hcju/proofwikiqa huggingface dataset:
Let $\xi$ be such that $\map k \xi \ne 0$. From Differentiable Function is Continuous, $k$ is continuous at $\xi$. It follows that there exists an $\epsilon > 0$, such that $\size h < \epsilon \implies \map k {\xi + h} \ne 0$. So let $\size h < \epsilon$. Then we have: {{begin-eqn}} {{eqn | l = \frac {\map f {\xi + h} - \map f \xi} h       | r = \frac 1 h \paren {\frac {\map j {\xi + h} } {\map k {\xi + h} } - \frac {\map j \xi} {\map k \xi } }       | c = as $\map k \xi, \map k {\xi + h} \ne 0$ }} {{eqn | r = \frac {\map j {\xi + h} \map k \xi - \map k {\xi + h} \map j \xi} {h \map k {\xi + h} \map k \xi} }} {{eqn | r = \frac 1 {\map k {\xi + h} \map k \xi} \paren {\frac {\map j {\xi + h} - \map j \xi} h \map k \xi - \map j \xi \frac {\map k {\xi + h} - \map k \xi} h}       | c = algebraic manipulation }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | r = \lim_{h \mathop \to 0} \frac {\map f {\xi + h} - \map f \xi} h       | o =  }} {{eqn | r = \lim_{h \mathop \to 0} \frac 1 {\map k {\xi + h} \map k \xi} \paren {\frac {\map j {\xi + h} - \map j \xi} h \map k \xi - \map j \xi \frac {\map k {\xi + h} - \map k \xi} h} }} {{eqn | r = \lim_{h \mathop \to 0} \frac 1 {\map k {\xi + h} \map k \xi} \cdot \lim_{h \mathop \to 0} \paren {\frac {\map j {\xi + h} - \map j \xi} h \map k \xi - \map j \xi \frac {\map  k {\xi + h} - \map k \xi} h}       | c = Product Rule for Limits of Functions }} {{end-eqn}} Thus by: :continuity of $k$ at $\xi$ :differentiability of $j$ and $k$ at $\xi$ :Combined Sum Rule for Limits of Functions: it is concluded that: :$\displaystyle \lim_{h \mathop \to 0} \frac {\map f {\xi + h} - \map f \xi} h = \frac 1 {\map k \xi^2} \paren {\map {j'} \xi \map k \xi - \map j \xi \map {k'} \xi}$ From the definition of differentiability, $f$ is differentiable at $\xi$, with stated value. {{qed}}
Let $k,l$ be natural numbers with $k \geq l$. By the Local Submersion Theorem, there exists coordinates in some open sets of $x,y$ such that $f(x_1, x_2, \ldots, x_k)=(x_1, \ldots,x_l)$ and $y$ corresponds to $(0, \ldots, 0)$. Let $V$ be that neighborhood of $x$. Then $f^{-1}(y) \cap V$ is the set of points where $x_1=0, \ldots, x_l=0$. The functions $x_{l+1}, \ldots, x_k$ therefore form a coordinate system on the set $f^{-1}(y) \cap V$, which is a relatively open subset of $f^{-1}(y)$. Together these functions then form a diffeomorphism to a Euclidean space. We also have, by the regular value properties of $y$, a surjection of tangent spaces from $x$ to $y$. This ensures smoothness of the solution set $f^{-1}(y)$. {{qed}}
{{AimForCont}} $\exists k \in \N$ such that $\map P k$ is false. From Power of Real Number greater than One is Unbounded Above, $\set {2^n: n \in \N}$ is unbounded above Therefore we can find: :$M = 2^N > k$ Now let us create the set: :$S = \set {n \in \N: n < M, \map P n \text { is false} }$ As $k < M$ and $\map P k$ is false, $S \ne \O$ and as $\forall x \in S: x < M$ it follows that $S$ is bounded above. So from Set of Integers Bounded Above by Integer has Greatest Element, $S$ has a greatest element. However, $\map P n$ holds for $m < n \le M$ and hence $\map P {m + 1}$ holds. But $\map P {m + 1} \implies \map P m$ and so $\map P m$ holds after all. So there can be no such $m$ and therefore $S = \O$, hence there can be no such $k \in \N$ such that $\map P k$ is false. Hence the result. {{qed}}
From Difference of Two Powers: :$\displaystyle a^n - b^n = \paren {a - b} \paren {a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \dotsb + a b^{n - 2} + b^{n - 1} } = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$ Let $a > b > 0$. Then: :$\dfrac {a^n - b^n} {a - b} = a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \dotsb + a b^{n - 2} + b^{n - 1}$ Also, note that: :$b^{n - 1} = b^{n - j - 1} b^j \le a^{n - j - 1} b^j \le a^{n - j -1} a^j \le a^{n - 1}$ Hence if $a > b > 0$: :$n b^{n - 1} \le \dfrac {a^n - b^n} {a - b} \le n a^{n - 1}$ Taking reciprocals: :$\dfrac 1 {n b^{n - 1} } \ge \dfrac {a - b} {a^n - b^n} \ge \dfrac 1 {n a^{n - 1} }$ Thus: :$\dfrac {a^n - b^n} {n a^{n - 1} } \le a - b \le \dfrac {a^n - b^n} {n b^{n - 1} } \implies \dfrac {a \paren {a^n - b^n} } {a^n} \le n \paren {a - b} \le \dfrac {b \paren {a^n - b^n} } {b^n}$ Now suppose $x > y$. Then: :$x - y = \size {x - y}$ and: :$x^{1/n} - y^{1/n} = \size {x^{1/n} - y^{1/n} }$ Then put $a = x^{1/n}$ and $b = y^{1/n}$ in the above: :$\dfrac {x^{1/n} \size {x - y} } x \le n \size {x^{1/n} - y^{1/n} } \le \dfrac {y^{1/n} \size {x - y} } y$ Multiplying through by $x y$: :$x^{1/n} y \size {x - y} \le n x y \size {x^{1/n} - y^{1/n} } \le x y^{1/n} \size {x - y}$ Similarly, if $y > x$: :$y^{1/n} x \size {x - y} \le n x y \size {x^{1/n} - y^{1/n} } \le y x^{1/n} \size {x - y}$ The result follows after some algebra. {{qed}}
First suppose that $\xi > 0$. Let $X, Y \in \R$ such that $0 < X < \xi < Y$. Let $x \in \R$ such that $X < x < Y$. From Inequalities Concerning Roots: :$X Y^{1/n} \ \size {x - \xi} \le n X Y \ \size {x^{1/n} - \xi^{1/n} } \le Y X^{1/n} \ \size {x - \xi}$ Thus: :$\dfrac 1 {n Y} Y^{1/n} \ \size {x - \xi} \le \size {x^{1/n} - \xi^{1/n} } \le \dfrac 1 {n X} X^{1/n} \ \size {x - \xi}$ The result follows by applying the Squeeze Theorem. Now we need to show that $\map f x \to 0$ as $x \to 0^+$. We need to show that: :$\forall \epsilon > 0: \exists \delta > 0: x^{1/n} = \size {x^{1/n} - 0} < \epsilon$ provided $0 < x < \delta$. Clearly, for any given $\epsilon$, we can choose $\delta = \epsilon^n$. Hence the result. {{qed}}
Let $M > 2 k + 1$ be a natural number such that $f: X \to \R^M$ is an injective immersion. Define a map $h: X \times X \times \R \to \R^M$ by: :$\map h {x, y, t} = \map t {\map f x - \map f y}$ Define a map $g: \map T X \to \R^M$ by: :$\map g {x, v} = \map {\d f_x} v$ where $\map T X$ is the tangent bundle of $X$. Since $M > 2 k + 1$, the Morse-Sard Theorem implies $\exists a \in \R^M$ such that $a$ is in the image of neither function. Let $\pi$ be the projection of $\R^M$ onto the orthogonal complement of $a, H$.   Suppose: :$\map {\paren {\pi \circ f} } x = \map {\paren {\pi \circ f} } y$ for some $x, y$. Then: :$\map f x - \map f y = t a$ for some scalar $t$. {{AimForCont}} $x \ne y$. Then as $f$ is injective: :$t \ne 0$ But then $\map h {x, y, 1/t} = a$, contradicting the choice of $a$. By Proof by Contradiction: :$x = y$ and so $\pi \circ f$ is injective. Let $v$ be a nonzero vector in $\map {T_x} X$ (the tangent space of $X$ at $x$) for which: :$\map {\d \paren {\pi \circ f}_x} v = 0$ Since $\pi$ is linear: :$\map {\d \paren {\pi \circ f}_x} = \pi \circ \d f_x$ Thus: :$\map {\pi \circ \d f_x} v = 0$ so: :$\map {\d f_x} v = t a$ for some scalar $t$. Because $f$ is an immersion, $t \ne 0$. Hence: :$\map g {x, 1/t} = a$ contradicting the choice of $a$. Hence $\pi \circ f: X \to H$ is an immersion.  $H$ is obviously isomorphic to $\R^{M-1}$. Thus whenever $M > 2 k + 1$ and $X$ admits of a one-to-one immersion in $\R^M$, it follows that $X$ also admits of a one-to-one immersion in $\R^{M-1}$.
Let $p \left({z}\right)$ be a polynomial in $\C$: : $p \left({z}\right) = z^m + a_1 z^{m-1} + \cdots + a_m$ where not all of $a_1, \ldots, a_m$ are zero. Define a homotopy: : $p_t \left({z}\right) = t p \left({z}\right) + \left({1-t}\right) z^m$ Then: : $\displaystyle \frac{p_t \left({z}\right)}{z^m} = 1 + t \left({a_1 \frac 1 z + \cdots + a_m \frac 1 {z^m}}\right)$ The terms in the parenthesis go to $0$ as $z \to \infty$. {{explain}} Therefore, there is an $r \in \R_{>0}$ such that: : $\forall z \in \C: \left|{z}\right| = r: \forall t \in \left[{0 \,.\,.\, 1}\right]: p_t \left({z}\right) \ne 0$ Hence the homotopy: : $\displaystyle \frac{p_t}{ \left|{p_t}\right|}:S \to \Bbb S^1$ is well-defined for all $t$. This shows that for any complex polynomial $p \left({z}\right)$ of order $m$, there is a circle $S$ of sufficiently large radius in $\C$ such that $\dfrac{p \left({z}\right)}{\left|{p \left({z}\right)}\right|}$ and $\dfrac{z^m}{\left|{z^m}\right|}$ are freely homotopic maps $S \to \Bbb S^1$.  Hence $\dfrac{p \left({z}\right)}{\left|{p \left({z}\right)}\right|}$ must have the same degree of $\left({z / r}\right)^m$, which is $m$.   When $m > 0$, that is $p$ is non-constant, this result and the Extendability Theorem for Intersection Numbers imply $\dfrac{p \left({z}\right)}{\left|{p \left({z}\right)}\right|}$ does not extend to the disk $\operatorname{int} \left({S}\right)$, implying $p \left({z}\right) = 0$ for some $z \in \operatorname{int} \left({S}\right)$. {{explain|$\operatorname{int} \left({S}\right)$}} {{qed}}
Let $p: \C \to \C$ be a complex polynomial with $p \left({z}\right) \ne 0$ for all $z \in \C$. Then $p$ extends to a continuous transformation of the Riemann sphere: :$\hat \C = \C \cup \left\{{\infty}\right\}$ This extension also has no zeroes. {{MissingLinks|Write and link to appropriate version of "continuous".}} {{explain|why this extension has no zeroes}} By Riemann Sphere is Compact, there is some $\epsilon \in \R_{>0}$ such that: :$\forall z \in \C: \left|{p \left({z}\right)}\right| \ge \epsilon$ Now consider the holomorphic function $g: \C \to \C$ defined by: : $g \left({z}\right) := \dfrac 1 {p \left({z}\right)}$ We have: : $\forall z \in \C: \left|{g \left({z}\right)}\right| \le \dfrac 1 \epsilon$ By Liouville's Theorem, $g$ is constant. Hence $p$ is also constant, as claimed. {{qed}}
Let $p: \C \to \C$ be a complex, non-constant polynomial. {{AimForCont}} that $\map p z \ne 0$ for all $z \in \C$. Now consider the closed contour integral: :$\displaystyle \oint \limits_{\gamma_R} \frac 1 {z \cdot \map p z} \rd z$ where $\gamma_R$ is a circle with radius $R$ around the origin. By Derivative of Complex Polynomial, the polynomial $z \cdot \map p z$ is holomorphic. Since $\map p z$ is assumed to have no zeros, the only zero of $z \cdot \map p z$ is $0 \in \C$. Therefore by Reciprocal of Holomorphic Function $\dfrac 1 {z \cdot \map p z}$ is holomorphic in $\C \setminus \set 0$. Hence the Cauchy-Goursat Theorem implies that the value of this integral is independent of $R > 0$. On the one hand, one can calculate the value of this integral in the limit $R \to 0$ (or use the Residue Theorem), using the parameterization $z = R e^{i \phi}$ of $\gamma_R$: {{begin-eqn}} {{eqn | l = \lim \limits_{R \mathop \to 0} \oint \limits_{\gamma_R} \frac 1 {z \cdot \map p z} \rd z       | r = \frac 1 {\map p 0} \lim \limits_{R \mathop \to 0} \int \limits_0^{2 \pi} \frac 1 {R e^{i \phi} } \, i R e^{i \phi} \rd \phi       | c = Real Polynomial Function is Continuous and Product Rule }} {{eqn | r = \lim_{R \mathop \to 0} \frac 1 {\map p 0} \int_0^{2 \pi} i \rd \phi }} {{eqn | r = \frac {2 \pi i} {\map p 0}       | c = }} {{end-eqn}} which is non-zero. On the other hand, we have the following upper bound for the absolute value of the integral: {{begin-eqn}} {{eqn | l = \size {\oint \limits_{\gamma_R} \frac 1 {z \cdot \map p z} \rd z}       | o = \le       | r = 2 \pi R \max \limits_{\size z \mathop = R} \paren {\frac 1 {\size {z \cdot \map p z} } }       | c = Estimation Lemma }} {{eqn | r = 2 \pi \max \limits_{\size z \mathop = R} \paren {\frac 1 {\size {\map p z} } } }} {{end-eqn}} But this goes to zero for $R \to \infty$. We arrive at a contradiction. Hence the assumption that $\map p z \ne 0$ for all $z \in \C$ must be wrong. {{qed}}
We examine each condition for equivalence. === Reflexivity === For any function $f: X \to Y$, define $H: X \times \closedint 0 1 \to Y$ by $\map H {x, t} := \map f x$. This yields a homotopy between $f$ and itself. Also, trivially, if $x \in K$ and $t \in \closedint 0 1$, then: :$\map f x = H \left({x, t}\right)$ so that $H$ is a homotopy relative to $K$. Thus $\sim$ is a reflexive relation. {{qed|lemma}} === Symmetry === Given a $K$-relative homotopy: :$H: X \times \closedint 0 1 \to Y$  from $\map f x = \map H {x, 0}$ to $\map g x = \map H {x, 1}$, the function: :$\map G {x, t} = \map H {x, 1 - t}$ is a $K$-relative homotopy from $g$ to $f$. Thus $\sim$ is a symmetric relation. {{qed|lemma}} === Transitivity === Suppose that $f \sim g$ and $g \sim h$. Let $F, G: X \times \closedint 0 1 \to Y$ be $K$-relative homotopies between $f$ and $g$, $g$ and $h$, respectively. Define $H: X \times \closedint 0 1 \to Y$ by: :$\map H {x, t} := \begin {cases} \map F {x, 2 t} & : 0 \le t \le \dfrac 1 2 \\ \map G {x, 2 t - 1} & : \dfrac 1 2 \le t \le 1 \end{cases}$ By Continuous Mapping on Finite Union of Closed Sets, $H$ is a $K$-relative homotopy between $f$ and $h$. Thus $\sim$ is a transitive relation. {{qed|lemma}} Having verified all three conditions, it follows that $\sim$ is an equivalence relation. {{qed}} Category:Homotopy Theory k4bkjkcd1kcgumsnpf4mlv4xrsm2po4
Let $\phi: X \to Y$ be a homeomorphism. We must show that: :$(1): \quad$ If $c: \left[{0 \,.\,.\, 1}\right]^n \to X$ is a continuous mapping, then $ \phi \circ c: \left[{0 \,.\,.\, 1}\right]^n \to Y$ is also continuous :$(2): \quad$ If $c, d: \left[{0 \,.\,.\, 1}\right]^n \to X$ are freely homotopic, then $\phi \circ c, \phi \circ d: \left[{0 \,.\,.\, 1}\right]^n \to Y$ are also freely homotopic :$(3): \quad$ If $c, d: \left[{0 \,.\,.\, 1}\right]^n \to X$ are not freely homotopic, there can be no free homotopy between $\phi \circ c$ and $\phi \circ d$ :$(4): \quad$ The image of the concatenation of two maps, $\phi \left({c * d}\right)$, is freely homotopic to the concatenation of the images, $\phi \left({c}\right) * \phi \left({d}\right)$. {{finish}} Category:Homotopy Theory Category:Algebraic Topology k56fmuxw5p3gs43xabyg5nndpghuet7
By definition of local maximum: :$(1): \quad \map f {\xi + \epsilon} - \map f \xi < 0$ for sufficiently small $\epsilon \in \R_{>0}$. Similarly, definition of local minimum: :$(2): \quad \map f {\xi + \epsilon} - \map f \xi > 0$ for sufficiently small $\epsilon \in \R_{>0}$. Let it be assumed that $\map f {\xi + \epsilon}$ can be expanded, using Taylor's Theorem, in positive integer powers of $\epsilon$ Then: :$(3): \quad \map f {\xi + \epsilon} - \map f \xi = \epsilon \map {f'} \xi + \dfrac {\epsilon^2 \map {f''} \xi} 2 + \map \OO {\epsilon^3}$ where $\map \OO {\epsilon^3}$ denotes big-$\OO$ notation. That is, the quantity $\dfrac {\map \OO {\epsilon^3} } {\epsilon^3}$ is bounded. From $(1)$ and $(2)$, at a local maximum or local minimum, the sign of $\map f {\xi + \epsilon} - \map f \xi$ is independent the sign of $\epsilon$ itself. So from $(3)$ it follows that $\map {f'} = 0$. {{qed}}
From the definition of limit of a function: :$\forall \epsilon > 0: \exists \delta > 0: 0 < \left|{x - \xi}\right| < \delta \implies \left|{f \left({x}\right) - l}\right| < \epsilon$ Let $l > 0$. Since this is true for all $\epsilon > 0$, it is also true for $\epsilon = l$. So let the value of $\delta$, for the above to be true, labelled $h$. Then: : $0 < \left|{x - \xi}\right| < h \implies \left|{f \left({x}\right) - l}\right| < l$ That is: : $\xi - h < x < \xi + h, x \ne \xi \implies 0 = l - l < f \left({x}\right) < l + l = 2 l$ Hence: : $\forall x: \xi - h < x < \xi + h, x \ne \xi: 0 < f \left({x}\right)$ Now let $l < 0$, and consider $\epsilon = -l$. A similar thread of reasoning leads us to: : $\xi - h < x < \xi + h, x \ne \xi \implies -2 l < f \left({x}\right) < 0$ and hence the second result. {{qed}}
Since $X$ is path-connected there exists a path $f$ connecting $y$ and $x$. We define $\phi_f: \pi_1 \left({X, x}\right) \to \pi_1 \left({X, y}\right)$ by $\phi_f \left({ \left[{g}\right] }\right) =\left[{ f^{-1} g f }\right]$. $\phi$ is a homomorphism of groups, as is seen from: :$\displaystyle \phi_f \left({ \left[{g h}\right] }\right) = \left[{ f^{-1} g h f }\right] = \left[{ f^{-1} g f f^{-1} h f }\right] = \left[{ f^{-1} g f }\right] \left[{ f^{-1} h f }\right] = \phi_f \left({ \left[{g}\right] }\right) \phi_f \left({ \left[{h}\right] }\right)$  Also, $f^{-1}$ is a path from $x$ to $y$. Then by the same argument as before, $\phi_{f^{-1}}: \pi_1 \left({X, y}\right) \to \pi_1 \left({X, x}\right)$ where $\phi_{f^{-1}} \left({ \left[{g}\right] }\right) = \left[{ f g f^{-1} }\right]$, is a homomorphism of groups. Trivially, $\phi_f \circ \phi_{f^{-1}} = I_{\pi_1 \left({X, y}\right)}$ and $\phi_{f^{-1}} \circ \phi_f=I_{\pi_1 \left({X, x}\right)}$. Then $\phi_f^{-1} = \phi_{f^{-1}}$, so $\phi_f$ is bijective and thus an isomorphism. {{qed}} Category:Algebraic Topology 9vnhgyxsw38bkybqgtelfb0suywyrkj
We have that $f$ is continuous on $\closedint a b$. It follows from Continuous Image of Closed Interval is Closed Interval that $f$ attains: :a maximum $M$ at some $\xi_1 \in \closedint a b$ and: :a minimum $m$ at some $\xi_2 \in \closedint a b$. Suppose $\xi_1$ and $\xi_2$ are both end points of $\closedint a b$. Because $\map f a = \map f b$ it follows that $m = M$ and so $f$ is constant on $\closedint a b$. Then, by Derivative of Constant, $\map {f'} \xi = 0$ for all $\xi \in \openint a b$. Suppose $\xi_1$ is not an end point of $\closedint a b$. Then $\xi_1 \in \openint a b$ and $f$ has a local maximum at $\xi_1$. Hence the result follows from Derivative at Maximum or Minimum. Similarly, suppose $\xi_2$ is not an end point of $\closedint a b$. Then $\xi_2 \in \openint a b$ and $f$ has a local minimum at $\xi_2$. Hence the result follows from Derivative at Maximum or Minimum. {{qed}}
From Continuous Real Function is Darboux Integrable, $f$ is Darboux integrable on $\closedint a b$. By the Extreme Value Theorem, there exist $m, M \in \closedint a b$ such that: :$\displaystyle \map f m = \min_{x \mathop \in \closedint a b} \map f x$ :$\displaystyle \map f M = \max_{x \mathop \in \closedint a b} \map f x$ Then, from Upper and Lower Bounds of Integral: :$\displaystyle \map f m \paren {b - a} \le \int_a^b \map f x \rd x \le \map f M \paren {b - a}$  Dividing all terms by $\paren {b - a}$ gives: :$\displaystyle \map f m \le \frac 1 {b - a}\int_a^b \map f x \rd x \le \map f M$ By the Intermediate Value Theorem, there exists some $k \in \openint a b$ such that: {{begin-eqn}} {{eqn | l = \frac 1 {b - a} \int_a^b \map f x \rd x       | r = \map f k       | c =  }} {{eqn | ll= \leadsto       | l = \int_a^b \map f x \rd x       | r = \map f k \paren {b - a}       | c =  }} {{end-eqn}} {{qed}}
From Continuous Real Function is Darboux Integrable, $f$ is Darboux integrable on $\closedint a b$. Let $F : \closedint a b \to \R$ be a real function defined by:  :$\displaystyle \map F x = \int_a^x \map f x \rd x$ We are assured that this function is well-defined, since $f$ is integrable on $\closedint a b$. From Fundamental Theorem of Calculus: First Part, we have:  : $F$ is continuous on $\closedint a b$ : $F$ is differentiable on $\openint a b$ with derivative $f$ By the Mean Value Theorem, there therefore exists $k \in \openint a b$ such that:  :$\map {F'} k = \dfrac {\map F b - \map F a} {b - a}$ As $F$ is differentiable on $\openint a b$ with derivative $f$:  :$\map {F'} k = \map f k$ We therefore have:  {{begin-eqn}} {{eqn	| l = \map f k 	| r = \frac {\map F b - \map F a} {b - a} }} {{eqn	| r = \frac 1 {b - a} \paren {\int_a^b \map f x \rd x - \int_a^a \map f x \rd x} }} {{eqn	| r = \frac 1 {b - a} \int_a^b \map f x \rd x 	| c = Definite Integral on Zero Interval }} {{end-eqn}} giving:  :$\displaystyle \int_a^b \map f x \rd x = \paren {b - a} \map f k$ as required. {{qed}}
Let $g : \closedint a b \to \R$ be a real function with:  :$\map g x = x$ for all $x \in \closedint a b$.  By Power Rule for Derivatives, we have:  :$g$ is differentiable with $\map {g'} x = 1$ for all $x \in \closedint a b$. Note that in particular:  :$\map {g'} x \ne 0$ for all $x \in \openint a b$. Since $f$ is continuous on $\closedint a b$ and differentiable on $\openint a b$, we can apply the Cauchy Mean Value Theorem. We therefore have that there exists $\xi \in \openint a b$ such that:  :$\dfrac {\map {f'} \xi} {\map {g'} \xi } = \dfrac {\map f b - \map f a} {\map g b - \map g a}$ Note that:  :$\map {g'} \xi = 1$ and:  :$\map g b - \map g a = b - a$ so this can be rewritten:  :$\map {f'} \xi = \dfrac {\map f b - \map f a} {b - a}$ {{qed}}
Let $y \in \left[{a \,.\,.\, b}\right]$. Then $f$ satisfies the conditions of the Mean Value Theorem on $\left[{a \,.\,.\, y}\right]$. Hence: :$\exists \xi \in \left({a \,.\,.\, y}\right): f' \left({\xi}\right) = \dfrac {f \left({y}\right) - f \left({a}\right)} {y - a}$ But: :$f' \left({\xi}\right) = 0$ which means: :$f \left({y}\right) - f \left({a}\right) = 0$ and hence: :$f \left({y}\right) = f \left({a}\right)$ As $y$ is any $y \in \left[{a \,.\,.\, b}\right]$, the result follows. {{qed}}
Let $\norm {\, \cdot \,}$ be a non-trivial norm on the rational numbers $\Q$. === Archimedean Norm Case === Let $\norm {\, \cdot \,}$ be an Archimedean norm. {{:Ostrowski's Theorem/Archimedean Norm}}{{qed|lemma}} === Non-Archimedean Norm Case === Let $\norm {\, \cdot \,}$ be a non-Archimedean Norm. {{:Ostrowski's Theorem/Non-Archimedean Norm}}{{qed}}
Let $F$ be the real function defined on $\left[{a \,.\,.\, b}\right]$ by $F \left({x}\right) = f \left({x}\right) + h g \left({x}\right)$, where $h \in \R$ is a constant. The plan is to choose the constant $h$ such that $F \left({a}\right) = F \left({b}\right)$ and so apply Rolle's Theorem. We need to make: :$f \left({a}\right) + h g \left({a}\right) = f \left({b}\right) + h g \left({b}\right)$ We have that: :$\forall x \in \left({a \,.\,.\, b}\right): g' \left({x}\right) \ne 0$ So, by Rolle's Theorem: :$g \left({a}\right) \ne g \left({b}\right)$ Thus we can let: :$h = - \dfrac {f \left({b}\right) - f \left({a}\right)} {g \left({b}\right) - g \left({a}\right)}$ So, by Rolle's Theorem: :$\exists \xi \in \left({a \,.\,.\, b}\right): 0 = F' \left({\xi}\right) = f' \left({\xi}\right) + h g' \left({\xi}\right)$ That is: : $h = - \dfrac {f' \left({\xi}\right)} {g' \left({\xi}\right)}$ Hence the result, from the definition of the derivative. {{qed}}
Let $\epsilon > 0$. Let $\delta = \epsilon$. From the definition of a limit of a function, we need to show that $\left\vert{f \left({x}\right) - 0}\right\vert < \epsilon$ provided that $0 < \left\vert{x - \xi}\right\vert < \delta$, where $f \left({x}\right) = \left\vert{x - \xi}\right\vert$. Thus, provided $0 < \left\vert{x - \xi}\right\vert < \delta$, we have: {{begin-eqn}} {{eqn | l=\left\vert{x - \xi}\right\vert - 0       | r=\left\vert{x - \xi}\right\vert       | c= }} {{eqn | o=<       | r=\delta       | c= }} {{eqn | r=\epsilon       | c= }} {{end-eqn}} Hence the result. {{qed}}
Note that $\size {\map f x - \xi} \le \size {\xi - x}$. From Limit of Absolute Value we have that $\size {x - \xi} \to 0$ as $x \to \xi$. The result follows from the Squeeze Theorem. {{qed}}
From Strictly Monotone Mapping with Totally Ordered Domain is Injective, $f$ is an injection. From Surjection by Restriction of Codomain, $f: I \to J$ is a surjection. {{qed}}
The function $f$ is a bijection from Strictly Monotone Real Function is Bijective. Hence from Bijection iff Inverse is Bijection, $f^{-1}$ always exists and is also a bijection. From the definition of strictly increasing: :$x < y \iff \map f x < \map f y$ Hence: :$\map {f^{-1} } x < \map {f^{-1} } y \iff \map {f^{-1} } {\map f x} < \map {f^{-1} } {\map f y}$ and so: :$\map {f^{-1} } x < \map {f^{-1} } y \iff x < y$ Similarly, from the definition of strictly decreasing: :$x < y \iff \map f x > \map f y$ Hence: :$\map {f^{-1} } x < \map {f^{-1} } y \iff \map {f^{-1} } {\map f x} > \map {f^{-1} } {\map f y}$ and so: :$\map {f^{-1} } x < \map {f^{-1} } y \iff x > y$ {{qed}}
=== Part 1 === By the Whitney Immersion Theorem, there exists an immersion of $M$ into $\R^7$.   Suppose $\exists M'$ such that $M'$ embeds in $\R^6$ and that $M'$ and $M$ are cobordant. By Images of Smooth Embeddings bound Oriented Manifolds, $M'$ must bound a $5$-manifold $W'$. {{explain|The above redlink was originally labelled as "a proof due to R. Thom". This appears to be {{AuthorRef|René Frédéric Thom}}.}} The union of the cobordism and $W'$ are necessarily a $5$-manifold $W$ which satisfy the theorem. Hence it suffices to show that for any smooth, orientable $4$-manifold, there exists a similar manifold which is cobordant to the original and embeds in $\R^6$. {{finish}} === Part 2 === {{ProofWanted}} {{Namedfor|Vladimir Abramovich Rokhlin|cat = Rokhlin}} Category:Topology ae6hp92f8o2b4nad37ji4iepc5gzpnc