Having studied open sets, closed sets, limit points, and convergence in metric spaces, we now study pointwise continuity. Pointwise continuity asks how a mapping behaves near one fixed point of its domain. This idea is basic in functional analysis because many important spaces are themselves spaces of functions, and mappings between them must be tested by the metric of the range. The focus keyword pointwise continuity means continuity at a specified point, not necessarily a uniform condition over the whole domain. Students often confuse the two; in pointwise continuity, the number $\delta$ may depend on the point $a$. :::definition[Continuity at a Point] Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces, let $A\subseteq X$, let $f:A\to Y$, and let $a\in A$. The function $f$ is called continuous at $a$ if for every $\epsilon>0$, there exists $\delta>0$ such that $$\begin{equation*} d_Y(f(x),f(a))<\epsilon \end{equation*}$$ whenever $x\in A$ and $$\begin{equation*} d_X(x,a)<\delta. \end{equation*}$$ ::: :::definition[Continuity on a Set] Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces, let $A\subseteq X$, and let $f:A\to Y$. The function $f$ is called continuous on $A$ if it is continuous at every point of $A$. ::: :::remark In the definition of pointwise continuity, the number $\delta$ may depend on the point $a$ and on $\epsilon$. If a positive number $\delta$ works for a fixed point and a fixed $\epsilon$, then every smaller positive number also works. ::: :::example[Isolated Points] Let $a$ be an isolated point of $A$. Then every function $f:A\to Y$ is continuous at $a$. Given that $a$ is isolated, there exists $r>0$ such that $$\begin{equation*} S(a,r)\cap A=\{a\}. \end{equation*}$$ Thus, whenever $x\in A$ and $d_X(x,a)<r$, we have $x=a$. Therefore, $$\begin{equation*} d_Y(f(x),f(a))=0. \end{equation*}$$ Hence, $f$ is continuous at $a$. ::: :::theorem[Sequential Criterion for Continuity] Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces, let $A\subseteq X$, and let $f:A\to Y$. Then $f$ is continuous at $a\in A$ if and only if for every sequence $\{x_n\}$ in $A$ such that $x_n\to a$, we have $f(x_n)\to f(a)$. ::: :::proof Given that $(X,d_X)$ and $(Y,d_Y)$ are metric spaces, $A\subseteq X$, $f:A\to Y$, and $a\in A$. To prove that $f$ is continuous at $a$ if and only if every sequence in $A$ converging to $a$ has image sequence converging to $f(a)$. First suppose that $f$ is continuous at $a$. Let $\{x_n\}$ be a sequence in $A$ such that $x_n\to a$. Let $\epsilon>0$. Since $f$ is continuous at $a$, there exists $\delta>0$ such that $d_Y(f(x),f(a))<\epsilon$ whenever $x\in A$ and $d_X(x,a)<\delta$. Since $x_n\to a$, there exists $n_0\in\mathbb N$ such that $d_X(x_n,a)<\delta$ for all $n\geq n_0$. Therefore $d_Y(f(x_n),f(a))<\epsilon$ for all $n\geq n_0$. Hence, $f(x_n)\to f(a)$. Conversely, suppose every sequence $\{x_n\}$ in $A$ with $x_n\to a$ satisfies $f(x_n)\to f(a)$. To prove that $f$ is continuous at $a$, if possible let $f$ not be continuous at $a$. Then there exists $\epsilon_0>0$ such that for every $\delta>0$, there exists $x\in A$ satisfying $d_X(x,a)<\delta$ and $d_Y(f(x),f(a))\geq\epsilon_0$. Taking $\delta=1/n$, choose $x_n\in A$ such that $d_X(x_n,a)<1/n$ and $d_Y(f(x_n),f(a))\geq\epsilon_0$. Then $x_n\to a$, but $f(x_n)$ does not converge to $f(a)$. A contradiction. Hence, $f$ is continuous at $a$. $\square$ ::: :::solved-problem Show that $f:\mathbb R^2\to\mathbb R$ defined by $$\begin{equation*} f(x,y)= \begin{cases} \dfrac{xy}{x^2+y^2}, & (x,y)\neq(0,0),\\ 0, & (x,y)=(0,0) \end{cases} \end{equation*}$$ is not continuous at $(0,0)$. ::: :::solution Let $p_n=(1/n,1/n)$. Then $p_n\to(0,0)$. But $$\begin{align*} f(p_n)&=\frac{(1/n)(1/n)}{1/n^2+1/n^2}\\ &=\frac12. \end{align*}$$ Therefore $f(p_n)\to1/2$, while $f(0,0)=0$. Hence, by the sequential criterion, $f$ is not continuous at $(0,0)$. ::: :::solved-problem Show that $f:\mathbb R^2\to\mathbb R$ defined by $$\begin{equation*} f(x,y)= \begin{cases} \dfrac{x^3-y^3}{x^2+y^2}, & (x,y)\neq(0,0),\\ 0, & (x,y)=(0,0) \end{cases} \end{equation*}$$ is continuous at $(0,0)$. ::: :::solution For $(x,y)\neq(0,0)$, $$\begin{align*} \left|\frac{x^3-y^3}{x^2+y^2}\right|&=\left|\frac{(x-y)(x^2+xy+y^2)}{x^2+y^2}\right|\\ &\leq |x-y|\left(1+\frac{|xy|}{x^2+y^2}\right). \end{align*}$$ Since $2|xy|\leq x^2+y^2$, we get $|xy|/(x^2+y^2)\leq1/2$. Therefore, $$\begin{equation*} |f(x,y)|\leq\frac32|x-y|\leq\frac32(|x|+|y|). \end{equation*}$$ As $(x,y)\to(0,0)$, the right hand side tends to $0$. Hence, $\lim_{(x,y)\to(0,0)}f(x,y)=0=f(0,0)$. ::: :::exercise 1. Prove that every constant mapping between metric spaces is continuous. 2. Prove that every function on a discrete metric space is continuous. 3. Use the sequential criterion to test continuity of $f(x)=|x|$ at $0$. ::: :::answer 1. If $f(x)=y_0$, then $d_Y(f(x),f(a))=0$ for all $x,a$. 2. In the discrete metric, choose $\delta=1/2$ at each point. 3. If $x_n\to0$, then $|x_n|\to0=|0|$. ::: :::faq[Frequently Asked Questions] Q: Does pointwise continuity require one $\delta$ for all points? A: No. For pointwise continuity at $a$, $\delta$ may depend on $a$ and $\epsilon$. Q: Why are sequences enough in metric spaces? A: Metric spaces are first countable, so sequences detect continuity at a point. :::