hcju/proofwikiqa huggingface dataset:
Let $m \in T$ be a smallest element of $T$. By definition: :$\forall y \in T: m \preceq y$ But as $T \subseteq S$, it follows that $m \in S$. Hence: :$\exists m \in S: \forall y \in T: m \preceq y$ Thus $T$ is bounded below by the lower bound $m$. {{qed}}
Let $M \in T$ be a greatest element of $T$. By definition: :$\forall y \in T: y \preceq M$ But as $T \subseteq S$, it follows that $M \in S$. Hence: :$\exists M \in S: \forall y \in T: y \preceq M$ Thus $T$ is bounded above by the upper bound $M$. {{qed}}
Suppose $T_* \cap T = \O$, where $\O$ denotes the empty set. That means $T$ contains none of its lower bounds, if indeed it has any. From Smallest Element is Lower Bound, if $T$ had a smallest element, it would be a lower bound contained in $T$. It follows that $T$ can have no smallest element. Otherwise $T_* \cap T \ne \O$. That means $T$ contains at least one of its lower bounds. Suppose $\exists a, b \in T_* \cap T$. From Intersection is Subset it follows that $a, b \in T$. Then: :$\forall y \in T: a \preceq y$ :$\forall y \in T: b \preceq y$ Thus both $a$ and $b$ fulfil the criteria for being a smallest element of $T$. From Smallest Element is Unique it follows that $a = b$ and so $T_* \cap T$ is a singleton containing the smallest element of $T$. {{qed}}
Suppose $T^* \cap T = \varnothing$, where $\varnothing$ denotes the empty set. That means $T$ contains none of its upper bounds, if indeed it has any. From Greatest Element is Upper Bound, if $T$ had a greatest element, it would be an upper bound contained in $T$. It follows that $T$ can have no greatest element. Otherwise $T^* \cap T \ne \varnothing$. That means $T$ contains at least one of its upper bounds. Suppose $\exists a, b \in T^* \cap T$. From Intersection is Subset it follows that $a, b \in T$. Then: :$\forall y \in T: y \preceq a$ :$\forall y \in T: y \preceq b$ Thus both $a$ and $b$ fulfil the criteria for being a greatest element of $T$. From Greatest Element is Unique it follows that $a = b$ and so $T^* \cap T$ is a singleton containing the greatest element of $T$. {{qed}}
{{begin-eqn}} {{eqn | l = \frac a {b + c} + \frac b {a + c} + \frac c {a + b}       | o = \ge       | r = \dfrac 3 2 }} {{eqn | ll= \leadstoandfrom       | l = \frac {a + b + c} {b + c} + \frac {a + b + c} {a + c} + \frac {a + b + c} {a + b}       | o = \ge       | r = \frac 9 2       | c = by adding 3 }} {{eqn | ll= \leadstoandfrom       | l = \frac {a + b + c} {b + c} + \frac {a + b + c} {a + c} + \frac {a + b + c} {a + b}       | o = \ge       | r = \frac {9 \paren {a + b + c} } {\paren {b + c} + \paren {a + c} + \paren {a + b} }       | c = as $\dfrac {a + b + c} {\paren {b + c} + \paren {a + c} + \paren {a + b} } = \dfrac 1 2$ }} {{eqn | ll= \leadstoandfrom       | l = \frac {\frac 1 {b + c} + \frac 1 {a + c} + \frac 1 {a + b} } 3       | o = \ge       | r = \frac 3 {\paren {b + c} + \paren {a + c} + \paren {a + b} }       | c = dividing by $3 \paren {a + b + c}$ }} {{end-eqn}} These are the arithmetic mean and the harmonic mean of $\dfrac 1 {b + c}$, $\dfrac 1 {a + c}$ and $\dfrac 1 {a + b}$. From Arithmetic Mean is Never Less than Harmonic Mean the last inequality is true. Thus Nesbitt's Inequality holds. {{qed}} {{Namedfor|A.M. Nesbitt|cat = Nesbitt, A.M.}}
By definition of greatest element: :$\forall y \in S: y \preceq M$ Suppose $M \preceq y$. As $\preceq$ is an ordering, $\preceq$ is by definition antisymmetric. Thus it follows by definition of antisymmetry that $M = y$. That is: :$M \preceq y \implies M = y$ which is precisely the definition of a maximal element. {{qed}}
By definition of smallest element: :$\forall y \in S: m \preceq y$ Suppose $y \preceq m$. As $\preceq$ is an ordering, $\preceq$ is by definition antisymmetric. Thus it follows by definition of  antisymmetry that $m = y$. That is: :$y \preceq m \implies m = y$ which is precisely the definition of a minimal element. {{qed}}
By definition of minimal element: :$\forall y \in S: y \preceq m \implies m = y$ As $\struct {S, \preceq}$ is a totally ordered set, by definition $\preceq$ is connected. That is: :$\forall x, y \in S: y \preceq x \lor x \preceq y$ It follows that: :$\forall y \in S: m = y \lor m \preceq y$ But as $m = y \implies m \preceq y$ by definition of $\preceq$, it follows that: :$\forall y \in S: m \preceq y$ which is precisely the definition of smallest element. Hence $(1)$ holds. {{qed|lemma}} Suppose $m_1$ and $m_2$ are both minimal elements of $S$. By $(1)$ it follows that both are smallest elements. It follows from Smallest Element is Unique that $m_1 = m_2$. That is, $(2)$ holds. {{qed}}
By definition of maximal element: :$\forall y \in S: M \preceq y \implies M = y$ As $\struct {S, \preceq}$ is a totally ordered set, by definition $\preceq$ is a connected. That is: :$\forall x, y \in S: y \preceq x \lor x \preceq y$ It follows that: :$\forall y \in S: M = y \lor y \preceq M$ But as $M = y \implies y \preceq M$ by definition of $\preceq$, it follows that: :$\forall y \in S: y \preceq M$ which is precisely the definition of greatest element. Hence $(1)$ holds. {{qed|lemma}} Suppose $M_1$ and $M_2$ are both maximal elements of $S$. By $(1)$ it follows that both are greatest elements. It follows from Greatest Element is Unique that $M_1 = M_2$. That is, $(2)$ holds. {{qed}}
The fact that $\struct {S, \max}$ is a semigroup is demonstrated in Max Operation on Toset is Semigroup. Then the max operation is idempotent: :$\forall x \in S: \max \set {x, x} = x$ The result follows by the definition of idempotent semigroup. {{qed}}
The Max Semigroup is Commutative and idempotent.  Hence the result, by definition of a semilattice. {{qed}}
As $\struct {S, \preceq}$ be a totally ordered set, all elements of $S$ are comparable by $\preceq$. Therefore there are two cases to consider: === Case 1: $x \preceq y$ === In this case: :$\max \set {x, y} = y = \sup \set {x, y}$ === Case 2: $y \preceq x$ === In this case: :$\max \set {x, y} = x = \sup \set {x, y}$ In either case, the result holds. {{qed}}
=== Proof of $(1)$ === Let $\epsilon > 0$. To verify the convergence, it is necessary to find a finite $F \subseteq I$ such that: :$\displaystyle d \left({\sum_{i \mathop\in G} z_i + w_i, z + w}\right) < \epsilon$ for all finite $G$ with $F \subseteq G \subseteq I$ Now let $F_z, F_w \subseteq I$ be finite subsets of $I$ such that: :$\displaystyle d \left({\sum_{i \mathop\in G} z_i, z}\right) < \frac \epsilon 2$ for all finite $G$ with $F_z \subseteq G \subseteq I$ :$\displaystyle d \left({\sum_{i \mathop\in G} w_i, w}\right) < \frac \epsilon 2$ for all finite $G$ with $F_w \subseteq G \subseteq I$ The set $F_z \cup F_w$ will be the sought $F$. Let $G$ be finite such that $F_z \cup F_w \subseteq G \subseteq I$. It follows that: {{begin-eqn}} {{eqn | l = d \left({\sum_{i \mathop\in G} z_i + w_i, z + w}\right)       | r = \left\vert{\left({\sum_{i \mathop\in G} z_i + w_i}\right) - \left({z + w}\right)}\right\vert       | c = {{Defof|Metric Induced by Norm}} }} {{eqn | r = \left\vert{\left({\sum_{i \mathop\in G} z_i - z}\right) + \left({\sum_{i \mathop\in G} w_i - w}\right)}\right\vert }} {{eqn | o = \le       | r = \left\vert{\left({\sum_{i \mathop\in G} z_i - z}\right)}\right\vert + \left\vert{\left({\sum_{i \mathop\in G} w_i, w}\right)}\right\vert       | c = Triangle Inequality for $\left\vert{\cdot}\right\vert$ }} {{eqn | r = d \left({\sum_{i \mathop\in G} z_i, z}\right) + d \left({\sum_{i \mathop\in G} w_i, w}\right)       | c = {{Defof|Metric Induced by Norm}} }} {{eqn | o = <       | r = \epsilon       | c = $F_z, F_w \subseteq G$ }} {{end-eqn}} From the definition of convergence, conclude that $\displaystyle \sum \left\{{ z_i + w_i: i \in I }\right\} = z+w$. {{qed|lemma}} === Proof of $(2)$ === Let $\epsilon > 0$. To verify the convergence, it is necessary to find a finite $F \subseteq I$ such that: :$\displaystyle d \left({\sum_{i \mathop\in G} \lambda z_i, \lambda z}\right) < \epsilon$ for all finite $G$ with $F \subseteq G \subseteq I$ Now let $F_z \subseteq I$ be a finite subset of $I$ such that: :$\displaystyle d \left({\sum_{i \mathop\in G} z_i, z}\right) < \frac \epsilon {\left\vert{\lambda}\right\vert}$ for all finite $G$ with $F_z \subseteq G \subseteq I$ The set $F_z$ will be the sought $F$. Let $G$ be finite such that $F_z \subseteq G \subseteq I$. It follows that: {{begin-eqn}} {{eqn | l = d \left({\sum_{i \mathop \in G} \lambda z_i, \lambda z}\right)       | r = \left\vert{\left({\sum_{i \mathop\in G} \lambda z_i}\right) - \lambda z}\right\vert       | c = {{Defof|Metric Induced by Norm}} }} {{eqn | r = \left\vert{\lambda \left({\sum_{i \mathop \in G} z_i - z}\right)}\right\vert }} {{eqn | r = \left\vert{\lambda}\right\vert \left\vert{\sum_{i \mathop \in G} z_i - z}\right\vert       | c = Multiplicativity for $\left\vert{\cdot}\right\vert$ }} {{eqn | r = \left\vert{\lambda}\right\vert d \left({\sum_{i \mathop \in G} z_i, z}\right)       | c = {{Defof|Metric Induced by Norm}} }} {{eqn | o = <       | r = \epsilon       | c = $F_z \subseteq G$ }} {{end-eqn}} From the definition of convergence, conclude that $\displaystyle \sum \left\{{ \lambda z_i: i \in I }\right\} = \lambda z$. {{qed}}
As $\left({S, \preceq}\right)$ be a totally ordered set, all elements of $S$ are comparable by $\preceq$. Therefore there are two cases to consider: === Case 1: $x \preceq y$ === In this case: : $\min \left({x, y}\right) = x = \inf \left({\left\{{x, y}\right\}}\right)$ === Case 2: $y \preceq x$ === In this case: : $\min \left({x, y}\right) = y = \inf \left({\left\{{x, y}\right\}}\right)$ In either case, the result holds. {{qed}}
=== $(2)$ implies $(1)$ === By Generalized Sum is Linear, the stated convergences lead to: {{begin-eqn}} {{eqn | l = z       | r = \operatorname{Re} z + i \operatorname{Im} z       | c = Definition of Real and Imaginary Parts }} {{eqn | r = \sum \left\{ {\operatorname{Re} z_i : i \in I}\right\} + i \sum \left\{ {\operatorname{Im} z_i : i \in I}\right\}       | c = Statement $(2)$ }} {{eqn | r = \sum \left\{ {\operatorname{Re} z_j + i \operatorname{Im} z_j: j \in I}\right\}       | c = Generalized Sum is Linear }} {{eqn | r = \sum \left\{ {z_i : i \in I}\right\}       | c = Definition of Real and Imaginary Parts }} {{end-eqn}} {{qed|lemma}} === $(1)$ implies $(2)$ === Statement $(1)$, according to the definition of convergence, amounts to the following: For every $\epsilon > 0$, there exists a finite $G \subseteq I$ such that: :For every finite $F \subseteq I$ with $G \subseteq F$: ::$\displaystyle \left\vert{z - \sum_{i \mathop \in F} z_i}\right\vert < \epsilon$ Now suppose that for $\epsilon > 0$, $G$ and $F$ are as above. Then observe that: {{begin-eqn}} {{eqn | l = \epsilon^2       | o = >       | r = \left\vert{z - \sum_{i \mathop \in F} z_i}\right\vert^2 }} {{eqn | r = \left({\operatorname{Re} z - \sum_{i \mathop \in F} \operatorname{Re} z_i}\right)^2 + \left({\operatorname{Im} z - \sum_{i \mathop \in F} \operatorname{Im} z_i}\right)^2       | c = Definition of Modulus }} {{end-eqn}} Hence, by Square of Real Number is Non-Negative, both of the terms on the right hand side are smaller than $\epsilon^2$. It follows that, taking square roots, $G$ satisfies, for any finite $F \supseteq G$: :$\displaystyle \left\vert{\operatorname{Re} z - \sum_{i \mathop \in F} \operatorname{Re} z_i}\right\vert < \epsilon$ :$\displaystyle \left\vert{\operatorname{Im} z - \sum_{i \mathop \in F} \operatorname{Im} z_i}\right\vert < \epsilon$ As $\epsilon > 0$ was arbitrary, using the definition of convergence, this implies precisely that: :$\displaystyle \sum \left\{{\operatorname{Re} z_i : i \in I}\right\}, \sum \left\{{\operatorname{Im} z_i : i \in I}\right\}$ converge to $\operatorname{Re} z, \operatorname{Im} z \in \R$, respectively. Hence, $(1)$ is shown to imply $(2)$. {{qed}}
Our assertion is that $\equiv$ cannot be defined in terms of $\mathsf B$. Seeking a contradiction, assume that it can. Call this assumption $\left({A}\right)$. If $\left({A}\right)$ holds, it must hold in all systems. Let one such system be $\left({\R^2, \mathsf B_1, \equiv_1}\right)$ where: :$\R^2$ is the cartesian product of the set of real numbers with itself :$\mathsf B_1$ is a ternary relation of betweenness :$\equiv_1$ is a quaternary relation of equidistance Let $\mathcal G$ be the discipline preceding the given discipline, where $\mathcal G$ is as defined above (excluding both $\equiv$ and $\mathsf B$). {{questionable|if $\mathcal G$ isn't strong enough to create $\R^2$, how can we use it? There is something incorrect in my presentation, particularly since we're going to use $\cdot$ and $\le$ -- GFP.}} Define $\mathsf B_1$ as follows: {{:Definition:Between (Geometry)}} Define $\equiv_1$ as follows: {{:Definition:Equidistance}} Now, define the isomorphism $\phi$ on $\left({\R^2, \mathsf B_2, \equiv_2}\right)$ as: :$\phi: \R^2 \to \R^2$ on $\left({\R^2, \mathsf B_1, \equiv_1}\right), \left({x_1, x_2}\right) \mapsto \left({x_1, 2 x_2}\right)$ {{finish|prove that $\phi$ is an isomorphism}} Now consider the system: :$\left({\R^2, \mathsf B_2, \equiv_2}\right)$ where $\mathsf B_2$ and $\equiv_2$ are the relations defined as above, but on the elements in the images of $\mathsf B_1$ and $\equiv_1$, respectively. Observe that $\mathsf B_1$ and $\mathsf B_2$ coincide, because in: :$\left({x_1 - y_1}\right) \cdot \left({2 y_2 - 2 z_2}\right) = \left({2 x_2 - 2 y_2}\right) \cdot \left({y_1 - z_1}\right) \land$ :$\left({0 \le \left({x_1 - y_1}\right) \cdot \left({y_1 - z_1}\right)}\right) \land \left({0 \le \left({2 x_2 - 2 y_2}\right) \cdot \left({2 y_2 - 2 z_2}\right)}\right)$ we can simply factor out the $2$ and divide both sides of the equality of inequality by $2$. But consider the elements: :$p_1 = \left({0, 0}\right)$ :$p_2 = \left({0, 1}\right)$ :$p_3 = \left({1, 0}\right)$ Observe that $p_1 p_2 \equiv_1 p_1 p_3$: :$\left({0 - 0}\right)^2 + \left({0 - 1}\right)^2 = \left({0 - 1}\right)^2 + \left({0 - 0}\right)^2$ But $\neg \left({p_1 p_2 \equiv_2 p_1 p_3}\right)$: :$\left({0 - 0}\right)^2 + \left({0 - 2}\right)^2 \ne \left({0 - 1}\right)^2 + \left({0 - 0}\right)^2$ But both $\left({\R^2, \mathsf B_1, \equiv_1}\right)$ and $\left({\R^2, \mathsf B_2, \equiv_2}\right)$ are both models of $\mathcal G$. {{explain|prove it}} Recall that if $\left({A}\right)$ holds, it must hold in all systems. But it does not. Hence $\left({A}\right)$ is false, from Proof by Contradiction. {{qed}} {{proofread}} {{help}}
Consider the sequence $\sequence {z_n}=$ defined by $z_n := y_n - x_n$. The Sum Rule for Real Sequences grants that $z_n \to m - l$ as $n \to \infty$. Furthermore, the assumption that $x_n \le y_n$ for all $n \in \N$ means that $z_n \ge 0$ for all $n \in \N$. Applying the main result to the sequence $\sequence {z_n}$ leads to the conclusion that $m - l \ge 0$. That is, $l \le m$. {{qed}} Category:Limits of Sequences o61xs1dfmo77y9rhq6olejsr2tjnnbe
This is an instance of Limit of Function by Convergent Sequences, as the reals form a metric space. {{qed}}
To show that the algebraic structure $\struct {S, \vee}$ is a semilattice, the following need to be verified: : Closure : Associativity : Commutativity : Idempotence In order: === Closure === By definition of an upper semilattice: :$\forall x, y \in S: \sup \set {x, y} \in S$ Since $x \vee y = \sup \set {x, y} \in S$ for all $x, y \in S$: :$\struct {S, \vee}$ is closed. {{qed|lemma}} === Associativity === Let $x, y, z \in S$. By definition of $\vee$: {{begin-eqn}} {{eqn | l = \paren {x \vee y} \vee z       | r = \sup \set {x, y} \vee z }} {{eqn | r = \sup \set {\sup \set {x, y}, z} }} {{eqn | r = \sup \set {x, y, z} }} {{eqn | r = \sup \set {x, \sup set {y, z} } }} {{eqn | r = x \vee \sup \set {y, z} }} {{eqn | r = x \vee \paren {y \vee z} }} {{end-eqn}} {{MissingLinks|The supremum equalities in the middle. Result might not yet exist on PW.}} Hence $\struct {S, \vee}$ is associative. {{qed|lemma}} === Commutativity === Let $x, y \in S$. By definition of $\vee$: {{begin-eqn}} {{eqn | l = x \vee y       | r = \sup \set {x, y} }} {{eqn | r = \sup \set {y, x}       | c = {{Defof|Supremum of Set}} }} {{eqn | r = y \vee x }} {{end-eqn}} Hence $\struct {S, \vee}$ is commutative. {{qed|lemma}} === Idempotence === For all $x \in S$: {{begin-eqn}} {{eqn | l = x \vee x       | r = \sup \set {x, x} }} {{eqn | r = x }} {{end-eqn}} Hence $\vee$ is idempotent. {{qed|lemma}} Having explicitly verified all prerequisites, it follows that $\struct {S, \vee}$ is a semilattice. {{qed}}
If $A = C$ and $B = D$, it is immediate that $A \times B = C \times D$. Now suppose that $A \times B = C \times D$. By definition of Cartesian product: {{begin-eqn}} {{eqn | l = x \in A, y \in B       | o = \leadstoandfrom       | r = \tuple {x, y} \in A \times B }} {{eqn | o = \leadstoandfrom       | r = \tuple {x, y} \in C \times D       | c = {{Defof|Set Equality}} }} {{eqn | o = \leadstoandfrom       | r = x \in C, y \in D }} {{end-eqn}} It follows that $x \in A$ {{iff}} $x \in C$, and so $A = C$. Similarly, $y \in B$ {{iff}} $y \in D$, hence $B = D$. The result follows. {{qed}}
=== Proof of $(1)$ === Let $\lambda \in \Bbb F$, $h \in H, k \in K$. Then: {{begin-eqn}} {{eqn|l = \left\langle{\left({\lambda A}\right)h, k}\right\rangle_K      |r = \lambda \left\langle{Ah, k}\right\rangle_K      |c = Property $(2)$ of inner products }} {{eqn|r = \lambda \left\langle{h, A^*k}\right\rangle_H      |c = Definition of adjoint }} {{eqn|r = \left\langle{h, \left({\overline \lambda A^*}\right) k}\right\rangle_H      |c = Properties $(1), (2)$ of inner products }} {{end-eqn}} Thus, by Existence and Uniqueness of Adjoint, $\left({\lambda A}\right)^* = \overline \lambda A^*$. {{qed|lemma}}
Let $h \in H, l \in L$. Then: {{begin-eqn}} {{eqn | l = \innerprod {\paren {AB} h} l_L       | r = \innerprod {B h} {A^* l}_K       | c = {{Defof|Adjoint Linear Transformation}} of $A$ }} {{eqn | r = \innerprod h {B^* A^* l}_H       | c = {{Defof|Adjoint Linear Transformation}} of $B$ }} {{end-eqn}} Thus, by Existence and Uniqueness of Adjoint: :$\paren {A B}^* = B^* A^*$ {{qed}}
Let $h \in H, k \in K$. Then: {{begin-eqn}} {{eqn|l = \left\langle{Ah, k}\right\rangle_K      |r = \left\langle{h, A^*k}\right\rangle_H      |c = Definition of adjoint of $A$ }} {{eqn|r = \left\langle{A^{**}h, k}\right\rangle_K      |c = Definition of adjoint of $A^*$ }} {{end-eqn}} Thus, by Existence and Uniqueness of Adjoint, $A^{**} = A$. {{qed}}
By definition of inverse, one has $AA^{-1} = I_K$, where $I_K$ is the identity operator on $K$. Now observe from Adjoint of Composition that: :$I_K = I_K^* = \left({AA^{-1}}\right)^* = \left({A^{-1}}\right)^*A^*$. Similarly, one has: :$I_H = I_H^* = \left({A^{-1}A}\right)^* = A^*\left({A^{-1}}\right)^*$ Hence, by definition of inverse, $\left({A^*}\right)^{-1} = \left({A^{-1}}\right)^*$. This also means that $A^*$ is invertible. {{qed}}
Let $a$ be a point with position $P_1$. Let $b$ be a point with position $P_2$. By hypothesis, $P_1 = P_2$. By Leibniz's Law, two objects are the same object {{iff}} they share every property in common. By the definition of point, the ''only'' property possessed by a point is position. We have: :$P_1 = P_2 \dashv \vdash a = b$ Hence the result. {{qed}}
=== Formulation 2 implies Formulation 1 === {{:Equivalence of Versions of Axiom of Choice/Formulation 2 implies Formulation 1}}{{qed|lemma}} === Formulation 1 implies Formulation 2 === The argument can be seen to be reversible. {{Handwaving}} {{qed|lemma}} === Formulation 1 implies Formulation 3 === {{:Equivalence of Versions of Axiom of Choice/Formulation 1 implies Formulation 3}}{{qed|lemma}} === Formulation 3 implies Formulation 1 === {{:Equivalence of Versions of Axiom of Choice/Formulation 3 implies Formulation 1}}{{qed}}
Let $h \in H$ such that $\norm h_H \le 1$. Then: {{begin-eqn}} {{eqn | l = \norm {A h}_K^2       | r = \innerprod {A h} {A h}_K       | c = {{Defof|Inner Product Norm}} }} {{eqn | r = \innerprod {A^* A h} h_H       | c = {{Defof|Adjoint Linear Transformation}} }} {{eqn | o = \le       | r = \norm {A^*A h}_H \norm h_H       | c = Cauchy-Bunyakovsky-Schwarz Inequality }} {{eqn | o = \le       | r = \norm {A^* A} \norm h_H^2       | c = Submultiplicativity of Operator Norm }} {{eqn | o = \le       | r = \norm {A^* A}       | c = Assumption on $h$ }} {{eqn | o = \le       | r = \norm {A^*} \norm A       | c = Submultiplicativity of Operator Norm }} {{end-eqn}} By definition $(1)$ for $\norm A$, it follows that: :$\norm A^2 \le \norm {A^* A} \le \norm {A^*} \norm A$ That is: :$\norm A \le \norm {A^*}$. By substituting $A^*$ for $A$, and using $A^{**} = A$ from Double Adjoint is Itself, the reverse inequality is obtained. Hence $\norm A^2 = \norm {A^* A} = \norm {A^*}^2$. {{qed}}
Let $g, h \in H$. Then by the definition of adjoint: :$\left\langle{Ag, Ah}\right\rangle_K = \left\langle{A^*Ag, h}\right\rangle_H$ From the uniqueness of the adjoint, it follows that: :$\left\langle{Ag, Ah}\right\rangle_K = \left\langle{g, h}\right\rangle_H$ holds iff $A^*A = I_H$. Hence the result by definition of isometry. {{qed}}
Let $x \in H$ be arbitrary. Then: {{begin-eqn}} {{eqn | l = x       | o = \in        | r = \map \ker A }} {{eqn | ll= \leadstoandfrom       | l = A x       | r = \mathbf 0_K       | c = {{Defof|Kernel of Linear Transformation}} }} {{eqn | ll= \leadstoandfrom       | l = A x        | o = \in        | r = K^\perp       | c = Linear Subspace Dense iff Zero Orthocomplement ($K$ is trivially dense) }} {{eqn | ll= \leadstoandfrom       | l = \innerprod {A x} K_K       | r = \set 0       | c = {{Defof|Orthocomplement}} }} {{eqn | ll= \leadstoandfrom       | l = \innerprod x {A^* K}_H       | r = \set 0       | c = {{Defof|Adjoint Linear Transformation}} }} {{eqn | ll= \leadstoandfrom       | l = \innerprod x {\Img {A^*} }_H       | r = \set 0       | c = {{Defof|Image of Mapping}} }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in        | r = \paren {\Img {A^*} }^\perp       | c = {{Defof|Orthocomplement}} }} {{end-eqn}} Hence by definition of set equality: :$\ker A = \paren {\Img {A^*} }^\perp$ {{qed}}
In $I - A$, $I$ is the identity operator on $H$. Hence: {{begin-eqn}} {{eqn|l = \left({I - A}\right)^2      |r = I^2 - IA - AI + A^2 }} {{eqn|r = I^2 - 2A + A^2      |c = $I$ is the identity operator }} {{eqn|r = I - A      |c = $I, A$ are idempotent }} {{end-eqn}} That is, $I - A$ is idempotent. {{qed}}
We have that the Natural Logarithm Function is Differentiable. The result follows from Differentiable Function is Continuous. {{qed}}
From Bounds of Natural Logarithm: :$\dfrac 1 2 < \map \ln 2 < 1$ Fix $x \in \R$. Consider $\dfrac x {\map \ln 2}$. From Rationals are Everywhere Dense in Topological Space of Reals: :$\forall \epsilon \in \R_{>0} \exists r \in \Q : \size {r - \dfrac x {\map \ln 2} } < \epsilon$ Thus: {{begin-eqn}} {{eqn | l = \size {r - \dfrac x {\map \ln 2} }       | o = <       | r = \epsilon }} {{eqn | ll= \leadsto       | l = \map \ln 2 \size {r - \dfrac x {\map \ln 2} }       | r = \size {\map \ln {2^r} - x }       | c = Natural Logarithm of Rational Power }} {{eqn | o = <        | r = \epsilon \, \map \ln 2       | c = Real Number Ordering is Compatible with Multiplication }} {{eqn | o = <        | r = \epsilon       | c = as $\map \ln 2 < 1$ }} {{eqn | ll= \leadsto       | l = \size {\map \ln t - x}       | o = <       | r = \epsilon       | c = substituting $t = 2^r$ }} {{end-eqn}} Thus: :$\forall \epsilon \in \R_{>0}: \exists t \in \R_{>0}: \size {\map \ln t - x} < \epsilon$ Thus, the image of $\R_{>0}$ under $\ln$ is everywhere dense in $\R$. From Monotone Real Function with Everywhere Dense Image is Continuous, $\ln$ is continuous on $\R_{>0}$. Hence the result. {{qed}}