Continuous real-valued functions on closed subsets of metric spaces can be extended to the whole space. The focus keyword Tietze extension theorem names the main result. The proof uses distance functions to separate closed sets, then constructs a uniformly convergent series of continuous corrections. :::theorem[Bounded Approximation Lemma for Extension] Let $Y$ be a nonempty closed subset of a metric space $(X,d)$, and let $f:Y\to\mathbb R$ be bounded and continuous with $|f(y)|\leq M$. Then there exists continuous $g:X\to\mathbb R$ such that $|g(x)|\leq M/3$ for all $x$ and $|f(y)-g(y)|\leq2M/3$ for all $y\in Y$. ::: :::proof Given that $f$ is bounded and continuous on closed $Y$. To construct $g$. Let $A=\{y\in Y:f(y)\leq-M/3\}$ and $B=\{y\in Y:f(y)\geq M/3\}$. These sets are closed and disjoint. Define $g(x)=\frac{M}{3}\cdot\frac{d(x,A)-d(x,B)}{d(x,A)+d(x,B)}$. The denominator is positive, and distance functions are continuous, so $g$ is continuous. The construction gives $|g|\leq M/3$ and $|f-g|\leq2M/3$ on $Y$. $\square$ ::: :::theorem[Tietze Extension Theorem] Let $Y$ be a closed subset of a metric space $X$. If $f:Y\to\mathbb R$ is bounded and continuous, then there exists continuous $g:X\to\mathbb R$ such that $g(y)=f(y)$ for every $y\in Y$. ::: :::proof Given that $f$ is bounded and continuous on closed $Y$. To prove the extension theorem. Assume $|f|\leq M$. Apply the bounded approximation lemma to obtain $g_1$ with $|g_1|\leq M/3$ and error at most $2M/3$. Apply the lemma repeatedly to the remaining error. This produces continuous $g_n$ with $|g_n|\leq(M/3)(2/3)^{n-1}$. By the Weierstrass M-test, $\sum g_n$ converges uniformly on $X$. Its sum $g$ is continuous. The error on $Y$ tends to $0$, so $g=f$ on $Y$. $\square$ ::: :::corollary Let $Y$ be a closed subset of a metric space $X$. If $f:Y\to\mathbb R$ is continuous, then $f$ has a continuous real-valued extension to $X$. ::: :::proof Given continuous $f:Y\to\mathbb R$. To prove extension. The function $\arctan f$ is bounded and continuous on $Y$. Extend it by Tietze's theorem to a continuous $H:X\to(-\pi/2,\pi/2)$. Define $g(x)=\tan H(x)$. Then $g$ is continuous and extends $f$. $\square$ ::: :::solved-problem Let $F$ be a closed subset of $\mathbb R$, and let $f:F\to\mathbb R$ be continuous. Prove that $f$ has a continuous extension to $\mathbb R$. ::: :::solution The complement $\mathbb R\setminus F$ is a union of disjoint open intervals. On each bounded component $(a,b)$, define the extension by linear interpolation between $f(a)$ and $f(b)$. On unbounded components, extend constantly from the finite endpoint when present. On $F$, keep the value $f$. This gives a continuous extension. ::: :::faq[Frequently Asked Questions] Q: Why must the subset be closed? A: Boundary points outside a nonclosed set can force incompatible limiting values. Q: Why does the proof use uniform convergence? A: The extension is built as an infinite sum of continuous corrections. :::