A continuous extension asks whether a function defined on a smaller set can be enlarged to a larger domain without losing continuity. The focus keyword continuous extension is important because many functions in analysis are first defined on dense sets. The main lesson is that agreement on a dense set determines a continuous map uniquely, but existence requires compatible limiting values at missing points. :::definition[Extension and Restriction] Let $X$ and $Y$ be sets, let $A\subset X$, and let $f:A\to Y$. A mapping $g:X\to Y$ is called an extension of $f$ if $g(x)=f(x)$ for every $x\in A$. In this case, $f$ is called the restriction of $g$ to $A$. ::: :::example[Reciprocal Function Without Extension] The function $f:(0,1)\to\mathbb R$ defined by $f(x)=1/x$ has no continuous extension to $[0,1)$, because $\lim_{x\to0^+}1/x$ does not exist as a real number. ::: :::theorem[Equality Set of Continuous Mappings] Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces, and let $f,g:X\to Y$ be continuous. Then $F=\{x\in X:f(x)=g(x)\}$ is closed in $X$. ::: :::proof Given that $f$ and $g$ are continuous. To prove that $F$ is closed. Let $x\in X\setminus F$. Then $d_Y(f(x),g(x))>0$. Put $\epsilon=d_Y(f(x),g(x))/2$. By continuity of $f$ and $g$, choose $\delta>0$ such that $d_Y(f(y),f(x))<\epsilon$ and $d_Y(g(y),g(x))<\epsilon$ whenever $d_X(x,y)<\delta$. Then $$\begin{equation*} d_Y(f(y),g(y))\geq d_Y(f(x),g(x))-d_Y(f(y),f(x))-d_Y(g(y),g(x))>0. \end{equation*}$$ Thus $S(x,\delta)\subseteq X\setminus F$, so $X\setminus F$ is open. Hence, $F$ is closed. $\square$ ::: :::corollary Let $f,g:X\to Y$ be continuous. If $f=g$ on a dense subset $A$ of $X$, then $f=g$ on $X$. ::: :::proof Given that $f=g$ on the dense subset $A$. To prove that $f=g$ on $X$. The equality set $F=\{x:f(x)=g(x)\}$ is closed and contains $A$. Hence $X=\overline A\subseteq F$. Therefore $F=X$, and $f=g$ on $X$. $\square$ ::: :::theorem[Continuous Extension from a Dense Subset] Let $A$ be dense in $(X,d_X)$ and let $(Y,d_Y)$ be a metric space. A mapping $f:A\to Y$ has a continuous extension $g:X\to Y$ if and only if for every limit point $x$ of $A$, the limit $\lim_{y\to x}f(y)$ exists in $Y$, and when $x\in A$, this limit equals $f(x)$. When the extension exists, it is unique. ::: :::proof Given that $A$ is dense in $X$. To prove the extension criterion. If $g$ is a continuous extension, then $\lim_{y\to x}f(y)=\lim_{y\to x}g(y)=g(x)$ at every limit point. Conversely, define $g(x)=f(x)$ for $x\in A$ and $g(x)=\lim_{y\to x}f(y)$ for $x\notin A$. The assumed limit condition gives continuity. Uniqueness follows because two continuous extensions agreeing on dense $A$ agree on all of $X$. $\square$ ::: :::example[Sine Reciprocal Function] The function $f(x)=\sin(1/x)$ for $x\neq0$ has no continuous extension to $0$, because sequences approaching $0$ can make the values approach $1$ and $-1$. ::: :::example[Sine Quotient Function] The function $f(x)=\sin x/x$ for $x\neq0$ has a continuous extension to $0$ by defining $f(0)=1$. ::: :::exercise 1. Prove uniqueness of continuous extensions from dense subsets. 2. Decide whether $x\sin(1/x)$ extends continuously to $0$. 3. Give a dense-domain function with no continuous extension. ::: :::answer 1. Two extensions agree on a dense set, hence everywhere. 2. Yes, with value $0$ at $0$. 3. $1/x$ on $(0,1)$ cannot extend continuously to $0$. :::