hcju/proofwikiqa huggingface dataset:
Let $x, y \in S_\alpha$ be distinct points. We want to find disjoint open sets $U, V \in \tau_\alpha$ containing $x$ and $y$ respectively. Since $f$ is injective the points $\map f x, \map f y \in S_\beta$ are distinct. By assumption $T_\beta$ is Hausdorff. Therefore we can choose disjoint open sets $U', V'$ of $T_\beta$ such that $\map f x \in U'$ and $\map f y \in V'$. Since $f$ is continuous, the sets $U = f^{-1} \sqbrk {U'}$ and $V = f^{-1} \sqbrk {V'}$ are open. Moreover we have $x \in U$ and $y \in V$. Finally we show that the sets $U$ and $V$ are disjoint. Suppose $z \in U \cap V$. Then by the definition of $U$, $V$ we have $\map f z \in U'$ and $\map f z \in V'$. This is a contradiction, since $U' \cap V' = \O$. {{Qed}}
=== Definition 1 iff Definition 2 === Let $f$ and $f^{-1}$ both be continuous. As $f$ is continuous, then by definition: : $V \in \tau_\beta \implies f^{-1} \left({V}\right) \in \tau_\alpha$ and as $f^{-1}$ is continuous, then by definition: : $U \in \tau_\alpha \implies \left({f^{-1}}\right)^{-1} \left({U}\right) = f \left({U}\right) \in \tau_\beta$ That is: : $\forall U \subseteq S_\alpha: U \in \tau_\alpha \iff f \left({U}\right) \in \tau_\beta$ Conversely, suppose that: : $\forall U \subseteq S_\alpha: U \in \tau_\alpha \iff f \left({U}\right) \in \tau_\beta$ By definition: : $U \in \tau_\alpha \implies f \left({U}\right) \in \tau_\beta$ iff $f^{-1}$ is continuous. Also by definition: : $f \left({U}\right) \in \tau_\beta \implies U = f^{-1} \left({f \left({U}\right)}\right) \in \tau_\alpha$ iff $f$ is continuous. That is, both $f$ and $f^{-1}$ are continuous. {{qed|lemma}} === Definition 1 iff Definition 3 === Let $f$ and $f^{-1}$ both be continuous. Then by Bijection is Open iff Inverse is Continuous, $f$ is open. Thus $f$ is both open and continuous. Conversely, let $f$ be both open and continuous. Then also by Bijection is Open iff Inverse is Continuous, $f^{-1}$ is continuous. Thus both $f$ and $f^{-1}$ are continuous. {{qed|lemma}} === Definition 3 iff Definition 4 === Let $f$ be both open and continuous. Then by Bijection is Open iff Closed it follows that $f$ is both closed and continuous. Conversely, let $f$ be both both closed and continuous. Then also by Bijection is Open iff Closed it follows that $f$ is both open and continuous. {{qed}} Category:Homeomorphisms tpsbuo6gn0rvywf6lq00ny7v7gw4urr
Consider the real number line with the Euclidean topology $\struct {\R, \tau}$. By Real Number Line satisfies all Separation Axioms, $\struct {\R, \tau}$ is a Hausdorff space. By Quotient Space of Real Line may not be Kolmogorov, there is a relation $\RR$ on $\R$ such that the quotient space $\struct {\R / \RR, \tau_\RR}$ is not a Kolmogorov space. Thus the theorem holds by Sequence of Implications of Separation Axioms. {{qed}}
Proof by induction: For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$\displaystyle \prod_{i \mathop = 1}^n T_i$ is compact {{iff}} all of $T_1, T_2, \ldots, T_n$ are compact === Basis for the Induction === $\map P 1$ is the case: :$T_1$ is compact {{iff}} $T_1$ is compact which is trivially true. Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\displaystyle \prod_{i \mathop = 1}^k T_i$ is compact {{iff}} all of $T_1, T_2, \ldots, T_k$ are compact from which it is to be shown that: :$\displaystyle \prod_{i \mathop = 1}^{k + 1} T_i$ is compact {{iff}} all of $T_1, T_2, \ldots, T_{k + 1}$ are compact === Induction Step === This is the induction step: We have: :$\displaystyle \prod_{i \mathop = 1}^{k + 1} T_i = \paren {\prod_{i \mathop = 1}^k T_i} \times T_{k + 1}$ Hence: {{begin-eqn}} {{eqn | l = \prod_{i \mathop = 1}^{k + 1} T_i       | o = \text {is}       | r = \text {compact} }} {{eqn | ll = \leadstoandfrom       | l = \prod_{i \mathop = 1}^k T_i, T_{k + 1}       | o = \text {are}       | r = \text {compact}       | c = Topological Product of Compact Spaces }} {{eqn | ll = \leadstoandfrom       | l = T_1, T_2, \ldots, T_k, T_{k + 1}       | o = \text {are}       | r = \text {compact}       | c = Induction Hypothesis }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and thus it follows by the Principle of Mathematical Induction that for any $n \ge 1$: :$\displaystyle \prod_{i \mathop = 1}^n T_i$ is compact {{iff}} all of $T_1, T_2, \ldots, T_n$ are compact {{qed}}
Let $M_1 = \left({A_1, d_1}\right)$ be any metric space which is not compact. Let $I_{M_1}: M_1 \to M_1$ be the identity mapping. From Identity Mapping is Uniformly Continuous, $I_{M_1}$ is uniformly continuous on $M_1$. Hence the result. {{qed}}
Let $f$ be uniformly continuous on $M_1$. Let $x \in M_1$. Let $\epsilon > 0$. As $f$ is uniformly continuous, $\exists \delta > 0$ such that: :$\forall y \in M_1: d_1 \left({x, y}\right) < \delta: d_2 \left({f \left({x}\right), f \left({y}\right)}\right) < \epsilon$ Thus by definition $f$ is continuous at $x$. {{qed}} Category:Metric Spaces Category:Continuous Mappings on Metric Spaces Category:Uniformly Continuous Mappings qdpnhtcjgrywzny4gfvtlykqmhh0bj9
From Real Number Line is Metric Space, $\R$ under the Euclidean metric is a metric space. The result follows by Uniformly Continuous Function is Continuous: Metric Space. {{qed}}
Let $x \in I$. Let $\epsilon \in \R_{>0}$. As $f$ is uniformly continuous:  :$\exists \delta \in \R_{>0}: \paren {x, y \in I, \size {x - y} < \delta \implies \size {\map f x - \map f y} < \epsilon}$ Then, for all $y \in I$ such that $\size {x - y} < \delta$: :$\size {\map f x - \map f y} < \epsilon$ Thus by definition $f$ is continuous at $x$. As $x$ was arbitrary, $f$ is continuous on all of $I$. {{qed}}
=== Necessary Condition === {{ProofWanted}} === Sufficient Condition === By Monotone Real Function is Darboux Integrable, $f$ and $f^{-1}$ are Darboux integrable. Let $b = \map f a$. Define: :$\displaystyle A = \int_0^a \map f u \rd u + \int_0^b \map {f^{-1} } v \rd v$ Consider any subdivision $P = \set {x_0, x_1, \ldots, x_n}$ of the closed real interval $\closedint 0 a$. Then: :$f P = \set {\map f {x_0}, \map f {x_1}, \ldots, \map f {x_n} }$ is a subdivision of $\closedint 0 b$. We have that: {{begin-eqn}} {{eqn | l = \map U {P, f} + \map L {f P, f^{-1} }       | r = \sum_{k \mathop = 1}^n \paren {x_k - x_{k - 1} } \map f {x_k} + \sum_{k \mathop = 1}^n \paren {\map f {x_k} - \map f {x_{k - 1} } } \map {f^{-1} } {\map f {x_{k - 1} } }       | c = because $f$ and $f^{-1}$ are increasing }} {{eqn | r = \sum_{k \mathop = 1}^n \paren {x_k \map f {x_k} - x_{k - 1} \map f {x_{k - 1} } }       | c = by Summation is Linear }} {{eqn | r = a b }} {{end-eqn}} Similarly: :$\map L {P, f} + \map U {f P, f^{-1} } = a b$ Let $\epsilon \in \R_{>0}$ be an arbitrary strictly positive real number. By the definition of the Darboux integral, there exist subdivisions $P_+$ and $P_-$ of $\closedint 0 a$ such that: :$a b = \map U {P_+, f} + \map L {f P_+, f^{-1} } < A + \epsilon$ :$a b = \map L {P_-, f} + \map U {f P_-, f^{-1} } > A - \epsilon$ The result follows from Real Plus Epsilon. {{qed}} {{Namedfor|William Henry Young}}
From Set with Relative Complement forms Partition: :$A = \complement_S \left({B}\right)$ and: :$B = \complement_S \left({A}\right)$ where $\complement_S$ denotes the complement relative to $S$. As $A$ and $B$ are both open, it follows by definition that $A$ and $B$ are also both closed. That is, by definition, they are clopen. {{qed}}
From Biconditional Equivalent to Biconditional of Negations it follows that the statement can be expressed as: :$T$ admits a separation {{iff}}: :there exist two closed sets of $T$ which form a (set) partition of $S$. By definition, a separation of $T$ is a (set) partition of $S$ by $A, B$ which are open in $T$. From Complements of Components of Two-Component Partition form Partition: :$A \mid B$ is a (set) partition of $S$ {{iff}} $\complement_S \left({A}\right) \mid \complement_S \left({B}\right)$ is a (set) partition of $S$. Hence the result by definition of closed set.
We proceed to verify the open set axioms for $\set \O$ to be a topology on $\O$. Let $\tau = \set \O$. === {{OpenSetAxiom|1}} === By Union of Empty Set: :$\displaystyle \bigcup \tau = \O \in \tau$ Thus open set axiom $\paren {\text O 1}$ is satisfied. {{qed|lemma}} === {{OpenSetAxiom|2}} === From Intersection with Empty Set: :$\O \cap \O = \O \in \tau$ and so open set axiom $\paren {\text O 2}$ is satisfied. {{qed|lemma}} === {{OpenSetAxiom|3}} === By definition $\O \in \tau$ and so open set axiom $\paren {\text O 3}$ is satisfied. {{qed}} Category:Topology Category:Empty Set gqv71ragp7zb18f9xx2eolvz3ta8dr1
=== Necessary Condition === Let $A \mid B$ be a partition of $S$. That is, by definition: :$(1): \quad A \cap B = \O$ :$(2): \quad A \cup B = S$ :$(3): \quad A, B \ne \O$ Thus: {{begin-eqn}} {{eqn | l = A \cap B        | r = \O       | c =  }} {{eqn | ll= \leadsto       | l = \relcomp S {A \cap B}       | r = S       | c = Relative Complement of Empty Set }} {{eqn | ll= \leadsto       | l = \relcomp S A \cup \relcomp S B       | r = S       | c = De Morgan's Laws: Complement of Intersection }} {{end-eqn}} {{begin-eqn}} {{eqn | l = A \cup B        | r = S       | c =  }} {{eqn | ll= \leadsto       | l = \relcomp S {A \cup B}       | r = \O       | c = Relative Complement with Self is Empty Set }} {{eqn | ll= \leadsto       | l = \relcomp S A \cap \relcomp S B       | r = \O       | c = De Morgan's Laws: Complement of Union }} {{end-eqn}} {{WLOG}}, now suppose that $\relcomp S A = \O$. {{begin-eqn}} {{eqn | l = \relcomp S A       | r = \O       | c =  }} {{eqn | ll= \leadsto       | l = A       | r = S       | c = Relative Complement of Empty Set }} {{eqn | ll= \leadsto       | l = B       | r = \O       | c = as $A \cap B = \O$ }} {{end-eqn}} Thus by Rule of Transposition: :$B \ne \O \implies \relcomp S A \ne \O$ Mutatis mutandis: :$A \ne \O \implies \relcomp S B \ne \O$ by hypothesis: :$A, B \ne \O$ and so: :$\relcomp S A, \relcomp S B \ne \O$ Thus all three conditions are satisfied for $\relcomp S A \mid \relcomp S B$ to be a partition of $S$. {{qed|lemma}} === Sufficient Condition === Let $\relcomp S A \mid \relcomp S B$ be a partition of $S$. By the necessary condition, it follows that $\relcomp S {\relcomp S A} \mid \relcomp S {\relcomp S B}$ is a partition of $S$. From Relative Complement of Relative Complement: :$\relcomp S {\relcomp S A} = A$ and $\relcomp S {\relcomp S B} = B$ and so $A \mid B$ is a partition of $S$. {{qed}} Category:Relative Complement nqeqddcvfi8knul9tva7cbh8tv0x4yp
Define for $k \in \N, k > n: A_k = X$. By axiom $(1)$ of a $\sigma$-algebra, it follows that $\forall k \in \N, k > n: A_k \in \Sigma$. From Sigma-Algebra Closed under Countable Intersection, it follows that $\displaystyle \bigcap_{k \mathop \in \N} A_k = \bigcap_{k \mathop = 1}^n A_k \in \Sigma$. {{qed}} Category:Sigma-Algebras 4cocq4ef5t089luotyuid8duajhg9f6
Let $H \subseteq S$ be a non-empty subset whose boundary $\partial H$ is empty. Thus: {{begin-eqn}} {{eqn | l = \partial H       | r = \varnothing       | c = by hypothesis }} {{eqn | ll= \leadsto       | l = H^- \cap \left({S \setminus H}\right)^-       | r = \varnothing       | c = Boundary is Intersection of Closure with Closure of Complement }} {{end-eqn}} From Topological Closure is Closed, both $H^-$ and $\left({S \setminus H}\right)^-$ are closed sets of $T$. From Union of Closure with Closure of Complement is Whole Space: :$H^- \cup \left({S \setminus H}\right)^- = S$ Thus $H^-$ and $\left({S \setminus H}\right)^-$ are two disjoint closed sets of $T$ whose union is $S$. Hence, by hypothesis, one of them must be empty. Suppose $H$ is not empty. It must therefore follow that: : $S \setminus H = \varnothing$ Therefore $H = S$. Thus the only subsets of $S$ whose boundary is empty are $S$ and $\varnothing$.
Let $H \subseteq S$ be a clopen set of $T$. From Set is Clopen iff Boundary is Empty, $H$ has an empty boundary. By hypothesis, $H = S$ or $H = \varnothing$. That is, the only clopen sets of $T$ are $S$ and $\varnothing$.
Suppose $A$ and $B$ are separated subsets of $T$ such that $A \cup B = S$. By definition of separated sets: :$A \cap B^- = \O$ Then: {{begin-eqn}} {{eqn | l = S       | r = A \cup B       | c =  }} {{eqn | o = \subseteq       | r = A \cup B^-       | c = Set is Subset of its Topological Closure }} {{eqn | o = \subseteq       | r = S       | c = by definition of $S$ }} {{end-eqn}} Hence $A = S \setminus B^-$. From Topological Closure is Closed, $B^-$ is closed in $T$. Thus $A$ is open in $T$. Also by definition of separated sets: :$A^- \cap B = \O$ Hence, by the same reasoning, $B$ must also be open. But: :$A \cap B \subseteq A \cap B^- = \O$ and $A \cup B = S$, by assumption. So: :$A = S \setminus B$ and $B = S \setminus A$ and we conclude that both $A$ and $B$ are clopen. Therefore, by hypothesis, one of them must be $S$ and the other must be $\O$. That is, there are no two non-empty separated sets of $T$ whose union is $S$.
Let $T = \struct {S, \tau}$ be a topological space such that there are no two non-empty separated sets whose union is $S$. Let $D = \struct {\set {0, 1}, \tau}$ be the discrete two-point space on $\set {0, 1}$. {{AimForCont}} $f: T \to \set {0, 1}$ is a continuous surjection. By definition of continuous mapping: :$\map {f^{-1} } 0$ and $\map {f^{-1} } 1$ are open sets of $T$. From the definition of a mapping: :$\map {f^{-1} } 0 \cup \map {f^{-1} } 1 = S$ and :$\map {f^{-1} } 0 \cap \map {f^{-1} } 1 = \O$ Then: :$\map {f^{-1} } 0 = S \setminus \map {f^{-1} } 1$ and: :$\map {f^{-1} } 1 = T \setminus \map {f^{-1} } 0$ are clopen. From Closed Set equals its Closure they are their respective closures. It follows from the definition that $\map {f^{-1} } 0$ and $\map {f^{-1} } 1$ are separated subsets of $T$ whose union is $S$. Hence, by hypothesis, one of them must be empty, and the other one must be $S$. Therefore $f$ is constant, and so is not a surjection. This contradicts the original hypothesis. That is, there exists no continuous surjection from $T$ onto a discrete two-point space.
Let $T = \left({S, \tau}\right)$ be a topological space such that there exists no continuous surjection from $T$ onto a discrete two-point space. Let $D = \left({\left\{{0, 1}\right\}, \tau}\right)$ be the discrete two-point space on $\left\{{0, 1}\right\}$. Let $A$ and $B$ be disjoint open sets of $T$ such that $A \cup B = S$. The aim is to show that one of them is empty. Let us define the mapping $f: S \to \left\{{0, 1}\right\}$ by: :$f \left({x}\right) = \begin{cases} 0 & : x \in A \\ 1 & : x \in B \end{cases}$ There are only four open sets in $\left\{{0, 1}\right\}$, namely: : $\varnothing$ : $\left\{{0}\right\}$ : $\left\{{1}\right\}$ : $\left\{{0, 1}\right\}$ We have that: :$f^{-1} \left[{\varnothing}\right] = \varnothing$ :$f^{-1} \left[{\left\{{0}\right\}}\right] = A$ :$f^{-1} \left[{\left\{{1}\right\}}\right] = B$ :$f^{-1} \left[{\left\{{0, 1}\right\}}\right] = S$ where $f^{-1} \left[{X}\right]$ denotes the preimage of the set $X$. But by hypothesis all of $\varnothing, A, B, S$ are open sets of $T$. So by definition $f$ is continuous. Also by hypothesis, $f$ cannot be surjective. It follows that $f$ must be constant. So either $A$ or $B$ must be empty, and the other one must be $S$. Hence the result.
Let $H_f$ be a set defined as follows: :$H_f = \set {\tuple {\sqbrk M, x}: \text {$M$ accepts $x$ in $\map f {\size x}$ steps} }$ where: :$M$ is a (deterministic) Turing machine :$x$ is its input (the initial contents of its tape) :$\sqbrk M$ denotes an input that encodes the Turing machine $M$ Let $m$ be the size of $\tuple {\sqbrk M, x}$. We know that we can decide membership of $H_f$ by way of a (deterministic) Turing machine that: :$(1): \quad$ calculates $f \left({\size x}\right)$ :$(2): \quad$ writes out a row of $0$s of that length :$(3): \quad$ uses this row of $0$s as a counter to simulate $M$ for at most that many steps. At each step, the simulating machine needs to look through the definition of $M$ to decide what the next action would be. It is safe to say that this takes at most $\map f m^3$ operations, so: :$ H_f \in \map {\mathsf{DTIME} } {\map f m^3}$ {{handwaving|"We know that ...", "It is safe to say ..."}} {{AimForCont}}: :$H_f \in \map {\mathsf{DTIME} } {\map f {\floor {\dfrac m 2} } }$ Then we can construct some machine $K$ which: :given some machine description $\sqbrk {M_K}$ and input $x$ :decides within $\map {\mathsf{DTIME} } {\map f {\floor {\dfrac m 2} } }$ whether $\tuple {\sqbrk {M_K}, x} \in H_f$. Construct another machine $N$ which: :takes a machine description $\sqbrk {M_N}$ :runs $K$ on $\left({ \sqbrk {M_N}, \sqbrk {M_N} }\right)$ :accepts only if $K$ rejects, and rejects if $K$ accepts. Let $m_n$ be the length of $\sqbrk {M_N}$. Then $m$ (the length of the input to $K$) is twice $m_n$ plus some delimiter symbol, so: :$m = 2m_n + 1$ $N$'s running time is thus: {{begin-eqn}} {{eqn | l = \map {\mathsf{DTIME} } {\map f {\floor {\frac m 2} } }       | r = \map {\mathsf{DTIME} } {\map f {\floor {\frac {2 m_n + 1} 2} } }       | c =  }} {{eqn | r = \map {\mathsf{DTIME} } {\map f {m_n} }       | c =  }} {{end-eqn}} Now consider the case $M_N = N$. That is we feed $\sqbrk N$ as input into $N$ itself). In this case $m_n$ is the length of $\sqbrk N$. * If $N$ '''accepts''' $\sqbrk N$ (which we know it does in at most $\map f {m_n}$ operations): ** By the definition of $N$, $K$ '''rejects''' $\tuple {\sqbrk N, \sqbrk N}$ ** Therefore, by the definition of $K$, $\tuple {\sqbrk N, \sqbrk N} \notin H_f$ ** Therefore, by the definition of $H_f$, $N$ does not accept $\sqbrk N$ in $\map f {m_n}$ steps -- a contradiction. * If $N$ '''rejects''' $\sqbrk N$ (which we know it does in at most $\map f {m_n}$ operations): ** By the definition of $N$, $K$ '''accepts''' $\tuple {\sqbrk N, \sqbrk N}$ ** Therefore, by the definition of $K$, $\tuple {\sqbrk N, \sqbrk N} \in H_f$ ** Therefore, by the definition of $H_f$, $N$ '''does''' accept $\sqbrk N$ in $\map f {m_n}$ steps -- a contradiction. Therefore, $K$ does not exist. So, by Proof by Contradiction: :$H_f \notin \map {\mathsf{DTIME} } {\map f {\floor {\dfrac m 2} } }$ Substituting $2 n + 1$ for $m$, we get: :$H_f \notin \map {\mathsf{DTIME} } {\map f n}$ and, from the earlier result: :$H_f \in \map {\mathsf{DTIME} } {\map f {2 n + 1}^3}$ {{qed}} Category:Complexity Theory Category:Computer Science Category:Named Theorems 23hffzr83e7hv17hpn8s4dsqf0rrlhl
Let $B$ be an object of $\mathbf C$. Let $\epsilon_B: B^A \times A \to B$ be the evaluation morphism at $B$. Then, since: {{begin-eqn}} {{eqn|l = \operatorname{id}_B \epsilon_B      |r = \epsilon_B }} {{eqn|r = \epsilon_B \operatorname{id}_{B^A \times A} }} {{eqn|r = \epsilon_B \left({\operatorname{id}_{B^A} \times \operatorname{id}_A}\right) }} {{end-eqn}} it follows that $\left({\operatorname{id}_B}\right)^A = \operatorname{id}_{B^A}$. Next, let $f: B \to C, g: C \to D$ be composable morphisms. Then: {{begin-eqn}} {{eqn|l = g f \epsilon_B      |r = g \epsilon_C \left({f^A \times \operatorname{id}_A}\right)      |c = Definition of $f^A$ }} {{eqn|r = \epsilon_D \left({g^A \times \operatorname{id}_A}\right) \left({f^A \times \operatorname{id}_A}\right)      |c = Definition of $g^A$ }} {{eqn|r = \epsilon_D \left({g^A f^A \times \operatorname{id}_A}\right) }} {{end-eqn}} Thus, it is seen that $g^A f^A$ is the morphism satisfying the UMP for $\left({g f}\right)^A$. That is to say: :$\left({g f}\right)^A = g^A f^A$ Hence $\left({-}\right)^A$ is a functor. {{qed}} {{expand|diagram}}
Recall the definition of the cosine function: {{begin-eqn}} {{eqn | l = \cos z       | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n} } {\paren {2 n}!}       | c =  }} {{eqn | r = 1 - \frac {z^2} {2!} + \frac {z^4} {4!} - \frac {z^6} {6!} + \cdots + \paren {-1}^n \frac {z^{2 n} } {\paren {2 n}!} + \cdots       | c =  }} {{end-eqn}} Recall the definition of the exponential as a power series: {{begin-eqn}} {{eqn | l = \exp z       | r = \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}       | c =  }} {{eqn | r = 1 + \frac z {1!} + \frac {z^2} {2!} + \frac {z^3} {3!} + \cdots + \frac {z^n} {n!} + \cdots       | c =  }} {{end-eqn}} Then, starting from the {{RHS}}: {{begin-eqn}} {{eqn | l = \frac {\exp \paren {i z} + \exp \paren {-i z} } 2       | r = \frac 1 2 \paren {\sum_{n \mathop = 0}^\infty \frac {\paren {i z}^n} {n!} + \sum_{n \mathop = 0}^\infty \frac {\paren {-i z}^n} {n!} } }} {{eqn | r = \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i z}^n + \paren {-i z}^n} {n!} }       | c = Cosine Function is Absolutely Convergent }} {{eqn | r = \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i z}^{2 n} + \paren {-i z}^{2 n} } {\paren {2 n}!} + \frac {\paren {i z}^{2 n + 1} + \paren {-i z}^{2 n + 1} } {\paren {2 n + 1}!} }       | c = split into even and odd $n$ }} {{eqn | r = \frac 1 2 \sum_{n \mathop = 0}^\infty \frac {\paren {i z}^{2 n} + \paren {-i z}^{2 n} } {\paren {2 n}!}       | c = $\paren {-i z}^{2 n + 1} = -\paren {i z}^{2 n + 1}$ }} {{eqn | r = \frac 1 2 \sum_{n \mathop = 0}^\infty \frac {2 \paren {i z}^{2 n} } {\paren {2 n}!}       | c = $\left({ -1 }\right)^{2n} = 1$ }} {{eqn | r = \sum_{n \mathop = 0}^\infty \frac {\paren {i z}^{2 n} } {\paren {2 n}!}       | c = cancel $2$ }} {{eqn | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n} } {\paren {2 n}!}       | c = $i^{2 n} = \paren {-1}^n$ }} {{eqn | r = \cos z }} {{end-eqn}} {{qed}}
Recall Euler's Formula: :$\exp \paren {i z} = \cos z + i \sin z$ Then, starting from the {{RHS}}: {{begin-eqn}} {{eqn | l = \frac {\exp \paren {i z} + \exp \paren {-i z} } 2       | r = \frac {\cos z + i \sin z + \cos \paren {-z} + i \sin \paren {-z} } 2 }} {{eqn | r = \frac {\cos z + \cos \paren {-z} } 2       | c = Sine Function is Odd }} {{eqn | r = \frac {2 \cos z} 2       | c = Cosine Function is Even }} {{eqn | r = \cos z }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | n = 1       | l = \exp \paren {i z}       | r = \cos z + i \sin z       | c = Euler's Formula }} {{eqn | n = 2       | l = \exp \paren {-i z}       | r = \cos z - i \sin z       | c = Euler's Formula: Corollary }} {{eqn | ll= \leadsto       | l = \exp \paren {i z} + \exp \paren {-i z}       | r = \paren {\cos z + i \sin z} + \paren {\cos z - i \sin z}       | c = $(1) + (2)$ }} {{eqn | r = 2 \cos z       | c = simplifying }} {{eqn | ll= \leadsto       | l = \frac {\exp \paren {i z} + \exp \paren {-i z} } 2       | r = \cos z       | c =  }} {{end-eqn}} {{qed}}
We start by proving the special case where $\forall x: g \left({x}\right) = 0$ and $L=0$, in which case $\displaystyle \lim_{x \to a} \ h \left({x}\right) = 0$. Let $\epsilon > 0$ be a positive real number. Then by the definition of the limit of a function: : $\exists \delta > 0: 0 < \left|{x - a}\right| < \delta \implies \left|{h \left({x}\right)}\right| < \epsilon$ Now: : $\forall x \ne a: 0 = g \left({x}\right) \le f \left({x}\right) \le h \left({x}\right)$ so that: :$\left|{f \left({x}\right)}\right| \le \left|{h \left({x}\right)}\right|$ Thus: : $0 < |x-a| < \delta \implies \left|{f \left({x}\right)}\right| \le \left|{h \left({x}\right)}\right| < \epsilon$ By the transitive property of $\le$, this proves that: : $\displaystyle \lim_{x \to a} \ f \left({x}\right) = 0 = L$ We now move on to the general case, with $g \left({x}\right)$ and $L$ arbitrary. For $x \ne a$, we have: : $g \left({x}\right) \le f \left({x}\right) \le h \left({x}\right)$ By subtracting $g \left({x}\right)$ from all expressions, we have: : $0 \le f \left({x}\right) - g \left({x}\right) \le h \left({x}\right) - g \left({x}\right)$ Since as $x \to a, h \left({x}\right) \to L$ and $g \left({x}\right) \to L$, we have: : $h \left({x}\right) - g \left({x}\right) \to L - L = 0$ From the special case, we now have: : $f \left({x}\right) - g \left({x}\right) \to 0$ We conclude that: : $f \left({x}\right) = \left({f \left({x}\right) - g \left({x}\right)}\right) + g \left({x}\right) \to 0 + L = L$ {{qed}}