[proofplan] The proof uses the $\varepsilon$-$\delta$ definition of continuity at a point. Since $a$ is isolated, there is a positive-radius ball around $a$ containing no point of $X$ except $a$ itself. Choosing this radius as the continuity radius forces every nearby point $x$ to equal $a$, so the image distance $d_Y(f(x),f(a))$ is zero and therefore smaller than any prescribed $\varepsilon>0$. [/proofplan] [step:Use isolation to choose a neighbourhood containing only $a$] Let $f:X \to Y$ be an arbitrary function. Since $a$ is an isolated point of $X$, there exists a number $\delta_0>0$ such that the open ball \begin{align*} B_X(a,\delta_0)=\{x\in X: d_X(x,a)<\delta_0\} \end{align*} satisfies $B_X(a,\delta_0)=\{a\}$. [guided] We first translate the phrase "$a$ is isolated" into the metric form needed for an $\varepsilon$-$\delta$ proof. The definition says that there is some positive radius $\delta_0>0$ for which the open ball around $a$ contains no points of $X$ other than $a$ itself. With the notation \begin{align*} B_X(a,\delta_0)=\{x\in X: d_X(x,a)<\delta_0\}, \end{align*} this means precisely that \begin{align*} B_X(a,\delta_0)=\{a\}. \end{align*} This radius is the whole reason the theorem is true: near an isolated point, there are no genuinely different nearby inputs whose images could create a continuity obstruction. [/guided] [/step] [step:Verify the epsilon-delta condition at $a$] Let $\varepsilon>0$ be arbitrary, and set $\delta:=\delta_0$. If $x\in X$ satisfies $d_X(x,a)<\delta$, then $x\in B_X(a,\delta_0)=\{a\}$, so $x=a$. Therefore $f(x)=f(a)$, and the metric identity axiom in $(Y,d_Y)$ gives \begin{align*} d_Y(f(x),f(a))=d_Y(f(a),f(a))=0. \end{align*} Since $\varepsilon>0$, we have $0<\varepsilon$, hence \begin{align*} d_Y(f(x),f(a))<\varepsilon. \end{align*} Thus for every $\varepsilon>0$ there exists $\delta>0$ such that every $x\in X$ with $d_X(x,a)<\delta$ satisfies $d_Y(f(x),f(a))<\varepsilon$. This is exactly continuity of $f$ at $a$. [guided] We now check the definition of continuity at $a$ directly. The definition asks us to prove that for every tolerance $\varepsilon>0$ in the target [metric space](/page/Metric%20Space) $(Y,d_Y)$, there is a positive radius $\delta>0$ in the source metric space $(X,d_X)$ such that \begin{align*} d_X(x,a)<\delta \end{align*} forces \begin{align*} d_Y(f(x),f(a))<\varepsilon. \end{align*} Let $\varepsilon>0$ be given. We choose $\delta:=\delta_0$, where $\delta_0$ is the isolation radius found above. This number is positive because isolation provides a positive-radius neighbourhood. Now take any point $x\in X$ satisfying $d_X(x,a)<\delta$. Since $\delta=\delta_0$, this says exactly that $x\in B_X(a,\delta_0)$. But $B_X(a,\delta_0)=\{a\}$, so $x=a$. Applying the function $f:X\to Y$ to both sides gives $f(x)=f(a)$. Therefore, by the identity property of the metric $d_Y$, \begin{align*} d_Y(f(x),f(a))=d_Y(f(a),f(a))=0. \end{align*} Because the chosen $\varepsilon$ is positive, $0<\varepsilon$, and hence \begin{align*} d_Y(f(x),f(a))<\varepsilon. \end{align*} This proves the required implication for the arbitrary $\varepsilon>0$. Therefore $f$ is continuous at $a$. [/guided] [/step]