hcju/proofwikiqa huggingface dataset:
It needs to be demonstrated that $M$ is: :$(1): \quad$ a closed algebraic structure under $+$ :$(2): \quad$ closed for scalar product $\circ$. So let $a, b \in M$. By definition of intersection, $a, b \in M_i$ for all $i \in I$. As the $M_i$ are subspaces of $V$, $a + b \in M_i$ for all $i \in I$. That is, by definition of intersection, $a + b \in M$. It follows that $M$ is closed under $+$. Now let $\lambda \in K, a \in M$. By definition of intersection, $a \in M_i$ for all $i \in I$. As the $M_i$ are subspaces of $V$, $\lambda \circ a \in M_i$ for all $i \in I$. That is, by definition of intersection, $\lambda \circ a \in M$. It follows that $M$ is closed under $\circ$. Hence the result, by definition of subspace. {{qed}} Category:Vector Subspaces n61jxciytzxpoqcg6jss1ihrwb7gbgb
By Set of Linear Subspaces is Closed under Intersection, $M$ is a linear subspace of $V$. By Topology Defined by Closed Sets, the intersection of closed sets is again closed. As the $M_i$ are all closed, it follows that $M$ is closed. Hence $M$ is a closed linear subspace of $V$. {{qed}} Category:Vector Subspaces 4qr0useq9yg5nggqucqyd0u0twbljkt
By definition, $M \ominus N = M \cap N^\perp$. By Orthocomplement is Closed Linear Subspace, $N^\perp$ is a closed linear subspace of $H$. Hence the result, by Closed Linear Subspaces Closed under Intersection. {{qed}} Category:Hilbert Spaces 2lj3gg359w9cwzc1kxa4wbfnzjwhsak
By Linear Subspaces Closed under Setwise Addition, $M + N$ is a linear subspace of $H$. Now to show that it is closed. Let $P: H \to H$ denote the orthogonal projection on $M$. Denote by $I - P$ the complementary projection of $P$. Define $N' := \left\{{ n - Pn: n \in N}\right\}$. $N'$ is a closed linear subspace of $H$. {{explain|prove the above claim}} Observe $m + n = \left({m + Pn}\right) + \left({n - Pn}\right) \in M + N'$; hence, $M + N \subseteq M + N'$.  By $m + \left({n - Pn}\right) = \left({m - Pn}\right) + n \in M + N$, conclude that $M + N' \subseteq M + N$, hence equality. Furthermore, $N' \subseteq \operatorname{ran} \left({I - P}\right) = \operatorname{ker} P$ by Range of Idempotent is Kernel of Complementary Idempotent. That is, $N' \subseteq M^\perp$ by Properties of Orthogonal Projection, and hence $M \perp N'$. Denote by $P'$ the orthogonal projection on $N'$. Suppose now that $h \in M + N'$. Then: {{begin-eqn}} {{eqn | l = h       | r = \left({P + \left({I - P}\right)}\right)h }} {{eqn | r = \left({P + \left({I - P}\right)}\right) \left({P' + \left({I - P'}\right)}\right)h       | c = $I$ is the identity operator }} {{eqn | r = PP'h + \left({I - P}\right)P'h + P\left({I - P'}\right)h \left({I - P}\right)\left({I - P'}\right)h }} {{eqn | r = 0 + P'h + Ph + 0 }} {{end-eqn}} For this last equality, observe $M \perp N'$, hence $M \subseteq N'^\perp$, $N' \subseteq M^\perp$ and $\left({M + N'}\right)^\perp \subseteq M^\perp \cup N'^\perp$. {{finish|link to results establishing the implications of these relations for the associated projections. These results don't exist at the moment}} The conclusion is that every $h \in M + N'$ can be uniquely decomposed as $P'h + Ph$, with $P'h \in N', Ph \in M$.  So suppose there is a sequence $h_n \to h$ in $M + N'$. Then $Ph_n$ and $P'h_n$ are sequences in $M, N'$, respectively. As those are closed linear subspaces of $H$, there are $m \in M, n \in N'$ with $Ph_n \to m, P'h_n \to n$. It follows that $h = m + n \in M + N'$. That is, $M + N' = M + N$ is a closed linear subspace of $H$. {{qed}} Category:Hilbert Spaces 0gbg8zzxrz1hypc60x8yl9c087m1opm
It needs to be demonstrated that $L$ is closed under $+$ and $\circ$. So let $m_1 + n_1, m_2 + n_2 \in L$. Then $\paren {m_1 + n_1} + \paren {m_2 + n_2} = \paren {m_1 + m_2} + \paren {n_1 + n_2} \in L$. It follows that $L$ is closed under $+$. Now let $\lambda \in K, m + n \in L$. Then $\lambda \circ \paren {m + n} = \paren {\lambda \circ m} + \paren {\lambda \circ n} \in L$. It follows that $L$ is closed under $\circ$. Hence the result, by definition of linear subspace. {{qed}} Category:Vector Subspaces fkp9zb9z6qnfaekdo3j3vmer3oxbvlp
By Characterization of Projections, $A$ is self-adjoint. Then $\left({I - A}\right)^* = I^* - A^* = I - A$ from Adjoining is Linear. So $I - A$ is also self-adjoint. From Complementary Idempotent is Idempotent, $I - A$ is idempotent. Hence, applying Characterization of Projections, $I - A$ is a projection. {{qed}} Category:Linear Transformations on Hilbert Spaces 0o5qci51vwo2q9gzp82zog9zl95dky6
Assert that $U: M \oplus M^\perp \to H: \left({m, m^\perp}\right) \mapsto m + m^\perp$ is an isomorphism. According to the definition of isomorphism, it is sufficient to check that $U$ is surjective and that: :$\left\langle{ U \left({m, m^\perp}\right), U \left({n, n^\perp}\right) }\right\rangle_H = \left\langle{ \left({m, m^\perp}\right), \left({n, n^\perp}\right) }\right\rangle_{M \oplus M^\perp}$ First the surjectivity: Denote by $P$ the orthogonal projection on $M$. Then for any $h \in H$, have $h = Ph + \left({h - Ph}\right)$. By definition of $P$, $Ph \in M$. Furthermore, by Properties of Orthogonal Projection, $h - Ph \perp M$; that is, $h - Ph \in M^\perp$. It follows that $h = U \left({Ph, h - Ph}\right)$, showing that $U$ is a surjection. It remains to check that $U$ preserves the inner product: {{begin-eqn}} {{eqn|l = \left\langle{ U \left({m, m^\perp}\right), U \left({n, n^\perp}\right) }\right\rangle_H      |r = \left\langle{ m + m^\perp, n + n^\perp}\right\rangle_H }} {{eqn|r = \left\langle{m, n}\right\rangle_H + \left\langle{m^\perp, n}\right\rangle_H + \left\langle{m, n^\perp}\right\rangle_H + \left\langle{m^\perp, n^\perp}\right\rangle_H      |c = Linearity of $\left\langle{\cdot, \cdot}\right\rangle_H$ }} {{eqn|r = \left\langle{m, n}\right\rangle_H + \left\langle{m^\perp, n^\perp}\right\rangle_H      |c = Defining property of orthocomplement }} {{eqn|r = \left\langle{ \left({m, m^\perp}\right), \left({n, n^\perp}\right) }\right\rangle_{M \oplus M^\perp}      |c = Definition of $\left\langle{\cdot, \cdot}\right\rangle_{M \oplus M^\perp}$ }} {{end-eqn}} Hence $U$ is an isomorphism, and so $M \oplus M^\perp$ is isomorphic to $H$. {{qed}} Category:Hilbert Spaces 5sfhxq1m9kc6yz967uacbnschftll3f
As $\left({S, \preceq}\right)$ is a totally ordered set, the relation $\preceq$ is a total ordering, and is by definition: * reflexive * antisymmetric * transitive * connected From Properties of Restriction of Relation, a restriction of a relation which has all those properties inherits them all. Thus $\preceq \restriction_T$ is also: * reflexive * antisymmetric * transitive * connected and so is also a total ordering. Hence the result, by definition of totally ordered set. {{Qed}}
{{NotZFC}} {{AimForCont}} that $\neg \On \subseteq A$. Then: :$\paren {\On \setminus A} \ne \O$ From Set Difference is Subset, $\On \setminus A$ is a subclass of the ordinals. By Epsilon Relation is Strongly Well-Founded on Ordinal Class, $\On \setminus A$ must have a $\in$-minimal element $y$. By Element of Ordinal is Ordinal, $y$ must be a subset of $\On$, the class of all ordinals. However, from the fact that $y$ is a $\in$-minimal element of $\On \setminus A$: :$\paren {\On \setminus A} \cap y = \O$ So by its subsethood of $\On$: :$\paren {\On \cap y} \setminus A = \paren {y \setminus A} = \O$ Therefore $y \subseteq A$. However, by the hypothesis, $y$ must also be an element of $A$. This contradicts the fact that $y$ is an element of $\On \setminus A$. Therefore $\On \subseteq A$. {{qed}} {{LEM|Reductio ad Absurdum|4}}
By hypothesis: :$f \left({x}\right) = g \left({x}\right)$ for all $x$ such that: :$x \in \R: 0 < \left \vert{\alpha - x}\right \vert < \epsilon$ Putting $c$ for $\alpha$ and $\delta$ for $\epsilon$, this is equivalent to: :$x \in \R: 0 < \left \vert{x - c}\right \vert < \delta$ By definition, that the limit of $f \left({x}\right)$ exists is to say: :$\exists L: \forall \varepsilon > 0: \exists \delta > 0: 0 < \left \vert{x - c}\right \vert < \delta \implies \left \vert{f \left({x}\right) - L}\right \vert < \varepsilon$ For all $x$ relevant, we can substitute $f \left({x}\right)$ with $g \left({x}\right)$ and get: :$\exists L: \forall \varepsilon > 0: \exists \delta > 0: 0 < \left \vert{x - c}\right \vert < \delta \implies \left \vert{g \left({x}\right) - L}\right \vert < \varepsilon$ The result follows by the definition of limit. {{qed}}
By Reciprocal Function is Strictly Decreasing, the reciprocal function is strictly decreasing. By Mapping from Totally Ordered Set is Dual Order Embedding iff Strictly Decreasing, the reciprocal function is a dual order embedding. That is: :$x \le y \iff \dfrac 1 y \le \dfrac 1 x$ {{qed}}
By Reciprocal Function is Strictly Decreasing, the reciprocal function is strictly decreasing. Thus  :$x \le y \implies \dfrac 1 y \le \dfrac 1 x$ Suppose then that $\dfrac 1 y \le \dfrac 1 x$. If $x, y > 0$, then from Reciprocal of Strictly Positive Real Number is Strictly Positive: $\dfrac 1 y, \dfrac 1 x > 0$ Similarly, if $x, y < 0$ or $x, y > 0$, then from Reciprocal of Strictly Negative Real Number is Strictly Negative: :$\dfrac 1 y, \dfrac 1 x < 0$ Thus we can apply the above to show that: :$\dfrac 1 {1/x} \le \dfrac 1 {1/y}$ By Inverse of Multiplicative Inverse: :$x \le y$ {{qed}}
This is the contrapositive of the Comparison Test.  Hence the result, from the Rule of Transposition. {{qed}}
{{AimForCont}} that for each $x \in X$ there is a $y \in X$ such that $x \prec y$. By the Axiom of Choice, there is a mapping $f: X \to X$ such that: :$\forall x \in X: x \prec \map f x$ Let $\mathcal C$ be the set of all chains in $X$. By the premise, each element of $\mathcal C$ has an upper bound in $X$. Thus by the Axiom of Choice, there is a mapping $g: \mathcal C \to X$ such that for each $C \in \mathcal C$, $\map g C$ is an upper bound of $C$. Let $p$ be an arbitrary element of $X$. Define a mapping $h: \operatorname {Ord} \to X$ by transfinite recursion thus: {{begin-eqn}} {{eqn | l = \map h 0       | r = p }} {{eqn | l = \map h {\alpha^+}       | r = \map f {\map h \alpha } }} {{eqn | l = \map h \lambda       | r = \map f {\map g {f \sqbrk \lambda} }       | c = if $\lambda$ is a limit ordinal }} {{end-eqn}} {{explain|The notation $f \sqbrk \lambda$ needs to be defined.}} Then $h$ is strictly increasing, and thus injective. {{explain|The above statement needs to be justified.}} Let $h'$ be the restriction of $h$ to $\operatorname {On} \times \map h {\operatorname{On} }$. Then ${h'}^{-1}$ is a surjection from $\map h {\operatorname{On} } \subseteq X$ onto $\operatorname{On}$. By the Axiom of Replacement, $\operatorname{On}$ is a set. By Burali-Forti Paradox, this is a contradiction. Thus we conclude that some element of $X$ has no strict successor, and is thus maximal. {{qed}}
We have that the Derivative of Identity Function is $1$. Hence, by Derivative of Monotone Function, $x$ is strictly increasing. Now, by the definition of infinite limit at infinity, the first assertion is: :$\forall M \in \R_{>0}: \exists N \in \R_{>0}: x > N \implies \map f x > M$ For every $M$, choose $N = M$. The second assertion is proved similarly. {{qed}} Category:Unbounded Mappings Category:Identity Mappings 1jh9mhvnfumhmsqe3h8jtjpnt84altt
Let $\displaystyle \lim_{x \mathop \to +\infty} \map f x = +\infty$. By the definition of infinite limit at infinity, this means: :$\forall M_1 \in \R_{>0}: \exists N_1 \in \R_{>0}: x > N_1 \implies \map f x > M_1$ Now, the assertion that $\map g x \to +\infty$ is: :$\forall M_1 \in \R_{>0}: \exists N_2 \in \R_{>0}: x > N_2 \implies \map g x > M_1$ By the premise that $\map g x > \map f x$ for sufficiently large $x$, there is an $N$ such that: :$x > N \implies \map g x > \map f x$ Now, given $M_1$, let $N_2$ be greater than both $N_1$ and $N$. Then since $N_2 > N$ and $N_2 > N_1$ respectively: :$\map g x > \map f x > M_1$ whence: :$\displaystyle \lim_{x \mathop \to +\infty} \map g x = +\infty$ {{qed}}
By definition of set equality, it suffices to prove the following two inclusions: :$\ds \bigcap_{i \mathop \in I} M_i^\perp \subseteq \paren {\vee \set {M_i : i \in I} }^\perp$ :$\paren {\vee \set {M_i : i \in I} }^\perp \subseteq \ds \bigcap_{i \mathop \in I} M_i^\perp$ === $\ds \bigcap_{i \mathop \in I} M_i^\perp$ is contained in $\paren {\vee \set {M_i : i \in I} }^\perp$ === By Orthocomplement is Closed Linear Subspace and Closed Linear Subspaces Closed under Intersection, both spaces considered are closed linear subspaces of $H$. By Orthocomplement Reverses Subset, the required containment is equivalent to: :$\vee \set {M_i : i \in I} \subseteq \ds \paren {\bigcap_{i \mathop \in I} M_i^\perp }^\perp$ For $h \in \ds \bigcap_{i \mathop \in I} M_i^\perp$, by definition one has: :$h \perp M_i$ for all $i \in I$ That is: :$M_i \perp \ds \bigcap_{i \mathop \in I} M_i^\perp$ This is equivalent to saying that: :$M_i \subseteq \ds \paren {\bigcap_{i \mathop \in I} M_i^\perp }^\perp$ Definition $(2)$ of closed linear span now grants the desired subset relation. {{qed|lemma}} === $\paren {\vee \set {M_i : i \in I} }^\perp$ is contained in $\ds \bigcap_{i \mathop \in I} M_i^\perp$ === By definition $(2)$ of closed linear span: :$\forall i \in I: M_i \subseteq \vee \set {M_i^\perp : i \in I}$ By Orthocomplement Reverses Subset, it follows that: :$\forall i \ni I: \paren {\vee \set {M_i : i \in I} }^\perp \subseteq M_i^\perp$ Therefore, by definition of set intersection: :$\paren {\vee \set {M_i : i \in I} }^\perp \subseteq \displaystyle \bigcap_{i \mathop \in I} M_i^\perp$ {{qed|lemma}} Thus we have established that: :$h \in \paren {\vee \set {M_i : i \in I} }^\perp \iff h \in \ds \bigcap_{i \mathop \in I} M_i^\perp$ From the definition of set equality, it follows that: :$\ds \bigcap_{i \mathop \in I} M_i^\perp = \paren {\vee \set {M_i : i \in I} }^\perp$ {{qed}}
Let $X$ be a set. Let $\mathcal F$ be the set of mappings defined as: :$f \in \mathcal F \iff \begin{cases} \operatorname{Dom} \left({f}\right) \subseteq \mathcal P \left({X}\right) & \ \\ \operatorname{Im} \left({f}\right) \subseteq X & \ \\ \forall A \in \operatorname{Dom} \left({f}\right): f \left({A}\right) \in A & \ \end{cases}$ Let $\preceq$ be the relation defined on $\mathcal F$ as: :$\forall f_1, f_2 \in \mathcal F: f_1 \preceq f_2 \iff f_2$ is an extension of $f_1$. Straightforwardly, $\preceq$ is a partial ordering on $\mathcal F$. Suppose Zorn's Lemma holds. Then there exists a maximal element of $\mathcal F$. {{explain|Check the chain condition (and nonemptiness of $\mathcal F$)}} We then show that if $g$ is such a maximal element, then: :$\operatorname{Dom} \left({g}\right) = \mathcal P \left({X}\right) \setminus \varnothing$ {{proof wanted}}
Suppose $h \in B^\perp$. That is, for all $b \in B$: $h \perp b$. As $A \subseteq B$, it follows that $h \perp a$ for all $a \in A$. Hence, by definition of orthocomplement, $h \in A^\perp$. Therefore, $B^\perp \subseteq A^\perp$. {{qed}} Category:Hilbert Spaces ci6vkbc5ru0wk49jgokohq7rlk6l14t
From Upper Bound of Natural Logarithm: :$\forall n > 0: n \ln x < x^n$ which, by Multiple Rule for Continuous Functions, implies: {{begin-eqn}} {{eqn | l = \lim_{x \mathop \to +\infty} n \ln x       | r = n \lim_{x \mathop \to +\infty} \ln x }} {{end-eqn}} From Logarithm Tends to Infinity: :$n \ln x \to +\infty$ as $x \to +\infty$ The result follows from the Push Theorem. {{qed}}
Consider the case where $D_x \map f x \ge 0$ Let $n \in \N$ be in the domain of $\sequence {a_n}$. From Derivative of Monotone Function, the sign of $D_x f$ is indicative of the monotonicity of $f$. Because Differentiable Function is Continuous and Continuous Real Function is Darboux Integrable, $D_x f$ is integrable. Hence: {{begin-eqn}} {{eqn | l = \map f {n + 1} - \map f n       | r = \int_n^{n + 1} D_x \map f x \rd x \ge 0       | c = Fundamental Theorem of Calculus }} {{eqn | o = \ge       | r = 0       | c = Relative Sizes of Definite Integrals }} {{eqn | ll= \leadsto       | l = \map f {n + 1}       | o = \ge       | r = \map f n       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | lo= \forall n \in \mathbb D:       | l = \map f n       | r = a_n       | c = by hypothesis }} {{eqn | ll= \leadsto       | lo= \forall n \in \mathbb D:       | l = a_{n + 1}       | o = \ge       | r = a_n       | c = as $n \in \mathbb D$ was arbitrary }} {{end-eqn}} Hence the result, by the definition of monotone. The proofs of the other cases are similar. {{qed}}
* Hausdorff Maximal Principle implies Well-Ordering Theorem * Well-Ordering Theorem is Equivalent to Axiom of Choice * Hausdorff Maximal Principle: Proof 3, from whence the Axiom of Choice is seen to imply the Hausdorff Maximal Principle {{qed}}
The result follows directly if $a = 0$ or $b = 0$. {{WLOG}}, assume that $a > 0$ and $b > 0$.  Then: {{begin-eqn}} {{eqn | l = a b       | r = \map \exp {\map \ln {a b} }       | c = Exponential of Natural Logarithm }} {{eqn | r = \map \exp {\ln a + \ln b}       | c = Sum of Logarithms }} {{eqn | r = \map \exp {\frac 1 p p \ln a + \frac 1 q q \ln b}       | c = {{Defof|Multiplicative Identity}} and {{Defof|Multiplicative Inverse}} }} {{eqn | r = \map \exp {\frac 1 p \map \ln {a^p} + \frac 1 q \map \ln {b^q} }       | c = Logarithms of Powers }} {{eqn | o = \le       | r = \frac 1 p \map \exp {\map \ln {a^p} } + \frac 1 q \map \exp {map \ln {b^q} }       | c = Exponential is Strictly Convex and the hypothesis that $\dfrac 1 p + \dfrac 1 q = 1$ }} {{eqn | r = \frac {a^p} p + \frac {b^q} q       | c = Exponential of Natural Logarithm }} {{end-eqn}} {{qed}}
200pxthumbrightThe blue colored region corresponds to $\displaystyle \int_0^\alpha t^{p-1} \mathrm d t$ and the red colored region to $\displaystyle \int_0^\beta u^{q-1} \mathrm d u$. In order for $\dfrac 1 p + \dfrac 1 q = 1$ it is necessary for both $p > 1$ and $q > 1$. {{begin-eqn}} {{eqn | l = \frac 1 p + \frac 1 q       | r = 1       | c =  }} {{eqn | ll= \implies       | l = p + q       | r = p q       | c = multiplying both sides by $p q$ }} {{eqn | ll= \implies       | l = p + q - p - q + 1       | r = p q - p - q + 1       | c = adding $1 - p - q$ to both sides }} {{eqn | ll= \implies       | l = 1       | r = \left({p - 1}\right) \left({q - 1}\right)       | c = elementary algebra }} {{eqn | ll= \implies       | l = \frac 1 {p - 1}       | r = q - 1       | c =  }} {{end-eqn}} Accordingly: : $u = t^{p-1} \iff t = u^{q-1}$ Let $a, b$ be any positive real numbers. Since $a b$ is the area of the rectangle in the given figure, we have: :$\displaystyle a b \le \int_0^a t^{p-1} \ \mathrm d t + \int_0^b u^{q-1} \ \mathrm d u = \frac {a^p} p + \frac {b^q} q$ Note that even if the graph intersected the side of the rectangle corresponding to $t = a$, this inequality would hold. Also note that if either $a = 0$ or $b = 0$ then this inequality holds trivially. {{qed}}
=== Definition 1 implies Definition 2 === Let $F \left({x}\right)$ be $\displaystyle \int_1^x \frac {\d t} t$. Let $f \left({t}\right)$ be $\displaystyle \int \frac {\d t} t$. Then: :$\dfrac {\d t} t = \dfrac 1 t$ Or: :$\dfrac {\d x} x = \dfrac 1 x$ Also: :$F \left({x}\right) = f \left({x}\right) - f \left({1}\right)$ Therefore: {{begin-eqn}} {{eqn | l = \frac {\d F \left({x}\right)} {\d x}       | r = \frac {\d f \left({x}\right)} {\d x} - \frac {\d F \left({1}\right)} {\d x}       | c =  }} {{eqn | r = \frac {\d f \left({x}\right)} {\d x}       | c = Derivative of Constant }} {{eqn | r = \frac 1 x       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d x} {\d F \left({x}\right)}       | r = x       | c = Derivative of Inverse Function }} {{end-eqn}} Furthermore: :$F \left({1}\right) = f \left({1}\right) - f \left({1}\right) = 0$ {{explain|Why is the reverse of $F \left({x}\right)$ a function?}} The result follows from the fifth definition of the exponential function: :$F \left({x}\right) \equiv e^x$ {{qed|lemma}} === Definition 2 implies Definition 1 === {{begin-eqn}} {{eqn | l = e^{F \left({x}\right)}       | r = x       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d x} {\d F \left({x}\right)}       | r = x       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d F \left({x}\right)} {\d x}       | r = \frac 1 x       | c = Derivative of Inverse Function }} {{end-eqn}} Let $f \left({t}\right)$ be $\displaystyle \int \frac 1 t \rd t$. Then: :$F \left({x}\right) = f \left({x}\right) + C$ When $F \left({x}\right) = 0$: :$x = e^{F \left({x}\right)} = 1$ :$F \left({1}\right) = f \left({1}\right) + C = 0 \implies f \left({1}\right) = - C$ Therefore: :$F \left({x}\right) = f \left({x}\right) - f \left({1}\right)$ Therefore: :$\displaystyle F \left({x}\right) = \int_1^x \frac {\d t} t$ {{qed|lemma}} Therefore: {{begin-eqn}} {{eqn | l = y = \int_1^x \frac {\d t} t       | o = \leadsto       | r = e^y = x       | c = Definition 1 implies Definition 2 }} {{eqn | l = e^y = x       | o = \leadsto       | r = y = \int_1^x \frac {\d t} t       | c = Definition 2 implies Definition 1 }} {{eqn | ll= \leadsto       | l = y = \int_1^x \frac {\d t} t       | o = \iff       | r = e^y = x       | c =  }} {{end-eqn}} {{qed}} {{ProofWanted|Add definition 3}} Category:Natural Logarithms 3we8f3g9j0l6sf4wzr3nydle0x0g01w
A restriction does not introduce any new arguments. Hence the result follows trivially from the definition of monotone mapping. {{qed}} {{handwaving}}
Consider $T_n$ where $n \in \N_{>0}$. It is clear that $\left({T_1, \preccurlyeq}\right)$ is order isomorphic to  $\left({S, \preceq}\right)$. Thus as $\preceq$ is a well-ordering on $S$, $\preccurlyeq$ is a well-ordering on $T_1$. Now, let us assume that $\preccurlyeq$ is a well-ordering on $T_k$ for some $k \in \N: k \ge 1$. Let $A$ be a non-empty subset of $T_{k+1}$. Let $A_1$ be the set of all of the first components of the ordered $n$-tuples that comprise $A$. Since $A_1$ is a non-empty subset of $S$, and $S$ is itself well-ordered by $\preceq$, it follows that $A_1$ contains a minimal element $x$ by $\preccurlyeq$. Let $A_x$ be the subset of $A$ in which the first component equals $x$. We may consider $A_x$ to be a subset of $T_k$ where this first component $x$ has been suppressed. But we assumed that $T_k$ is well-ordered by $\preccurlyeq$. So $A_x$ contains a minimal element $\left({x, x_2, x_3, \ldots, x_{k+1}}\right)$ by $\preccurlyeq$. This element $\left({x, x_2, x_3, \ldots, x_{k+1}}\right)$ is the minimal element of $A$ by $\preccurlyeq$. Hence, by definition, $T_{k+1}$ is well-ordered by $\preccurlyeq$. The result follows by induction. {{qed}}
Consider a set $S = \set {a, b}$ such that $a \prec b$. Then the set $\set {\tuple b, \tuple {a, b}, \tuple {a, a, b},  \tuple {a, a, a, b}, \ldots}$ has no minimal element by $\preccurlyeq$. {{handwaving}} Thus $T$ is not well-ordered by $\preccurlyeq$. {{qed}}
From Absolutely Convergent Series is Convergent, the convergence of: {{refactor|We really want to be linking to Absolutely Convergent Series is Convergent/Real Numbers here, because the student who is working through basic undergraduate analysis won't know (or even care) what a Banach space is. Consequently, Absolutely Convergent Series is Convergent/Real Numbers can be given a proof which basically goes "the real numbers form a Banach space, the result follows from Absolutely Convergent Series is Convergent iff Normed Vector Space is Banach" which gives us an in from both undergrad analysis and this more rarefied topological / functional-analytical level.}} :$\displaystyle \sum_{n \mathop = 1}^\infty \size {a_n}$ implies that of: :$\displaystyle \sum_{n \mathop = 1}^\infty a_n$ By Negative of Absolute Value: :$\displaystyle -\size {\sum_{n \mathop = 1}^j a_n} \le \sum_{n \mathop = 1}^j a_n \le \size {\sum_{n \mathop = 1}^j a_n}$ By repeated application of Triangle Inequality: :$\displaystyle -\sum_{n \mathop = 1}^j \size {a_n} \le \sum_{n \mathop = 1}^j a_n \le \sum_{n \mathop = 1}^j \size {a_n}$ By hypothesis, the leftmost and rightmost terms converge as $j \to +\infty$. Hence the result, from Squeeze Theorem for Real Sequences. {{qed}} Category:Series 9jr4c5grr3ojim3o4cb1ep9tiqu2no3