解 引理： R \mathbb{R} R具有稠密性，即 ( ∀ x , y ∈ R , x < y ) → ( ∃ a ∈ Q , b ∈ Q c , s . t . x < a < y , x < b < y ) \left( \forall x,y\in \mathbb{R},\text{ }x<y \right)\to \left( \exists a\in \mathbb{Q},\text{ }b\in {{\mathbb{Q}}^{c}},\text{ }s.t.\text{ }x<a<y,\text{ }x<b<y \right) (∀x,y∈R, x<y)→(∃a∈Q, b∈Qc, s.t. x<a<y, x<b<y)。
已知 ∀ r ( r ∈ I ∧ r ∈ Q → f ( r ) = 0 ) \forall r\left( r\in I\wedge r\in \mathbb{Q}\to f\left( r \right)=0 \right) ∀r(r∈I∧r∈Q→f(r)=0) 欲证明 ∀ r ( r ∈ I → f ( r ) = 0 ) \forall r\left( r\in I\to f\left( r \right)=0 \right) ∀r(r∈I→f(r)=0) 只需证明 ∀ r ( r ∈ I ∧ r ∈ Q c → f ( r ) = 0 ) \forall r\left( r\in I\wedge r\in {{\mathbb{Q}}^{c}}\to f\left( r \right)=0 \right) ∀r(r∈I∧r∈Qc→f(r)=0) 任取区间 I I I上的一个无理数 r 0 ∈ Q c {{r}_{0}}\in {{\mathbb{Q}}^{c}} r0∈Qc，假设 f ( r 0 ) ≠ 0 f\left( {{r}_{0}} \right)\ne 0 f(r0)=0。 f ∈ C 0 ( I ) , r 0 ∈ I ⇒ f ∈ C 0 ( r 0 ) ⇒ lim r → r 0 f ( r ) = f ( r 0 ) ≠ 0 ⇒ ∀ ε > 0 , ∃ δ > 0 , s . t . ∀ r ∈ U ( r 0 ; δ ) , ∣ f ( r ) − f ( r 0 ) ∣ < ε \begin{aligned} & f\in {{C}^{0}}\left( I \right),\text{ }{{r}_{0}}\in I\text{ }\Rightarrow \text{ }f\in {{C}^{0}}\left( {{r}_{0}} \right)\text{ }\Rightarrow \text{ }\underset{r\to {{r}_{0}}}{\mathop{\lim }}\,f\left( r \right)=f\left( {{r}_{0}} \right)\ne 0 \\ & \Rightarrow \forall \varepsilon >0,\text{ }\exists \delta >0,\text{ }s.t.\text{ }\forall r\in U\left( {{r}_{0}};\delta \right),\text{ }\left| f\left( r \right)-f\left( {{r}_{0}} \right) \right|<\varepsilon \\ \end{aligned} f∈C0(I), r0∈I ⇒ f∈C0(r0) ⇒ r→r0limf(r)=f(r0)=0⇒∀ε>0, ∃δ>0, s.t. ∀r∈U(r0;δ), ∣f(r)−f(r0)∣<ε 取 ε 0 = ∣ f ( r 0 ) ∣ 2 > 0 {{\varepsilon }_{0}}=\frac{\left| f\left( {{r}_{0}} \right) \right|}{2}>0 ε0=2∣f(r0)∣>0，则 ∃ δ 0 > 0 , s . t . ∀ r , r 0 − δ 0 < r < r 0 + δ 0 \exists {{\delta }_{0}}>0,\text{ }s.t.\text{ }\forall r,\text{ }{{r}_{0}}-{{\delta }_{0}}<r<{{r}_{0}}+{{\delta }_{0}} ∃δ0>0, s.t. ∀r, r0−δ0<r<r0+δ0， ∣ ∣ f ( r ) ∣ − ∣ f ( r 0 ) ∣ ∣ ≤ ∣ f ( r ) − f ( r 0 ) ∣ < ε 0 = ∣ f ( r 0 ) ∣ 2 ⇒ ∣ f ( r ) ∣ > ∣ f ( r 0 ) ∣ 2 > 0 \left| \text{ }\left| f\left( r \right) \right|-\left| f\left( {{r}_{0}} \right) \right|\text{ } \right|\le \left| f\left( r \right)-f\left( {{r}_{0}} \right) \right|<{{\varepsilon }_{0}}=\frac{\left| f\left( {{r}_{0}} \right) \right|}{2}\text{ }\Rightarrow \text{ }\left| f\left( r \right) \right|>\frac{\left| f\left( {{r}_{0}} \right) \right|}{2}>0 ∣ ∣f(r)∣−∣f(r0)∣ ∣≤∣f(r)−f(r0)∣<ε0=2∣f(r0)∣ ⇒ ∣f(r)∣>2∣f(r0)∣>0 由引理， ∃ q ( q ∈ ( r 0 − δ 0 , r 0 + δ 0 ) ∧ q ∈ Q ) \exists q\left( q\in \left( {{r}_{0}}-{{\delta }_{0}},{{r}_{0}}+{{\delta }_{0}} \right)\text{ }\wedge \text{ }q\in \mathbb{Q} \right) ∃q(q∈(r0−δ0,r0+δ0) ∧ q∈Q)， ∣ f ( q ) ∣ > 0 ⇒ f ( q ) ≠ 0 \left| f\left( q \right) \right|>0\text{ }\Rightarrow \text{ }f\left( q \right)\ne 0 ∣f(q)∣>0 ⇒ f(q)=0 这与 ∀ r ( r ∈ I ∧ r ∈ Q → f ( r ) = 0 ) \forall r\left( r\in I\wedge r\in \mathbb{Q}\to f\left( r \right)=0 \right) ∀r(r∈I∧r∈Q→f(r)=0) 矛盾。由反证法，即有 ∀ r ( r ∈ I ∧ r ∈ Q c → f ( r ) = 0 ) \forall r\left( r\in I\wedge r\in {{\mathbb{Q}}^{c}}\to f\left( r \right)=0 \right) ∀r(r∈I∧r∈Qc→f(r)=0)
f ( x ) f\left( x \right) f(x)在 I I I上严格单调递增 等价于 ∀ r 1 ∀ r 2 ( ( r 1 ∈ I ) ∧ ( r 2 ∈ I ) ∧ ( r 1 < r 2 ) → f ( r 1 ) < f ( r 2 ) ) \forall {{r}_{1}}\forall {{r}_{2}}\left( \left( {{r}_{1}}\in I \right)\wedge \left( {{r}_{2}}\in I \right)\wedge \left( {{r}_{1}}<{{r}_{2}} \right)\to f\left( {{r}_{1}} \right)<f\left( {{r}_{2}} \right) \right) ∀r1∀r2((r1∈I)∧(r2∈I)∧(r1<r2)→f(r1)<f(r2)) r 1 , r 2 ∈ Q {{r}_{1}},{{r}_{2}}\in \mathbb{Q} r1,r2∈Q的情况已由条件给出。以下我们将逐个考虑 r 1 , r 2 {{r}_{1}},{{r}_{2}} r1,r2一个有理数一个无理数 和 r 1 , r 2 ∈ Q c {{r}_{1}},{{r}_{2}}\in {{\mathbb{Q}}^{c}} r1,r2∈Qc的情况。
r 1 , r 2 {{r}_{1}},{{r}_{2}} r1,r2其中一个是有理数，一个是无理数。不妨令 r 1 ∈ Q , r 2 ∈ Q c , r 1 < r 2 {{r}_{1}}\in \mathbb{Q},\text{ }{{r}_{2}}\in {{\mathbb{Q}}^{c}},\text{ }{{r}_{1}}<{{r}_{2}} r1∈Q, r2∈Qc, r1<r2。令 d = r 2 − r 1 > 0 d={{r}_{2}}-{{r}_{1}}>0 d=r2−r1>0。在引理的支撑下，按照以下步骤构造序列 { a n } \left\{ {{a}_{n}} \right\} {an}和序列 { f ( a n ) } \left\{ f\left( {{a}_{n}} \right) \right\} {f(an)}。 ① 在 ( r 1 , r 2 ) ⋂ I \left( {{r}_{1}},{{r}_{2}} \right)\bigcap I (r1,r2)⋂I中任取一个有理数 a 1 {{a}_{1}} a1，有 f ( a 1 ) > f ( r 1 ) f\left( {{a}_{1}} \right)>f\left( {{r}_{1}} \right) f(a1)>f(r1) ② 在 ( max ( a 1 , r 1 + ∑ k = 1 1 d 2 k ) , r 2 ) ⋂ I \left( \max \left( {{a}_{1}},{{r}_{1}}+\sum\limits_{k=1}^{1}{\frac{d}{{{2}^{k}}}} \right),\text{ }{{r}_{2}} \right)\bigcap I (max(a1,r1+k=1∑12kd), r2)⋂I中任取一个有理数 a 2 {{a}_{2}} a2，有 f ( a 2 ) > f ( a 1 ) f\left( {{a}_{2}} \right)>f\left( {{a}_{1}} \right) f(a2)>f(a1) ③在 ( max ( a 2 , r 1 + ∑ k = 1 2 d 2 k ) , r 2 ) ⋂ I \left( \max \left( {{a}_{2}},{{r}_{1}}+\sum\limits_{k=1}^{2}{\frac{d}{{{2}^{k}}}} \right),\text{ }{{r}_{2}} \right)\bigcap I (max(a2,r1+k=1∑22kd), r2)⋂I中任取一个有理数 a 3 {{a}_{3}} a3，有 f ( a 3 ) > f ( a 2 ) f\left( {{a}_{3}} \right)>f\left( {{a}_{2}} \right) f(a3)>f(a2) … 重复以上过程得到序列 { a n } & { f ( a n ) } \left\{ {{a}_{n}} \right\}\And \left\{ f\left( {{a}_{n}} \right) \right\} {an}&{f(an)}。 （注：以上取法是可行的，因为 r 2 ∈ I , f ∈ C 0 ( r 2 ) ⇒ ∃ δ > 0 , f ∈ C 0 ( r 2 − δ , r 2 + δ ) {{r}_{2}}\in I,\text{ }f\in {{C}^{0}}\left( {{r}_{2}} \right)\Rightarrow \exists \delta >0,\text{ }f\in {{C}^{0}}\left( {{r}_{2}}-\delta ,{{r}_{2}}+\delta \right) r2∈I, f∈C0(r2)⇒∃δ>0, f∈C0(r2−δ,r2+δ)，而在 ( r 2 − δ , r 2 + δ ) \left( {{r}_{2}}-\delta ,{{r}_{2}}+\delta \right) (r2−δ,r2+δ)中由引理，一定总能取到有理数）
对序列 { a n } \left\{ {{a}_{n}} \right\} {an}，有 r 2 = r 1 + d ← r 1 + ∑ k = 1 n − 1 d 2 k ≤ max ( a n − 1 , r 1 + ∑ k = 1 n − 1 d 2 k ) < a n < r 2 → r 2 ( n → ∞ ) ⇒ a n → r 2 ( n → ∞ ) \begin{matrix} {{r}_{2}}={{r}_{1}}+d\leftarrow {{r}_{1}}+\sum\limits_{k=1}^{n-1}{\frac{d}{{{2}^{k}}}}\le \max \left( {{a}_{n-1}},{{r}_{1}}+\sum\limits_{k=1}^{n-1}{\frac{d}{{{2}^{k}}}} \right)<{{a}_{n}}<{{r}_{2}}\to {{r}_{2}}\left( n\to \infty \right) \\ \Rightarrow {{a}_{n}}\to {{r}_{2}}\left( n\to \infty \right) \\ \end{matrix} r2=r1+d←r1+k=1∑n−12kd≤max(an−1,r1+k=1∑n−12kd)<an<r2→r2(n→∞)⇒an→r2(n→∞)
对序列 { f ( a n ) } \left\{ f\left( {{a}_{n}} \right) \right\} {f(an)}，有
{ f ( a n ) } \left\{ f\left( {{a}_{n}} \right) \right\} {f(an)}是单调递增序列。 { f ( a n ) } \left\{ f\left( {{a}_{n}} \right) \right\} {f(an)}有上界。因为 f ∈ C 0 ( r 2 ) ⇒ ∃ δ > 0 , s . t . f ∈ C 0 ( r 2 − δ , r 2 + δ ) ⊇ ( r 2 , r 2 + δ ) f\in {{C}^{0}}\left( {{r}_{2}} \right)\text{ }\Rightarrow \text{ }\exists \delta >0,\text{ }s.t.f\in {{C}^{0}}\left( {{r}_{2}}-\delta ,{{r}_{2}}+\delta \right)\supseteq \left( {{r}_{2}},{{r}_{2}}+\delta \right) f∈C0(r2) ⇒ ∃δ>0, s.t.f∈C0(r2−δ,r2+δ)⊇(r2,r2+δ)，在 ( r 2 , r 2 + δ ) \left( {{r}_{2}},{{r}_{2}}+\delta \right) (r2,r2+δ)上由引理，任取一有理数 p p p，则有 ∀ a n , a n , p ∈ Q , a n < p ⇒ f ( a n ) < f ( p ) \forall {{a}_{n}},\text{ }{{a}_{n}},p\in \mathbb{Q},\text{ }{{a}_{n}}<p\text{ }\Rightarrow \text{ }f\left( {{a}_{n}} \right)<f\left( p \right) ∀an, an,p∈Q, an<p ⇒ f(an)<f(p) 因此可取 f ( p ) f\left( p \right) f(p)为 { f ( a n ) } \left\{ f\left( {{a}_{n}} \right) \right\} {f(an)}上界。 由定理：单调有界数列必有极限，有 lim n → ∞ f ( a n ) \underset{n\to \infty }{\mathop{\lim }}\,f\left( {{a}_{n}} \right) n→∞limf(an)存在，且有 lim n → ∞ f ( a n ) > f ( a 1 ) > f ( r 1 ) \underset{n\to \infty }{\mathop{\lim }}\,f\left( {{a}_{n}} \right)>f\left( {{a}_{1}} \right)>f\left( {{r}_{1}} \right) n→∞limf(an)>f(a1)>f(r1)结合序列 { a n } & { f ( a n ) } \left\{ {{a}_{n}} \right\}\And \left\{ f\left( {{a}_{n}} \right) \right\} {an}&{f(an)}的性质，有 lim n → ∞ f ( a n ) = lim a n → r 2 f ( a n ) \underset{n\to \infty }{\mathop{\lim }}\,f\left( {{a}_{n}} \right)=\underset{{{a}_{n}}\to {{r}_{2}}}{\mathop{\lim }}\,f\left( {{a}_{n}} \right) n→∞limf(an)=an→r2limf(an) 又函数 f ( x ) ∈ C 0 ( r 2 ) f\left( x \right)\in {{C}^{0}}\left( {{r}_{2}} \right) f(x)∈C0(r2)，因此有 lim a n → r 2 f ( a n ) = f ( lim a n → r 2 a n ) = f ( r 2 ) \underset{{{a}_{n}}\to {{r}_{2}}}{\mathop{\lim }}\,f\left( {{a}_{n}} \right)=f\left( \underset{{{a}_{n}}\to {{r}_{2}}}{\mathop{\lim }}\,{{a}_{n}} \right)=f\left( {{r}_{2}} \right) an→r2limf(an)=f(an→r2liman)=f(r2) 于是就有 f ( r 1 ) < f ( r 2 ) f\left( {{r}_{1}} \right)<f\left( {{r}_{2}} \right) f(r1)<f(r2)
r 1 , r 2 ∈ Q c {{r}_{1}},{{r}_{2}}\in {{\mathbb{Q}}^{c}} r1,r2∈Qc。
由于 f ∈ C 0 ( r 1 ) f\in {{C}^{0}}\left( {{r}_{1}} \right) f∈C0(r1)，因此 ∃ δ > 0 , s . t . f ∈ C 0 ( r 1 − δ , r 1 + δ ) \exists \delta >0,\text{ }s.t.f\in {{C}^{0}}\left( {{r}_{1}}-\delta ,{{r}_{1}}+\delta \right) ∃δ>0, s.t.f∈C0(r1−δ,r1+δ)， 在区间 ( r 1 , r 1 + δ ) \left( {{r}_{1}},{{r}_{1}}+\delta \right) (r1,r1+δ)上由引理，任取一有理数 q ∈ C 0 ( r 1 , r 1 + δ ) q\in {{C}^{0}}\left( {{r}_{1}},{{r}_{1}}+\delta \right) q∈C0(r1,r1+δ)。这样就有了 q ∈ Q , r 2 ∈ Q c q\in \mathbb{Q},\text{ }{{r}_{2}}\in {{\mathbb{Q}}^{c}} q∈Q, r2∈Qc，基于第一种情况的讨论，立刻就有 f ( q ) < f ( r 2 ) f\left( q \right)<f\left( {{r}_{2}} \right) f(q)<f(r2)同样，在区间 ( r 1 , q ) \left( {{r}_{1}},q \right) (r1,q)上由引理，任取一有理数 p ∈ C 0 ( r 1 , q ) p\in {{C}^{0}}\left( {{r}_{1}},q \right) p∈C0(r1,q)。这样就有了 r 1 ∈ Q c , p ∈ Q {{r}_{1}}\in {{\mathbb{Q}}^{c}},\text{ }p\in \mathbb{Q} r1∈Qc, p∈Q，基于第一种情况的讨论，立刻就有 f ( r 1 ) < f ( p ) f\left( {{r}_{1}} \right)<f\left( p \right) f(r1)<f(p)同时 p , q ∈ Q , f ( p ) < f ( q ) p,q\in \mathbb{Q},\text{ }f\left( p \right)<f\left( q \right) p,q∈Q, f(p)<f(q)，因此最终就有 f ( r 1 ) < f ( r 2 ) f\left( {{r}_{1}} \right)<f\left( {{r}_{2}} \right) f(r1)<f(r2)