hcju/proofwikiqa huggingface dataset:
Let $\sequence {f_n}$ be the  sequence of mappings $f_n : \R_{>0} \to \R$ defined as: :$\map {f_n} x = n \paren {\sqrt [n] x - 1}$ Let $\map M t = \max \set {\size {t - 1}, \size {\dfrac {t - 1} t} }$ From Bounds of Natural Logarithm: :$\forall t \in \R_{>0}: \size {\map {f_n} t} \le \map M t$ Fix $x, y \in \R_{>0}$. Then: {{begin-eqn}} {{eqn | l = \size {\map {f_n} {x y} - \paren {\map {f_n} x + \map {f_n} x} }       | r = \size {n \paren {\sqrt [n] {x y} - 1 - \sqrt [n] x + 1 - \sqrt [n] y + 1} }       | c = Definition of $\map {f_n} x$ }} {{eqn | r = n \size {\sqrt [n] {x y} - \sqrt [n] x - \sqrt [n] y + 1}       | c = Absolute Value Function is Completely Multiplicative }} {{eqn | r = n \size {\sqrt [n] x - 1} \size {\sqrt [n] y - 1}       | c = Absolute Value Function is Completely Multiplicative }} {{eqn | r = n \frac {\size {\map {f_n} x} } n \frac {\size {\map {f_n} y} } n       | c = Definition of $\map {f_n} x$ }} {{eqn | r = \frac 1 n \size {\map {f_n} x} \size {\map {f_n} y} }} {{eqn | o = \le       | r = \frac {\map M x \map M y} n       | c = Inequality of Product of Unequal Numbers }} {{eqn | ll= \leadsto       | l = \size {\map \ln {x y} - \paren  {\map \ln x + \map \ln y} }       | r = \size {\lim_{n \mathop \to \infty} \map {f_n} {x y} - \paren {\lim_{n \mathop \to \infty} \map {f_n} x + \lim_{n \mathop \to \infty} \map {f_n} y} } }} {{eqn | r = \lim_{n \mathop \to \infty} \size {\map {f_n} {x y} - \paren {\map {f_n} x + \map {f_n} y} }       | c = Modulus of Limit }} {{eqn | o = \le       | r = \lim_{n \mathop \to \infty} \frac {\map M x \map M y} n       | c = Limit of Bounded Convergent Sequence is Bounded }} {{eqn | r = \map M x \map M y \lim_{n \mathop \to \infty} \frac 1 n       | c = Multiple Rule for Real Sequences }} {{eqn | r = 0 }} {{end-eqn}} Thus: $\displaystyle \lim_{n \mathop \to \infty} \map {f_n} {x y} = \lim_{n \mathop \to \infty} \paren {\map {f_n} x + \map {f_n} y}$       Hence the result, from the definition of $\ln$. {{qed}}
{{AimForCont}} that $x < a$. Let $\epsilon = \dfrac{a - x} 2 > 0$ From the convergence of $\left \langle {x_n} \right \rangle$: :$\exists M_1 \in \N : n \ge M \implies x - \epsilon < x_n < x + \epsilon$ Or, equivalently: :$\exists M_1 \in \N : n \ge M \implies \dfrac {3x - a} 2 < x_n < \dfrac{x + a} 2$ From the convergence of $\left \langle {a_n} \right \rangle$: :$\exists M_2 \in \N : n \ge M \implies a - \epsilon < a_n < a + \epsilon$ Or, equivalently: :$\exists M_2 \in \N : n \ge M \implies \dfrac {x + a} 2 < a_n < \dfrac{3a - x} 2$ Let $M = \max\left\{ {N, M_1, M_2} \right\}$ Then, for any $n \ge M$: {{begin-eqn}} {{eqn | l = x_n       | o = <       | r = \dfrac {x + a} 2 }} {{eqn | o = <       | r = a_n }} {{end-eqn}} This contradicts the hypothesis that: :$\forall n \ge N : a_n \le x_n$ The same argument, mutatis mutandis, brings us to a contradiction if we suppose $x > b$. Hence the result, by Proof by Contradiction. {{qed}} Category:Limits of Sequences Category:Real Analysis 20djjxluat320apfe4oqvxm3pt6pi9b
Proof by Mathematical Induction: For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$\forall x \in \R_{>0}: \ln \left({ x^n }\right) = n\ln \left({ x }\right)$ === Basis for the Induction === $P \left({0}\right)$ is the case: :$\forall x \in \R_{>0}: \ln \left({ 1 }\right) = 0 $ from Logarithm of 1 is 0 This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0 $, then it logically follows that $P \left({k + 1}\right)$ is true. So this is our induction hypothesis: :$\forall x \in \R_{>0}: \ln \left({ x^k }\right) = k\ln \left({ x }\right)$ Then we need to show: :$\forall x \in \R_{>0}: \ln \left({ x^{k+1} }\right) = \left({ k + 1 }\right) \ln \left({ x }\right)$ === Inductive Step === This is our induction step: Fix $x \in \R_{>0}$. Then: {{begin-eqn}} {{eqn | l = \ln \left({ x^{k+1} }\right)       | r = \ln \left({ x^{k}x }\right) }} {{eqn | r = \ln \left({ x^{k} }\right) + \ln \left({ x }\right)       | c = Sum of Logarithms/Natural Logarithm }} {{eqn | r = k\ln \left({ x }\right) + \ln \left({ x }\right)       | c = induction step }} {{eqn | r = \left({ k + 1 }\right) \ln \left({ x }\right) }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 0}: \forall x \in \R_{>0}: \ln \left({ x^n }\right) = n\ln \left({ x }\right)$ {{qed}} Category:Logarithm of Power aymbmtbcoj8pq4ppxvizt7oih7bi9el
From Logarithm of Power/Natural Logarithm/Natural Power, the theorem is already proven for positive integers. Let $j \in \Z_{<0}$. Let $-j = k \in Z_{>0}$. Then: {{begin-eqn}} {{eqn | l = 0       | r = \ln \left({ 1 }\right)       | c = Logarithm of 1 is 0 }} {{eqn | r = \ln \left({ x^{k}x^{-k} }\right) }} {{eqn | r = \ln \left({ x^{k} }\right) + \ln \left({ x^{-k} }\right) }} {{eqn | r = k\ln \left({ x }\right) + \ln \left({ x^{-k} }\right)       | c = Logarithm of Power/Natural Logarithm/Natural Power }} {{eqn | ll = \implies       | l = \ln \left({ x^{-k} }\right)       | r = -k\ln \left({ x }\right) }} {{eqn | ll = \implies       | l = \ln \left({ x^{j} }\right)       | r = j\ln \left({ x }\right) }} {{end-eqn}} {{qed}} Category:Logarithm of Power qs8ggz27tpoxo5iyij4hi21m0oymec7
Let $r = \dfrac s t$, where $s \in \Z$ and $t \in \Z_{>0}$ First: {{begin-eqn}} {{eqn | l = \ln \left({ x }\right)       | r = \ln \left({ x^{t/t} }\right) }} {{eqn | r = \ln \left({ \left({ x^{1/t} }\right)^{t} }\right)       | c = Product of Indices of Real Number/Rational Numbers }} {{eqn | r = t\ln \left({ x^{1/t} }\right)       | c = Logarithm of Power/Natural Logarithm/Integer Power }} {{eqn | ll= \implies       | l = \ln \left({ x^{1/t} }\right)       | r = \frac 1 t \ln \left({ x }\right)       | c = Dividing both sides by $t$ }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \ln \left({x^{s/t} }\right)       | r = \ln \left({\left({x^{1/t} }\right)^s}\right)       | c = Product of Indices of Real Number/Rational Numbers }} {{eqn | r = s \ln \left({ x^{1/t} }\right)       | c = Logarithm of Power/Natural Logarithm/Integer Power }} {{eqn | r = \frac s t \ln \left({ x }\right)       | c = From the result above }} {{eqn | r = r \ln \left({ x }\right)       | c = From the definition of $s$ and $t$ }} {{end-eqn}} {{qed}} Category:Logarithm of Power j9zx0kfhcn7erra58301rrstw4qzbh0
=== Sufficient Condition === Let $x \ll y$ Let $I \in {\it Ids}\left({\mathscr S}\right)$ such that :$y = \sup I$ By definition of reflexivity: :$y \preceq \sup I$ By definition of meet-continuous: :$\mathscr S$ is up-complete. Thus by Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal: :$x \in I$ {{qed|lemma}} === Necessary Condition === Assume that :$\forall I \in {\it Ids}\left({\mathscr S}\right): y = \sup I \implies x \in I$ We will prove that :$\forall I \in {\it Ids}\left({\mathscr S}\right): y \preceq \sup I \implies x \in I$ Let $I \in {\it Ids}\left({\mathscr S}\right)$ such that :$y \preceq \sup I$ By definition of ideal: :$I$ is directed and lower. Define $Y := \left\{ {y \wedge d: d \in I}\right\}$ By Set of Finite Suprema is Directed: :$\left\{ {\sup A: A \in {\it Fin}\left({Y}\right) \land A \ne \varnothing}\right\}$ is directed. where ${\it Fin}\left({Y}\right)$ denotes the set of all finite subsets of $Y$. By definition of up-complete: :$\left\{ {\sup A: A \in {\it Fin}\left({Y}\right) \land A \ne \varnothing}\right\}$ admits a supremum. By definition of union: :$Y = \bigcup {\it Fin}\left({Y}\right)$ {{begin-eqn}} {{eqn | l = y       | r = y \wedge \sup I       | c = Preceding iff Meet equals Less Operand }} {{eqn | r = \sup Y       | c = definition of meet-continuous }} {{eqn | r = \sup \left\{ {\sup A: A \in {\it Fin}\left({Y}\right) \land A \ne \varnothing}\right\}       | c = Supremum of Suprema }} {{eqn | r = \sup \left\{ {\sup A: A \in {\it Fin}\left({Y}\right) \land A \ne \varnothing}\right\}^\preceq       | c = Supremum of Lower Closure of Set }} {{end-eqn}} where $^\preceq$ denotes the lower closure of set. By Lower Closure of Directed Subset is Ideal: :$\left\{ {\sup A: A \in {\it Fin}\left({Y}\right) \land A \ne \varnothing}\right\}^\preceq$ is an ideal in $\mathscr S$ By assumption: :$x \in \left\{ {\sup A: A \in {\it Fin}\left({Y}\right) \land A \ne \varnothing}\right\}^\preceq$ By definition of lower closure of set: :$\exists z \in \left\{ {\sup A: A \in {\it Fin}\left({Y}\right) \land A \ne \varnothing}\right\}: x \preceq z$ Then :$\exists A \in {\it Fin}\left({Y}\right): z = \sup A \land A \ne \varnothing$ We will prove as a sublemma that: :$A \subseteq I$ Let $a \in A$. By definition of subset: :$a \in Y$ By definition of $Y$: :$\exists i \in I: a = y \wedge i$ By Meet Precedes Operands: :$a \preceq i$ Thus by definition of lower set: :$a \in I$ This ends the proof of sublemma. By Directed iff Finite Subsets have Upper Bounds: :$\exists h \in I: \forall a \in A: a \preceq h$ By definition: :$h$ is upper bound for $A$ By definition of supremum: :$z = \sup A \preceq h$ By definition of transitivity: :$x \preceq h$ Thus by definition of lower set: :$x \in I$ {{qed|lemma}} Thus by Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal: :$x \ll y$ {{qed}}
{{begin-eqn}} {{eqn | l = \lg x        | r = \frac {\ln x} {\ln 2}       | c = Change of Base of Logarithm }} {{eqn | r = \frac {\log_{10} x} {\log_{10} 2}       | c = Change of Base of Logarithm }} {{eqn | ll= \leadsto       | l = \lg x       | r = \frac {\ln x + \log_{10} x} {\ln 2 + \log_{10} 2}       | c =  }} {{eqn | r = \frac {\ln x + \log_{10} x} {\ln 2 + \frac {\ln 2} {\ln 10} }       | c = Change of Base of Logarithm }} {{eqn | r = \frac {\ln x + \log_{10} x} {\ln 2 \left({1 + \frac 1 {\ln 10} }\right)}       | c =  }} {{eqn | r = \frac {\ln x + \log_{10} x} {\ln 2 \left({1 + \log_{10} e}\right)}       | c = Reciprocal of Logarithm }} {{end-eqn}} We have that: :$\ln 2 = 0 \cdotp 69347 \ 1805 \ldots$ :$\log_{10} e = 0 \cdotp 43429 \ 44819 \ldots$ Thus: {{begin-eqn}} {{eqn | l = \ln 2 \left({1 + \log_{10} e}\right)       | r = 0 \cdotp 69347 \ 1805 \ldots \left({1 + 0 \cdotp 43429 \ 44819 \ldots}\right)       | c =  }} {{eqn | r = 0 \cdotp 99462 \ldots       | c =  }} {{end-eqn}} Hence: :$\lg x + \log_{10} x \approx 0 \cdotp 994 \ln x$ That is, the approximation is $99.4 \%$ accurate. {{qed}}
Recall the definition of the natural logarithm as the definite integral of the reciprocal function: :$\displaystyle \ln x := \int_1^x \frac {\d t} t$ :800px Consider the diagram above. The value of $\ln x$ is represented by the area between the points: :$\left({1, 0}\right), \left({1, 1}\right), \left({x, \dfrac 1 x}\right), \left({x, 0}\right)$ which is represented by the yellow region above. Similarly, the value of $\ln y$ is represented by the area between the points: :$\left({1, 0}\right), \left({1, 1}\right), \left({y, \dfrac 1 y}\right), \left({y, 0}\right)$ Let the second of these areas be transformed by dividing its height by $x$ and multiplying its length by $x$. This will preserve its area, while making it into the area between the points: :$\left({x, 0}\right), \left({x, 1 / x}\right), \left({x y, \dfrac 1 {x y}}\right), \left({x y, 0}\right)$ which is exactly the green area. The total of the green and yellow areas represents the value of $\ln x y$. {{qed}}
From Derivative at Maximum or Minimum, when $f$ is at a minimum, its derivative $\dfrac \d {\d b} f$ will be zero. Let $y = f \left({b}\right)$. We have: {{begin-eqn}} {{eqn | l = y       | r = b \log_b x       | c =  }} {{eqn | r = \frac {b \ln x} {\ln b}        | c = Change of Base of Logarithm }} {{eqn | ll= \leadsto       | l = \frac {\d y} {\d b}       | r = \frac {\ln b \frac \d {\d b} \left({b \ln x}\right) - b \ln x \frac \d {\d b} \ln b} {\left({\ln b}\right)^2}       | c = Quotient Rule for Derivatives }} {{eqn | r = \frac {\ln b \ln x - b \ln x \frac 1 b} {\left({\ln b}\right)^2}       | c = Derivative of Natural Logarithm, Derivative of Identity Function }} {{eqn | r = \frac {\ln x} {\ln b} \left({1 - \frac 1 {\ln b} }\right)       | c = simplifying }} {{eqn | r = \frac {\ln x} {\ln b^2} \left({\ln b - 1}\right)       | c = simplifying }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \dfrac {\d y} {\d b}        | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\ln x} {\ln b}       | r = \frac {\ln x} {\left({\ln b}\right)^2}       | c =  }} {{eqn | ll= \leadsto       | l = \ln b       | r = 1       | c = simplifying }} {{eqn | ll= \leadsto       | l = b       | r = e       | c = Definition of Natural Logarithm }} {{end-eqn}} To determine that $f$ is a minimum at this point, we differentiate again {{WRT|Differentiation}} $b$: {{begin-eqn}} {{eqn | l = \frac {\d^2 y} {\d b^2}       | r = \frac \d {\d b} \left({\frac {\ln x} {\ln b^2} \left({\ln b - 1}\right) }\right)       | c =  }} {{eqn | r = \frac {\ln x} b \left({\frac {\ln b - 2 \left({\ln b - 1}\right)} {\left({\ln b}\right)^3} }\right)       | c =  }} {{end-eqn}} Setting $b = e$ gives: :$\dfrac {\d^2 y} {\d b^2} \Bigg \vert_{b \mathop = e} = \dfrac {\ln x} e \dfrac {\left({1 - 2 \left({0}\right)}\right)} 1$ which works out to be (strictly) positive. From Second Derivative of Strictly Convex Real Function is Strictly Positive, $f$ is strictly convex at this point. Thus $f$ is a minimum. {{qed}}
Let $f$ be increasing. {{AimForCont}} $f$ is discontinuous at $c \in I$. Let: :$L^-$ denote $\displaystyle \lim_{x \mathop \to c^-} \map f x$ :$L$ denote $\map f c$ :$L^+$ denote $\displaystyle \lim_{x \mathop \to c^+} \map f x$ From Discontinuity of Monotonic Function is Jump Discontinuity, $L^-$ and $L^+$ are (finite) real numbers. Since $f$ is increasing: :$L^- \le L \le L^+$ Let $S = \openint {L^-} {L^+}$. If $L^- = L = L^+$, then $c$ is not a discontinuity, a contradiction. Thus $S$ is non-degenerate. From Non-Degenerate Real Interval is Infinite, $S$ is infinite. From Relative Difference between Infinite Set and Finite Set is Infinite, $S \setminus \set L$ is  non-empty. From Finite Subset of Metric Space is Closed, $\set L$ is closed. From Open Set minus Closed Set is Open, $S \setminus \set L$ is open. === Lemma === {{:Monotone Real Function with Everywhere Dense Image is Continuous/Lemma}}{{qed|lemma}} Thus: {{begin-eqn}} {{eqn | l = S \cap f \sqbrk I       | o = \subseteq       | r = \set L       | c =  Lemma }} {{eqn | ll= \leadsto       | l = \paren {S \cap f \sqbrk I} \setminus \set L       | o = \subseteq       | r = \set L \setminus \set L       | c = Set Difference over Subset }} {{eqn | r = \O       | c = Set Difference with Self is Empty Set }} {{eqn | ll= \leadsto       | l = \paren {S \cap f \sqbrk I} \setminus \set L       | r = \O       | c = Subset of Empty Set iff Empty }} {{eqn | ll= \leadsto       | l = \O       | r = \paren {S \cap f \sqbrk I \setminus \set L}       | c = Equality is Symmetric }} {{eqn | r = \paren { S \setminus \set L} \cap f \sqbrk I       | c = Intersection with Set Difference is Set Difference with Intersection }} {{eqn | r = \paren {\paren {\openint {L^-} L \cup \set L \cup \openint L {L^+} } \setminus \set L} \cap f \sqbrk I       | c = Break $S$ into $3$  disjoint open intervals }} {{eqn | r = \paren {\paren {\openint {L^-} L \cup \openint L {L^+} } \setminus \set L} \cap f \sqbrk I       | c = Set Difference with Union is Set Difference }} {{eqn | r = \paren {\openint {L^-} L \cup \openint L {L^+} } \cap f \sqbrk I       | c = Set Difference with Disjoint Set }} {{end-eqn}} Thus there exists a non-empty open subset of $f \sqbrk I$ that is disjoint from $Y$. From Open Set Characterization of Denseness, $f \sqbrk I$ is not everywhere dense in $J$. This contradicts the hypothesis that $f \sqbrk I$ is everywhere dense in $J$. Therefore $I$ contains no discontinuities, by Proof by Contradiction. The same argument, mutatis mutandis, proves the case where $f$ is decreasing Hence the result, by Proof by Cases. {{qed}} Category:Real Analysis s0183mib65aiwfu78bwvbiwqtjxc830
{{AimForCont}} $S \setminus T$ is a finite set. Then: {{begin-eqn}} {{eqn | l = S       | o = \subseteq       | r = S \cup T       | c = Set is Subset of Union }} {{eqn | r = \paren {S \setminus T} \cup T       | c = Set Difference Union Second Set is Union }} {{end-eqn}} By Union of Finite Sets is Finite, $S \cup T$ is a finite set. By Subset of Finite Set is Finite, $S$ must also be a finite set. But $S$ is an infinite set. This is a contradiction. Therefore $S \setminus T$ is an infinite set. {{qed}} Category:Set Difference Category:Infinite Sets ktntts6ukzwzcluwrgimxns9p2ils5a
{{begin-eqn}} {{eqn | l = A \setminus S       | r = A \cap \map \complement S       | c = Set Difference as Intersection with Complement }} {{eqn | o = \subseteq       | r = B \cap \map \complement S       | c = Corollary to Set Intersection Preserves Subsets }} {{eqn | r = B \setminus S       | c = Set Difference as Intersection with Complement }} {{end-eqn}} {{qed}} Category:Set Difference jn9zvcha8xojzr8k7s5voxtb7nljqts
From Metric Space is Hausdorff, $M$ is Hausdorff. From Finite Subspace of Hausdorff Space is Closed, $S$ is closed. {{qed}} Category:Closed Sets Category:Metric Spaces o2ifexi1cunsg6pq68v9xoyi184umf8
From Discontinuity of Monotonic Function is Jump Discontinuity, $\displaystyle \lim_{x \mathop \to c^-} \map f x$ and $\displaystyle \lim_{x \mathop \to c^+} \map f x$ are finite. Since $f$ is increasing: :$\displaystyle \lim_{x \mathop \to c^-} \map f x < \lim_{x \mathop \to c^+} \map f x$ Suppose $z \in \displaystyle \openint {\lim_{x \mathop \to c^-} \map f x} {\lim_{x \mathop \to c^+} \map f x} \cap f \sqbrk I$. Then: :$\displaystyle \exists t \in I : \map f t \in \openint {\lim_{x \mathop \to c^-} \map f x} {\lim_{x \mathop \to c^+} \map f x}$ === Case 1 : $t < c$ === Suppose that $t < c$. {{begin-eqn}} {{eqn | l = t       | o = <       | r = c }} {{eqn | ll= \leadsto       | l = \map f t       | o = \le       | r = \lim_{x \mathop \to c^-} \map f x       | c = {{Defof|Increasing Real Function}} }} {{eqn | ll= \leadsto       | l = \map f t       | o = \notin       | r = \openint {\lim_{x \mathop \to c^-} \map f x} {\lim_{x \mathop \to c^+} \map f x}       | c = {{Defof|Open Real Interval}} }} {{end-eqn}} So we may discard this case. {{qed|lemma}} === Case 2 : $t = c$ === Suppose that $t = c$. {{begin-eqn}} {{eqn | l = t       | r = c }} {{eqn | ll= \leadsto       | l = \map f t       | r = \map f c       | c = {{Defof|Mapping}} }} {{eqn | ll= \leadsto       | l = \map f t       | o = \in       | r = \openint {\lim_{x \mathop \to c^-} \map f x} {\lim_{x \mathop \to c^+} \map f x}       | c = {{Defof|Open Real Interval}} }} {{end-eqn}} So the proposition holds in this case. {{qed|lemma}} === Case 3 : $c < t$ === Suppose that $t > c$. {{begin-eqn}} {{eqn | l = t       | o = >       | r = c }} {{eqn | ll= \leadsto       | l = \map f t       | o = \ge       | r = \lim_{x \mathop \to c^+} \map f x       | c = {{Defof|Increasing Real Function}} }} {{eqn | ll= \leadsto       | l = \map f t       | o = \notin       | r = \openint {\lim_{x \mathop \to c^-} \map f x} {\lim_{x \mathop \to c^+} \map f x}       | c = {{Defof|Open Real Interval}} }} {{end-eqn}} So we may discard this case. {{qed|lemma}} So $\map f t = c$, by Proof by Cases. Thus: {{begin-eqn}} {{eqn | l = z       | r = \map f t       | c = Definition of $t$ }} {{eqn | r = c       | c = from above }} {{eqn | ll = \leadsto       | l = z       | o = \in       | r = \set {\map f c} }} {{end-eqn}} Thus: :$\displaystyle z \in \openint {\lim_{x \mathop \to c^-} \map f x} {\lim_{x \mathop \to c^+} \map f x} \cap f \sqbrk I \implies z \in \set {\map f c}$ Hence the result, by definition of subset. {{qed}} Category:Real Analysis 4ghkaxwvk86zqyav0esgqfzxchrawyp
From Continuity of Root Function and Combination Theorem for Continuous Functions: :$\forall n \in \N : \map {f_n} x$ is continuous on $K$ From Defining Sequence of Natural Logarithm is Convergent, $\sequence {f_n}$ is pointwise convergent to $\ln$. From Natural Logarithm Function is Continuous, $\ln$ is continuous on $K$. From Defining Sequence of Natural Logarithm is Strictly Decreasing: :$\forall x \in K: \sequence {\map {f_n} x}$ is monotone. From Dini's Theorem, $\sequence {f_n}$ is uniformly convergent to $\ln$. Hence the result. {{qed}} Category:Natural Logarithms ezp28yikfargl93un1yx8ymacc4q5zj
{{begin-eqn}} {{eqn | l = x       | o = <       | r = 1 }} {{eqn | ll= \leadsto       | l = \frac 1 x       | o = <       | r = 1       | c = Ordering of Reciprocals }} {{eqn | ll= \leadsto       | l = \frac y x       | o = <       | r = y       | c = Real Number Ordering is Compatible with Multiplication }} {{end-eqn}} {{qed}} Category:Real Multiplication Category:Inequalities aylhlcergqss4bgexgl50e85n07nkrf
By Existence of Non-Empty Finite Suprema in Join Semilattice: :for every $A \in {\it Fin}\left({S}\right)$ if $A \ne \varnothing$, then $A$ admits a supremum. By definition of non-empty set: :$\exists a: a \in X$ By definitions of subset and singleton: :$\left\{ {x}\right\} \subseteq X$ By Singleton is Finite: :$\left\{ {x}\right\}$ is finite By definitions of non-empty set and singleton: :$\left\{ {x}\right\} \ne \varnothing$ By definition of ${\it Fin}$: :$\sup \left\{ {x}\right\} \in \left\{ {\sup A: A \in {\it Fin}\left({X}\right) \land A \ne \varnothing}\right\}$ Thus by definition: :$\left\{ {\sup A: A \in {\it Fin}\left({X}\right) \land A \ne \varnothing}\right\}$ is a non-empty set. Let $x, y \in \left\{ {\sup A: A \in {\it Fin}\left({X}\right) \land A \ne \varnothing}\right\}$ Then :$\exists A \in {\it Fin}\left({X}\right): x = \sup A$ and $A \ne \varnothing$ and :$\exists B \in {\it Fin}\left({X}\right): y = \sup B$ and $B \ne \varnothing$ By Finite Union of Finite Sets is Finite: :$A \cup B$ is finite By Union of Subsets is Subset: :$A \cup B \subseteq X$ By definitions of non-empty set and union: :$A \cup B \ne \varnothing$ By definition of $\it Fin$: :$\sup \left({A \cup B}\right) \in \left\{ {\sup A: A \in {\it Fin}\left({X}\right) \land A \ne \varnothing}\right\}$ By Set is Subset of Union: :$A \subseteq A \cup B$ and $B \subseteq A \cup B$ Thus by Supremum of Subset: :$x \preceq \sup \left({A \cup B}\right)$ and $y \preceq \sup \left({A \cup B}\right)$ Thus by definition: :$\left\{ {\sup A: A \in {\it Fin}\left({X}\right) \land A \ne \varnothing}\right\}$ is directed. {{qed}}
Define a mapping $f: \set x \to \N_{< 1}$: :$\map f x = 0$ By definition of singleton: :$\forall y, z \in \set x: \map f y = \map f z \implies y = z$ By definition: :$f$ is an injection. By definition of initial segment of natural numbers: :$\N_{< 1} = \set 0$ By definition of $f$: :$\forall n \in N_{< 1}: \exists z \in \set x: \map f z = n$ By definition: :$f$ is a surjection. By definition: :$f$ is a bijection. By definition of set equivalence: :$\set x \sim \N_{< 1}$ Thus by definition: :$\set x$ is a finite set. {{qed}}
From the definition, $T = \struct {S, \tau}$ is a normal space {{iff}}: :$\struct {S, \tau}$ is a $T_4$ space :$\struct {S, \tau}$ is a $T_1$ (Fréchet) space. From Separation Properties Preserved in Subspace, any subspace of a $T_1$ space is also a $T_1$ space. From T4 Property Preserved in Closed Subspace, any closed subspace of a $T_4$ space is also a $T_4$ space. Hence the result. {{qed}}