hcju/proofwikiqa huggingface dataset:
Let $f, g, h$ be real functions defined on an open interval $\left({a \,.\,.\, b}\right)$, except possibly at the point $c \in \left({a \,.\,.\, b}\right)$. Let: : $\displaystyle \lim_{x \mathop \to c} \ g \left({x}\right) = L$ : $\displaystyle \lim_{x \mathop \to c} \ h \left({x}\right) = L$ : $g \left({x}\right) \le f \left({x}\right) \le h \left({x}\right)$ except perhaps at $x = c$. Let $\left \langle {x_n} \right \rangle$ be a sequence of points of $\left({a \,.\,.\, b}\right)$ such that: : $\forall n \in \N_{>0}: x_n \ne c$ and: : $\displaystyle \lim_{n \mathop \to \infty} \ x_n = c$ By Limit of Function by Convergent Sequences: : $\displaystyle \lim_{n \mathop \to \infty} \ g \left({x_n}\right) = L$ and: : $\displaystyle \lim_{n \mathop \to \infty} \ h \left({x_n}\right) = L$ Since: : $g \left({x_n}\right) \le f \left({x_n}\right) \le h \left({x_n}\right)$ it follows from the Squeeze Theorem for Real Sequences that: : $\displaystyle \lim_{n \mathop \to \infty} \ f \left({x_n}\right) = L$ The result follows from Limit of Function by Convergent Sequences. {{qed}}
By the definition of the limit of a real function, we have to prove that: : $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \left({\left|{x - a}\right| < \delta \implies \left|{f \left({x}\right) - L}\right| < \epsilon}\right)$ Let $\epsilon \in \R_{>0}$ be given. We have: : $\displaystyle \lim_{x \mathop \to a} \ g \left({x}\right) = \lim_{x \mathop \to a} \ h \left({x}\right)$ Hence by Sum Rule for Limits of Functions: :$\displaystyle \lim_{x \mathop \to a} \ h \left({x}\right) - g \left({x}\right) = 0$ By the definition of the limit of a real function: : $(1): \quad \forall \epsilon' \in \R_{>0}: \exists \delta \in \R_{>0}: \left({\left|{x - a}\right| < \delta \implies \left|{h \left({x}\right) - L}\right| < \epsilon'}\right)$ : $(2): \quad \forall \epsilon' \in \R_{>0}: \exists \delta \in \R_{>0}: \left({\left|{x - a}\right| < \delta \implies \left|{g \left({x}\right) - L}\right| < \epsilon'}\right)$ : $(3): \quad \forall \epsilon' \in \R_{>0}: \exists \delta \in \R_{>0}: \left({\left|{x - a}\right| < \delta \implies \left|{h \left({x}\right) - g \left({x}\right)}\right| < \epsilon'}\right)$ Take $\epsilon' = \dfrac {\epsilon} 3$ in $(1)$, $(2)$, $(3)$. Then there exists $\delta_1, \delta_2, \delta_3$ that satisfies $(1)$, $(2)$, $(3)$ with $\epsilon' = \dfrac \epsilon 3$.  Take $\delta = \min\{\delta_1, \delta_2, \delta_3\}$. Then: {{begin-eqn}} {{eqn | l = \left\vert{x - a}\right\vert       | o = <       | r = \delta       | c =  }} {{eqn | ll= \implies       | l = \left\vert{h \left({x}\right) - L}\right\vert       | o = <       | r = \frac {\epsilon} 3       | c = }} {{eqn | lo= \land       | l = \left\vert{g \left({x}\right) - L}\right\vert       | o = <       | r = \frac {\epsilon} 3       | c =  }} {{eqn | lo= \land       | l = \left\vert{h \left({x}\right) - g \left({x}\right)}\right\vert       | o = <       | r = \frac {\epsilon} 3       | c =  }} {{end-eqn}} So, if $\left\vert{x - a}\right\vert<\delta$: {{begin-eqn}} {{eqn | l = \left\vert{f \left({x}\right) - L}\right\vert       | r = \left\vert{f \left({x}\right) - g \left({x}\right) + h \left({x}\right) - L + g \left({x}\right) - h \left({x}\right)}\right\vert       | c =  }} {{eqn | o = \le       | r = \left\vert{f \left({x}\right) - g \left({x}\right)}\right\vert + \left\vert{h \left({x}\right) - L}\right\vert + \left\vert{h \left({x}\right) - g \left({x}\right)}\right\vert       | c =  }} {{eqn | o = \le       | r = \left\vert{h \left({x}\right) - g \left({x}\right)}\right\vert + \left\vert{h \left({x}\right) - L}\right\vert + \left\vert{h \left({x}\right) - g \left({x}\right)}\right\vert       | c =  }} {{eqn | o = \le       | r = \frac {\epsilon} 3 + \frac {\epsilon} 3 + \frac {\epsilon} 3       | c =  }} {{eqn | r = \epsilon       | c =  }} {{end-eqn}} {{qed}}
Let $\alpha \in \R$ be an irrational number. By definition, $\alpha \notin \Q$. Consider the sets: :$S := \Q \cap \left({-\infty \,.\,.\, \alpha}\right)$ :$T := \Q \cap \left({\alpha \,.\,.\, \infty}\right)$ Let $x \in S$. Let $B_\epsilon \left({x}\right)$ be the open $\epsilon$-ball of $x$ in $\Q$. Then: :$\forall x \in S: \exists \epsilon \in \R_{>0}: B_\epsilon \left({x}\right) \subseteq S$ by setting $\epsilon = \alpha - x$. Similarly: :$\forall x \in T: \exists \epsilon \in \R_{>0}: B_\epsilon \left({x}\right) \subseteq T$ by setting $\epsilon = x - \alpha$. Thus $S$ and $T$ are open sets of $\Q$. Then: :$S \cup T = \Q$ :$S \cap T = \O$ :$S, T \ne \O$ So $S$ and $T$ fulfil the conditions for $S \mid T$ to be a separation of $\Q$. Hence the result by definition of connected topological space. {{qed}}
By Set is Subset of Itself, the result follows by setting $K = H^-$ in Set between Connected Set and Closure is Connected. {{qed}}
Since the open interval $\openint 0 \infty$ is connected, then so is $G$ by Continuous Image of Connected Space is Connected. It is enough, from Set between Connected Set and Closure is Connected, to show that $J \subseteq \map \cl G$. Let $p \in J$, say, $\tuple {0, y}$ where $-1 \le y \le 1$. We need to show that: :$\forall \epsilon > 0: \map {N_\epsilon} p \cap G \ne \O$ where $\map {N_\epsilon} p$ is the $\epsilon$-neighborhood of $p$. Let us choose $n \in \N: \dfrac 1 {2 n \pi} < \epsilon$. From Sine of Half-Integer Multiple of Pi: :$\map \sin {\dfrac {\paren {4 n + 1} \pi} 2} = 1$ and: :$\map \sin {\dfrac {\paren {4 n + 3} \pi} 2} = -1$ So by the Intermediate Value Theorem, $\map \sin {\dfrac 1 x}$ takes every value between $-1$ and $1$ in the closed interval $\closedint {\dfrac 2 {\paren {4 n + 3} \pi} } {\dfrac 2 {\paren {4 n + 1} \pi} }$. In particular, $\map \sin {\dfrac 1 {x_0} } = y$ for some $x_0$ in this interval. The distance between the points $\tuple {0, y}$ and $\tuple {x_0, \map \sin {\dfrac 1 {x_0} } } = \tuple {x_0, y}$ is $x_0 < \epsilon$. So: :$\tuple {x_0, \map \sin {\dfrac 1 {x_0} } } \in \map {N_\epsilon} p \cap G$ as required. {{qed}}
By hypothesis, $A$ and $B$ are separated: :$A \cap B^- = A^- \cap B = \O$ Then: {{begin-eqn}} {{eqn | l = H \cap B^-       | r = \paren {A \cup B} \cap B^- }} {{eqn | r = \paren {A \cap B^-} \cup \paren {B \cap B^-}       | c = Intersection Absorbs Union }} {{eqn | r = \O \cup B       | c = Set is Subset of its Topological Closure and Intersection with Subset is Subset }} {{eqn | r = B       | c = Union with Empty Set }} {{end-eqn}} Since the intersection of a closed set with a subspace is closed in the subspace, $B$ is closed in $H$. {{explain|Link to the above result}} Since $A = H \setminus B$ and $B$ is closed in $H$, $A$ is open in $H$. {{explain|$A {{=}} H \setminus B$: it may be trivial but all statements are linked to or explained.}} By the same argument with the roles of $A$ and $B$ reversed, $A$ is closed in $H$ and $B$ is open in $H$. Hence the result. {{qed}} Category:Separated Sets m1f02h1s1lcidd77jo0wfdkarkqzuml
It is to be shown that $\sim$ is reflexive, symmetric and transitive. === Reflexive === Let $\mathcal F \in \mathcal A$ be a $C^k$-atlas. That $\mathcal F$ is compatible with itself is precisely condition $(2)$ of the definition of $C^k$-atlas. {{qed|lemma}} === Symmetric === Let $\mathcal F$ and $\mathcal G$ be $C^k$-atlases and suppose that $\mathcal F \sim \mathcal G$. Let $\left({U, \phi}\right) \in \mathcal F$ and let $\left({V, \psi}\right) \in \mathcal G$ be charts. Then by hypothesis: :$\phi \circ \psi^{-1}: \psi \left({U \cap V}\right) \to \phi \left({U \cap V}\right)$ is a $C^k$ mapping. Because $\phi$ and $\psi$ are homeomorphisms, we have that: :$\psi \circ \phi^{-1}: \phi \left({U \cap V}\right) \to \psi \left({U \cap V}\right)$ is also a homeomorphism, and in particular continuous. By the Inverse Function Theorem, $\psi \circ \phi^{-1}$ is also a $C^k$ mapping. Since the charts were arbitrary, we conclude that $\mathcal G \sim \mathcal F$. {{qed|lemma}} === Transitive === Let $\mathcal F$, $\mathcal G$ and $\mathcal H$ be $C^k$-atlases, and suppose that $\mathcal F \sim \mathcal G$ and $\mathcal G \sim \mathcal H$. {{finish|That's messy; probably easiest to use partition of unity compatible with $\mathcal G$, or via common refinement}} {{explain|Clarification is needed as to why this result should be categorised in :Category:Manifolds.}} Category:Manifolds 3mb3byl8pdxtdl55kkak3fht2cf2sff
Assume that $w$ is wholly real. Then: {{begin-eqn}} {{eqn | l = w z       | r = \map \Re w \, \map \Re z - \map \Im w \, \map \Im z + i \paren {\map \Re w \, \map \Im z + \map \Im w \, \map \Re z}       | c = {{Defof|Complex Multiplication}} }} {{eqn | r = w \, \map \Re z + i w \, \map \Im z        | c = as $\map \Re w = w$ and $\map \Im w = 0$ }} {{end-eqn}} This equation shows that $\map \Re {w z} = w \, \map \Re z$, and $\map \Im {w z} = w \, \map \Im z$. This proves $(1)$. Now, assume that $w$ is wholly imaginary. Then: {{begin-eqn}} {{eqn | l = w z       | r = \map \Re w \, \map \Re z - \map \Im w \, \map \Im z + i \paren {\map \Re w \, \map \Im z + \map \Im w \, \map \Re z} }} {{eqn | r = -\map \Im w \, \map \Im z) + i \, \map \Im w \, \map \Re z       | c = as $\map \Re w = 0$ }} {{end-eqn}} This equation shows that $\map \Re {w z} = -\map \Im w \, \map \Im z$, and $\map \Im {w c} = \map \Im w \, \map \Re z$. This proves $(2)$. {{qed}} Category:Complex Multiplication 9aoaf7jbh49ghocleog40chvkz0qv08
Let $x, y \in \Q$ such that $x \ne y$. From Between two Rational Numbers exists Irrational Number, there exists $\alpha \in \R \setminus \Q$ such that $x < \alpha < y$. From Rational Numbers are not Connected, it follows that $x$ and $y$ belong to different components of $\Q$. As $x$ and $y$ are arbitrary, it follows that no rational number is in the same component as any other rational number. That is, the components of $\Q$ are singetons. Hence the result, by definition of a totally disconnected space. {{qed}}
Follows from: : Rational Number Space is Totally Separated : Totally Separated Space is Totally Disconnected {{qed}}
Follows from: * Discrete Space is Extremally Disconnected * Extremally Disconnected Space is Totally Separated * Totally Separated Space is Totally Disconnected {{qed}}
Follows from: :Rational Numbers are not Extremally Disconnected :Discrete Space is Extremally Disconnected. {{qed}}
The definition of extremally disconnected space can be stated as: :A $T_2$ (Hausdorff) topological space $T = \struct {S, \tau}$ is '''extremally disconnected''' {{iff}} the closure of every open set of $T$ is open. Hence the existence will be demonstrated of an open set in $\struct {\Q, \tau_d}$ whose closure is not open. First we establish that $\struct {\Q, \tau_d}$ is indeed a Hausdorff space. Indeed, we have: :Rational Numbers form Metric Space :Metric Space is Hausdorff We have that $\openint 0 1$ is open in $\struct {\R, \tau_d}$. Thus $\openint 0 1 \cap \Q$ is open in $\struct {\Q, \tau_d}$. We show that $0$ and $1$ are limit points of $\openint 0 1 \cap \Q$. For any $\epsilon \in \R{>0}$: :$\openint {-\epsilon} \epsilon \cap \Q \cap \openint 0 1 = \openint 0 \epsilon \cap \Q$ From Between two Real Numbers exists Rational Number: :$\openint 0 \epsilon \cap \Q \ne \O$ Similarly: :$\openint {1 - \epsilon} 1 \cap \Q \ne \O$ Thus $0$ and $1$ are limit points of $\openint 0 1 \cap \Q$. Hence: :$0, 1 \in \paren {\openint 0 1 \cap \Q}'$ Thus: :$\closedint 0 1 \cap \Q \subseteq \paren {\openint 0 1 \cap \Q}^-$ where $\paren {\openint 0 1 \cap \Q}^-$ denotes the closure of $\openint 0 1 \cap \Q$. Now we show that for any $x \notin \closedint 0 1$, $x$ is not a limit point of $\openint 0 1 \cap \Q$. Suppose $x < 0$. Then $\dfrac x 2 < 0$. Thus: :$\openint {x + \dfrac x 2} {x - \dfrac x 2} \cap \Q \cap \openint 0 1 = \O$ Similarly for $y > 1$: :$\openint {y - \dfrac {y - 1} 2} {y + \dfrac {y - 1} 2} \cap \Q \cap \openint 0 1 = \O$ Thus: :$\map \complement {\closedint 0 1 \cap \Q} \subseteq \map \complement {\paren {\openint 0 1 \cap \Q}^-}$ Hence we have: :$\closedint 0 1 \cap \Q = \paren {\openint 0 1 \cap \Q}^-$ From Between two Real Numbers exists Rational Number: :$\openint {-\epsilon} \epsilon \cap \Q \cap \map \complement {\closedint 0 1} = \openint {-\epsilon} 0 \cap \Q \ne \O$ A neighborhood of $0$ must intersect $\map \complement {\closedint 0 1}$. So $\closedint 0 1 \cap \Q$ is not open in $\struct {\Q, \tau_d}$. Hence $\openint 0 1$ is that open set in $\struct {\Q, \tau}$ whose closure is not open of which we were to demonstrate the existence. Thus the result follows from definition of extremally disconnected space. {{qed}} Category:Rational Numbers Category:Extremally Disconnected Spaces pj3e3go1uulkw6i5pzacjqxou6r42t6
Let $q$ be any point in $U$. The topology on the product space $X$ is the Tychonoff topology which has the natural basis as a (synthetic) basis. Then: :$\exists Q \in \BB : q \in Q \subseteq U$ where $\BB$ is the natural basis for Tychonoff topology on $X$. From Natural Basis of Tychonoff Topology: :$\displaystyle Q = \prod_{i \mathop \in I} Q_i$ where: ::for all $i \in I : Q_i$ is open in $X_i$ ::$J = \set {i \in I : Q_i \ne X_i}$ is finite Let $u$ be the point defined by: :$u_i = \begin {cases} q_i & i \in J \\ x_i & i \notin J \end {cases}$ By definition of $u$: :$\forall i \in I : u_i \in Q_i$ By definition of the Cartesian product: :$u \in Q \subseteq U$ By definition of $u$: :$\forall i \in I: i \notin J \implies u_i = x_i$ So by Rule of Transposition: :$\forall i \in I: u_i \ne x_i \implies i \in J$ Thus: :$\map K u \subseteq J$  From Subset of Finite Set is Finite, and because $J$ is finite, $\map K u$ is also finite. {{qed}} Category:Product Spaces 2b0la43c8y2wvbel7lntglyjkor0rw5
Let $z \in \C$, and let $\epsilon \in \R_{>0}$. Put $\delta = \epsilon$. For all $w \in \C$ with $\cmod {w - z} < \delta$: {{begin-eqn}} {{eqn | l = \cmod {\map \Re w - \map \Re z} + \cmod {\map \Im w - \map \Im z}       | r = \cmod {\map \Re w - \map \Re z} + \cmod {i \paren {\map \Im w - \map \Im z} }       | c = as $\cmod i = 1$ }} {{eqn | o = \le       | r = \cmod {\map \Re w - \map \Re z + i \map \Im w - i \map \Re z}       | c = Triangle Inequality for Complex Numbers }} {{eqn | r = \cmod {w - z} }} {{eqn | o = <       | r = \epsilon }} {{end-eqn}} {{Mistake|$\cmod {\map \Re w - \map \Re z} + \cmod {i \paren {\map \Im w - \map \Im z} } \le \cmod {\map \Re w - \map \Re z + i \map \Im w - i \map \Im z}$ is wrong -- triangle inequality works the other way}} This equation shows that $\cmod {\map x w - \map x z} < \epsilon$, and $\cmod {\map y w - \map y z} < \epsilon$. It follows by definition that $x$ and $y$ are both continuous. {{qed}}
Define the real function $x: \closedint a b \to \R$ by: :$\forall t \in \closedint a b : \map x t = \map \Re {\map f t}$ Define the real function $y: \closedint a b \to \R$ by: :$\forall t \in \closedint a b : \map y t = \map \Im {\map f t}$ where: :$\map \Re {\map f t}$ denotes the real part of the complex number $\map f t$ :$\map \Im {\map f t}$ denotes the imaginary part of $\map f t$. From Real and Imaginary Part Projections are Continuous, it follows that $\Re: \C \to \R$ and $\Im: \C \to \R$ are continuous functions. From Continuity of Composite Mapping, it follows that $x$ and $y$ are continuous. From Continuous Real Function is Darboux Integrable, it follows that $x$ and $y$ are Darboux integrable over $\closedint a b$. By definition, it follows that $f$ is complex Riemann integrable. {{qed}}
Let $C \subseteq M$ be compact. Let $\sequence {x_n}$ be a sequence in $C$. Let $S$ be the range of $\sequence {x_n}$. Thus $S \subseteq C$ and $S$ may be either finite or infinite. === Finite Range === Let $S$ be finite. Then at least one $x \in S$ must be repeated infinitely often in $\sequence {x_n}$. {{MissingLinks|Infinite Pigeonhole Principle}} Its occurrences form a subsequence converging to $x$. {{qed|lemma}} === Infinite Range === Let $S$ be infinite. From Subsequence of Sequence in Metric Space with Limit, it is enough to show that $S$ has a limit point in $C$. {{AimForCont}} $S$ has no limit point in $C$. Then for each $x \in C$, there exists $\epsilon \in \R_{>0}$ such that the open $\epsilon$-ball $\map {B_\epsilon} x$ contains no $y \in S: y \ne x$. That is: :$\map {B_\epsilon} x \cap S = \set x$ or :$\map {B_\epsilon} x \cap S = \O$ Because $C$ is compact, the open cover $\mathcal B := \set {\map {B_\epsilon} x: x \in C}$ has a finite subcover. But each $\map {B_\epsilon} x \in \mathcal B$ contains at most one point of $S$. Therefore the union of any finite subset of $\mathcal B$ contains only finitely many points of $S$. So no finite subset of $\mathcal B$ covers $S$, let alone $C$. Thus $S$ must have a limit point in $C$. {{qed}}
From Continuous Complex Function is Complex Riemann Integrable, it follows that all three complex Riemann integrals are well defined. From Real and Imaginary Part Projections are Continuous, it follows that $\operatorname{Re}: \C \to \R$ and $\operatorname{Im}: \C \to \R$ are continuous functions. {{explain|Revisit the above link -- see if there is a more appropriate one to use so as not to invoke the concept of metric spaces}} From Continuity of Composite Mapping, it follows that $\operatorname{Re} \circ f: \R \to \R$ and $\operatorname{Im} \circ f: \R \to \R$ are continuous real functions. Then: {{begin-eqn}} {{eqn | l = \int_a^b f \left({t}\right) \, \mathrm d t       | r = \int_a^b \operatorname{Re} \left({f \left({t}\right) }\right) \, \mathrm d t + i \int_a^b \operatorname{Im} \left({f \left({t}\right) }\right) \, \mathrm d t       | c = Definition of Complex Riemann Integral }} {{eqn | r = \int_a^c \operatorname{Re} \left({f \left({t}\right) }\right) \, \mathrm d t + \int_c^b \operatorname{Re} \left({f \left({t}\right) }\right) \, \mathrm d t + i \left({\int_a^c \operatorname{Im} \left({f \left({t}\right) }\right) \, \mathrm d t + \int_c^b \operatorname{Im} \left({f \left({t}\right) }\right) \, \mathrm d t}\right)       | c = Sum of Integrals on Adjacent Intervals for Continuous Functions }} {{eqn | r = \int_a^c \operatorname{Re} \left({f \left({t}\right) }\right) \, \mathrm d t + i \int_a^c \operatorname{Im} \left({f \left({t}\right) }\right) \, \mathrm d t + \int_c^b \operatorname{Re} \left({f \left({t}\right) }\right) \, \mathrm d t + i \int_c^b \operatorname{Im} \left({f \left({t}\right) }\right) \, \mathrm d t }} {{eqn | r = \int_a^c f \left({t}\right) \, \mathrm d t + \int_c^b f \left({t}\right) \, \mathrm d t }} {{end-eqn}} {{qed}} Category:Complex Analysis m40944pv7962zynma06z15sdjap8lli
=== $\epsilon$-$\delta$ Definition iff Definition by Limits === This is proved in Metric Space Continuity by Epsilon-Delta. {{qed|lemma}} === $\epsilon$-$\delta$ Definition iff $\epsilon$-Ball Definition === This is proved in Metric Space Continuity by Open Ball. {{qed}} === $\epsilon$-Ball Definition iff Definition by Neighborhood === This is proved in Metric Space Continuity by Neighborhood. {{qed}} Category:Continuity nq08azkwsibqfx0z1ch9d3jk75i17qu
=== $\epsilon$-$\delta$ Condition implies $\epsilon$-Ball Condition === Suppose that $f$ satisfies the $\epsilon$-$\delta$ condition: :$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: 0 < d_1 \left({x, c}\right) < \delta \implies d_2 \left({f \left({x}\right), L}\right) < \epsilon$ Let $y \in f \left({B_\delta \left({c; d_1}\right) \setminus \left\{{c}\right\}}\right)$. By definition of open $\epsilon$-ball, this means: :$\exists x \in A_1: 0 < d_1 \left({x, c}\right) < \delta: y = f \left({x}\right)$ By hypothesis, it follows that $d_2 \left({y, L}\right) < \epsilon$ That is, that $y \in B_\epsilon \left({L; d_2}\right)$. By definition of subset: :$f \left({B_\delta \left({c; d_1}\right) \setminus \left\{{c}\right\}}\right) \subseteq B_\epsilon \left({L; d_2}\right)$ Thus it follows that $f$ satisfies the $\epsilon$-ball condition. {{qed|lemma}} === $\epsilon$-Ball Condition implies $\epsilon$-$\delta$ Condition === Suppose that $f$ satisfies the $\epsilon$-ball condition: :$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: f \left({B_\delta \left({c; d_1}\right) \setminus \left\{{c}\right\}}\right) \subseteq B_\epsilon \left({L; d_2}\right)$. Let $0 < d_1 \left({x, c}\right) < \delta$. Then by definition of open $\epsilon$-ball, this means: :$x \in B_\delta \left({c; d_1}\right) \setminus \left\{{c}\right\}$ By hypothesis, it follows that: :$f \left({x}\right) \in B_\epsilon \left({L; d_2}\right)$ Thus by definition of open $\epsilon$-ball, this means: :$d_2 \left({f \left({x}\right), L}\right) < \epsilon$ That is, $f$ satisfies the $\epsilon$-$\delta$ condition. {{qed}}
By hypothesis, let $\epsilon \in \R_{>0}$ be a Lebesgue number for $M$. Then by definition: :$\forall x \in A: \exists U \left({x}\right) \in \mathcal U: B_\epsilon \left({x}\right) \subseteq U \left({x}\right)$ where $B_\epsilon \left({x}\right)$ is the open $\epsilon$-ball of $x$ in $M$. Let $\epsilon' \in \R_{>0}: \epsilon' < \epsilon$. Let $y \in B_{\epsilon'} \left({x}\right)$. {{begin-eqn}} {{eqn | l = d \left({y, x}\right)       | o = <       | r = \epsilon'       | c = {{Defof|Open Ball of Metric Space}} }} {{eqn | ll= \implies       | l = d \left({y, x}\right)       | o = <       | r = \epsilon       | c = as $\epsilon' < \epsilon$ }} {{eqn | ll= \implies       | l = y       | o = \in       | r = B_\epsilon \left({x}\right)       | c = {{Defof|Open Ball of Metric Space}} }} {{eqn | ll= \implies       | l = y       | o = \in       | r = U \left({x}\right)       | c = {{Defof|Subset}}: $B_\epsilon \left({x}\right) \subseteq U \left({x}\right)$ }} {{end-eqn}} That is: :$B_{\epsilon'} \left({x}\right) \subseteq U \left({x}\right)$ The result follows by definition of Lebesgue number. {{qed}}
Let $M := \left({0 \,.\,.\, 1}\right) \subseteq \R$ under the Euclidean metric. Let $\mathcal C := \left\{{\left({\dfrac 1 n \,.\,.\, 1}\right): n \ge 2}\right\}$. For any $\epsilon \in \R_{>0}$, take $n > \dfrac 1 \epsilon$ and $x = \dfrac 1 n$. There is no $\left({\dfrac 1 m \,.\,.\, 1}\right)$ such that $B_\epsilon \left({x}\right) \subseteq \left({\dfrac 1 m \,.\,.\, 1}\right)$, since $B_\epsilon \left({x}\right) = B_\epsilon \left({\dfrac 1 n}\right) = \left({0 \,.\,.\, \dfrac 1 n + \epsilon}\right)$. {{qed}}
Let $f_c: A_1 \to A_2$ be the constant mapping between two metric spaces $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$. Let $\epsilon \in \R: \epsilon > 0$. Pick any $\delta \in \R$ such that $\delta > 0$. Let $x, y \in A_1$ such that $d_1 \left({x, y}\right) < \delta$. Now we have: :$f_c \left({x}\right) = c = f_c \left({y}\right)$ Hence: :$d_2 \left({f_c \left({x}\right), f_c \left({y}\right)}\right) = d_2 \left({c, c}\right)$ By definition of a metric: :$d_2 \left({c, c}\right) = 0 < \epsilon$. Thus we have that: :$\forall \epsilon > 0: \exists \delta \in \R: d_1 \left({x, y}\right) < \delta \implies d_2 \left({f_c \left({x}\right), f_c \left({y}\right)}\right) < \epsilon$ Hence by definition of uniform continuity on a metric space, $f_c$ is uniformly continuous throughout $A_1$ with respect to $d_1$ and $d_2$. {{qed}} Category:Constant Mappings Category:Metric Spaces Category:Uniformly Continuous Mappings dwll4poxn4ev0ljeemh7zgf578t7aky
Follows directly from: :Constant Function is Uniformly Continuous: Metric Space :Real Number Line is Metric Space. {{qed}}