Let $x \in S$. Let $\epsilon > 0$. {{WLOG}}, assume that $\map f x \le \map g x$.  ==== Case 1 : $\map f x = \map g x$ ==== Let $ y = \map f x = \map g x$ By definition of the min operation, $\min \set {\map f x, \map g x} = y$. From Continuity Test for Real-Valued Functions applied to $f$: :$\exists U \in \tau : x \in U : \forall z \in U : \map f z \in \openint {y - \epsilon} {y + \epsilon}$ Similarly, Continuity Test for Real-Valued Functions applied to $g$: :$\exists V \in \tau : x \in V : \forall z \in V : \map g z \in \openint {y - \epsilon} {y + \epsilon}$ Thus for all $z \in U \cap V$: :$\map f z \in \openint {y - \epsilon} {y + \epsilon}$ :$\map g z \in \openint {y - \epsilon} {y + \epsilon}$ By definition of the min operation, for all $z \in U \cap V$: :$\min \set {\map f z, \map g z} \in \openint {y - \epsilon} {y + \epsilon}$ By the open set axiom $(\text O2)$: :$U \cap V \in \tau$ Since $\epsilon > 0$ was arbitrary then: :$\forall \epsilon > 0 : \exists W \in \tau : x \in W : \forall z \in W : \min \set {\map f z, \map g z} \in \openint {y - \epsilon} {y + \epsilon}$ From Continuity Test for Real-Valued Functions, $\min \set {f, g}$ is continuous at $x$. {{qed|lemma}} ==== Case 2 : $\map f x < \map g x$ ==== Let $\map f x < \map g x$. Then: :$\min \set {\map f x, \map g x} = \map f x$ Let $\delta = \min \set {\epsilon, \dfrac 1 2 \paren {\map g x - \map f x} }$. Then $\delta > 0$. From Continuity Test for Real-Valued Functions applied to $f$: :$\exists U \in \tau : x \in U : \forall z \in U : \map f z \in \openint {\map f x - \delta} {\map f x + \delta}$ Similarly, Continuity Test for Real-Valued Functions applied to $g$: :$\exists V \in \tau : x \in V : \forall z \in V : \map g z \in \openint {\map g x - \delta} {\map g x + \delta}$ Thus for all $z \in U \cap V$: :$\map f z \in \openint {\map f x - \delta} {\map f x + \delta}$ :$\map g z \in \openint {\map g x - \delta} {\map g x + \delta}$ Now: {{begin-eqn}} {{eqn | l = \map f z       | o = <       | r = \map f x + \delta       | c = as $\map f z \in \openint {\map f x - \delta} {\map f x + \delta}$ }} {{eqn | o = \le       | r = \map f x + \frac 1 2 \paren {\map g x - \map f x}       | c = as $\delta \le \dfrac 1 2 \paren {\map g x - \map f x}$ }} {{eqn | r = \frac 1 2 \paren {2 \map f x + \map g x - \map f x} }} {{eqn | r = \frac 1 2 \paren {\map g x + \map f x} }} {{eqn | r = \frac 1 2 \paren {\map g x + \map g x - \map g x + \map f x} }} {{eqn | r = \frac 1 2 \paren {2 \map g x - \paren{\map g x - \map f x} } }} {{eqn | r = \map g x - \frac 1 2 \paren {\map g x - \map f x} }} {{eqn | o = \le       | r = \map g x - \delta       | c = as $-\dfrac 1 2 \paren {\map g x - \map f x} \le -\delta$ }} {{eqn | o = <       | r = \map g z       | c = as $\map g z \in \openint {\map g x - \delta} {\map g x + \delta}$ }} {{end-eqn}} Thus: :$\min \set {\map f z, \map g z} = \map f z$ As :$\map f z \in \openint {\map f x - \delta} {\map f x + \delta}$ and :$\map f x = \min \set {\map f x, \map g x}$ then :$\min \set {\map f z, \map g z} \in \openint {\min \set {\map f x, \map g x} - \delta} {\min \set {\map f x, \map g x} + \delta}$ By the open set axiom $(\text O 2)$: :$U \cap V \in \tau$ Since $\epsilon > 0$ was arbitrary then: :$\forall \epsilon > 0 : \exists W \in \tau : x \in W : \forall z \in W : \min \set{\map f z, \map g z} \in \openint {\min \set {\map f x, \map g x} - \delta} {\min \set {\map f x, \map g x} + \delta}$ From Continuity Test for Real-Valued Functions, $\min \set {f, g}$ is continuous at $x$. {{qed|lemma}} In either case, $\min \set {f, g}$ is continuous at $x$. Since $x \in S$ was arbitrary, then $\min \set {f, g}$ is everywhere continuous. {{qed}}