| < r \implies |f(x)-f(a)| < \varepsilon. \end{align*} Now let $x \in E \cap (a-r,a+r)$. Then $x \in E$ and $|x-a| < r$, so $|f(x)-f(a)| < \varepsilon$. Hence \begin{align*} f(x) \in (f(a)-\varepsilon,f(a)+\varepsilon) \subset V. \end{align*} Therefore $x \in f^{-1}(V)$. Since $x$ was arbitrary, \begin{align*} E \cap (a-r,a+r) \subset f^{-1}(V). \end{align*} [guided] Assume that $f$ is continuous at $a$. We must prove the stated neighbourhood property for an arbitrary open interval $V \subset \mathbb{R}$ satisfying $f(a) \in V$. The first task is to convert the interval condition into the kind of numerical tolerance that appears in the $\varepsilon$-$\delta$ definition of continuity. Because $V$ is an open interval and $f(a) \in V$, there exists a symmetric interval around $f(a)$ contained in $V$. Thus we may choose $\varepsilon > 0$ such that \begin{align*} (f(a)-\varepsilon,f(a)+\varepsilon) \subset V. \end{align*} Now use continuity of $f: E \to \mathbb{R}$ at the point $a \in E$. This is continuity relative to the domain $E$: for this chosen $\varepsilon > 0$, there exists $r > 0$ such that every $x \in E$ with $|x-a| < r$ satisfies \begin{align*} |f(x)-f(a)| < \varepsilon. \end{align*} We claim this same $r$ gives the required inclusion. Let $x \in E \cap (a-r,a+r)$. Then $x \in E$, and membership in the interval $(a-r,a+r)$ is exactly the inequality $|x-a| < r$. Therefore the continuity implication gives \begin{align*} |f(x)-f(a)| < \varepsilon. \end{align*} This inequality is equivalent to \begin{align*} f(x) \in (f(a)-\varepsilon,f(a)+\varepsilon). \end{align*} Since this smaller interval is contained in $V$, we obtain $f(x) \in V$, which means precisely that $x \in f^{-1}(V)$. Hence every element of $E \cap (a-r,a+r)$ belongs to $f^{-1}(V)$, so \begin{align*} E \cap (a-r,a+r) \subset f^{-1}(V). \end{align*} [/guided] [/step] [step:Recover the epsilon-delta condition from the interval condition] Conversely, assume that for every open interval $V \subset \mathbb{R}$ with $f(a) \in V$, there exists $r > 0$ such that \begin{align*} E \cap (a-r,a+r) \subset f^{-1}(V). \end{align*} Let $\varepsilon > 0$ be arbitrary, and define the open interval $V \subset \mathbb{R}$ by \begin{align*} V := (f(a)-\varepsilon,f(a)+\varepsilon). \end{align*} Then $f(a) \in V$, so the assumed neighbourhood condition gives $r > 0$ such that \begin{align*} E \cap (a-r,a+r) \subset f^{-1}(V). \end{align*} If $x \in E$ and $|x-a| < r$, then $x \in E \cap (a-r,a+r)$, hence $x \in f^{-1}(V)$. Therefore $f(x) \in V$, which is equivalent to \begin{align*} |f(x)-f(a)| < \varepsilon. \end{align*} Since this holds for every $\varepsilon > 0$, $f$ is continuous at $a$. [/step]