hcju/proofwikips huggingface dataset:
Euclid's Lemma for Prime Divisors/General Result/Proof 2
== Lemma == {{:Euclid's Lemma for Prime Divisors/General Result}} == Proof ==  Proof by induction: For all $r \in \N_{>0}$, let $\map P r$ be the proposition: :$\displaystyle p \divides \prod_{i \mathop = 1}^r a_i \implies \exists i \in \closedint 1 r: p \divides a_i$ $\map P 1$ is true, as this just says $p \divides a_1 \implies p \divides a_1$. === Basis for the Induction === $\map P 2$ is the case: :$p \divides a_1 a_2 \implies p \divides a_2$ or $p \divides a_2$ which is proved in Euclid's Lemma for Prime Divisors. This is our basis for the induction. === Induction Hypothesis === Now we need to show 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 our induction hypothesis: :$\displaystyle p \divides \prod_{i \mathop = 1}^k a_i \implies \exists i \in \closedint 1 k: p \divides a_i$ Then we need to show: :$\displaystyle p \divides \prod_{i \mathop = 1}^{k + 1} a_i \implies \exists i \in \closedint 1 {k + 1}: p \divides a_i$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = p       | o = \divides       | r = a_1 a_2 \ldots a_{k + 1}       | c =  }} {{eqn | ll= \leadsto       | l = p       | o = \divides       | r = \paren {a_1 a_2 \ldots a_k} \paren {a_{k + 1} }       | c = }} {{eqn | ll= \leadsto       | l = p       | o = \divides       | r = a_1 a_2 \ldots a_k \lor p \divides a_{k + 1}       | c = Basis for the Induction }} {{eqn | ll= \leadsto       | l = p       | o = \divides       | r = a_1 \lor p \divides a_2 \lor \ldots \lor p \divides a_k \lor p \divides a_{k + 1}       | c = Induction Hypothesis }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall r \in \N: p \divides \prod_{i \mathop = 1}^r a_i \implies \exists i \in \closedint 1 r: p \divides a_i$   {{Namedfor|Euclid}} == Sources == * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Euclid's Lemma for Prime Divisors/Proof 2|next = Fundamental Theorem of Arithmetic}}: Chapter $3$: The Integers: $\S 12$. Primes: Theorem $21 \ \text{(ii)}$ * {{BookReference|Number Theory|1971|George E. Andrews|prev = Euclid's Lemma for Prime Divisors/Proof 2|next = Euclidean Algorithm/Examples/527 and 765}}: $\text {2-2}$ Divisibility: Corollary $\text {2-4}$ * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Euclid's Lemma for Prime Divisors/Proof 2|next = Fundamental Theorem of Arithmetic}}: $\S 12.6$: Highest common factors and Euclid's algorithm * {{BookReference|Computability and Unsolvability|1982|Martin Davis|ed = 2nd|edpage = Second Edition|prev = Euclid's Lemma for Prime Divisors/Proof 2|next = Fundamental Theorem of Arithmetic}}: Appendix $1$: Some Results from the Elementary Theory of Numbers: Corollary $9$
Euclid's Lemma for Irreducible Elements/General Result
== Lemma == Let $\struct {D, +, \times}$ be a Euclidean domain whose unity is $1$.  Let $p$ be an irreducible element of $D$. Let $n \in D$ such that: :$\displaystyle n = \prod_{i \mathop = 1}^r a_i$ where $a_i \in D$ for all $i: 1 \le i \le r$. If $p$ divides $n$, then $p$ divides $a_i$ for some $i$. That is: :$p \divides a_1 a_2 \ldots a_n \implies p \divides a_1 \lor p \divides a_2 \lor \cdots \lor p \divides a_n$  == Proof == Proof by induction: For all $r \in \N_{>0}$, let $\map P r$ be the proposition: :$\displaystyle p \divides \prod_{i \mathop = 1}^r a_i \implies \exists i \in \closedint 1 r: p \divides a_i$ $\map P 1$ is true, as this just says $p \divides a_1 \implies p \divides a_1$. === Basis for the Induction === $\map P 2$ is the case: :$p \divides a_1 a_2 \implies p \divides a_2$ or $p \divides a_2$ which is proved in Euclid's Lemma for Irreducible Elements. This is our basis for the induction. === Induction Hypothesis === Now we need to show 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 our induction hypothesis: :$\displaystyle p \divides \prod_{i \mathop = 1}^k a_i \implies \exists i \in \closedint 1 k: p \divides a_i$ Then we need to show: :$\displaystyle p \divides \prod_{i \mathop = 1}^{k + 1} a_i \implies \exists i \in \closedint 1 {k + 1}: p \divides a_i$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = p       | o = \divides       | r = a_1 a_2 \ldots a_{k + 1}       | c =  }} {{eqn | ll= \leadsto       | l = p       | o = \divides       | r = \paren {a_1 a_2 \ldots a_k} \paren {a_{k + 1} }       | c =  }} {{eqn | ll= \leadsto       | l = p       | o = \divides       | r = a_1 a_2 \ldots a_k \lor p \divides a_{k + 1}       | c = Basis for the Induction }} {{eqn | ll= \leadsto       | l = p       | o = \divides       | r = a_1 \lor p \divides a_2 \lor \ldots \lor p \divides a_k \lor p \divides a_{k + 1}       | c = Induction Hypothesis }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall r \in \N: p \divides \prod_{i \mathop = 1}^r a_i \implies \exists i \in \closedint 1 r: p \divides a_i$  {{Namedfor|Euclid}} == Also see == * Euclid's Lemma for Prime Divisors, for the usual statement of this result, which is this lemma as applied specifically to the integers. == Sources == * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Euclid's Lemma for Irreducible Elements|next = Polynomial Forms over Field form Principal Ideal Domain/Corollary 2}}: Chapter $6$: Polynomials and Euclidean Rings: $\S 29$. Irreducible elements: Theorem $56 \ \text{(ii)}$
Common Divisor Divides Integer Combination
== Corollary to Common Divisor in Integral Domain Divides Linear Combination ==  Let $c$ be a common divisor of two integers $a$ and $b$. That is: :$a, b, c \in \Z: c \divides a \land c \divides b$ Then $c$ divides any integer combination of $a$ and $b$: :$\forall p, q \in \Z: c \divides \paren {p a + q b}$  === Corollary === {{:Common Divisor Divides Integer Combination/Corollary}} === General Result === {{:Common Divisor Divides Integer Combination/General Result}} == Proof 1 == {{:Common Divisor Divides Integer Combination/Proof 1}} == Proof 2 == {{:Common Divisor Divides Integer Combination/Proof 2}} == Sources == * {{BookReference |An Introduction to the Theory of Numbers|1979|G.H. Hardy|author2 = E.M. Wright|ed = 5th|edpage = Fifth Edition|prev = Multiple of Divisor Divides Multiple|next = Definition:Prime Number/Definition 4}}: $\text I$: The Series of Primes: $1.1$ Divisibility of integers * {{BookReference|Elementary Number Theory|1980|David M. Burton|ed = revised|edpage = Revised Printing|prev = Absolute Value of Integer is not less than Divisors|next = Common Divisor Divides Integer Combination/General Result}}: Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Theorem $2 \text{-} 2 \ (7)$ * {{BookReference|A Course in Number Theory|1994|H.E. Rose|ed = 2nd|edpage = Second Edition|prev = Divisor Relation is Transitive|next = Definition:Greatest Common Divisor/Integers/Definition 2}}: $1$ Divisibility: $1.1$ The Euclidean algorithm and unique factorization: $\text {(v)}$
Divisor Relation is Antisymmetric/Corollary/Proof 2
== Corollary to Divisor Relation is Antisymmetric == {{:Divisor Relation is Antisymmetric/Corollary}} == Proof ==  Let $a \divides b$ and $b \divides a$. Then by definition of divisor: :$\exists c, d \in \Z: a c = b, b d = a$ Thus: {{begin-eqn}} {{eqn | l = a c d       | r = a       | c =  }} {{eqn | ll= \leadsto       | l = c d       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = d       | r = \pm 1       | c = Divisors of One }} {{eqn | ll= \leadsto       | l = a       | r = \pm b       | c = as $a = b d$ }} {{end-eqn}}   == Sources == * {{BookReference|Lectures in Abstract Algebra|1951|Nathan Jacobson|volume = I|subtitle = Basic Concepts|prev = Divisor Relation is Transitive|next = Definition:Greatest Common Divisor/Integers/Definition 2}}: Introduction $\S 6$: The division process in $I$ * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Divisors of One|next = Absolute Value of Integer is not less than Divisors/Corollary}}: Chapter $3$: The Integers: $\S 10$. Divisibility: Theorem $17 \ \text{(ii)}$ * {{BookReference|Algebra Volume 1|1982|P.M. Cohn|edpage = Second Edition|ed = 2nd|prev = Divisor Divides Multiple|next = Definition:Associate of Integer}}: Chapter $2$: Integers and natural numbers: $\S 2.2$: Divisibility and factorization in $\mathbf Z$
Polynomial Forms over Field form Principal Ideal Domain/Corollary 3
== Corollary to Polynomial Forms over Field form Principal Ideal Domain == Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$. Let $X$ be transcendental over $F$. Let $F \sqbrk X$ be the ring of polynomials in $X$ over $F$. Then $F \sqbrk X$ is a unique factorization domain.  == Proof == We have the result Principal Ideal Domain is Unique Factorization Domain. The result then follows from Polynomial Forms over Field form Principal Ideal Domain.  == Sources == * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Polynomial Forms over Field form Principal Ideal Domain/Proof 1|next = Polynomial Forms over Field form Principal Ideal Domain/Corollary 1}}: $\S 65.2$ Some properties of $F \sqbrk X$, where $F$ is a field
Polynomial Forms over Field form Principal Ideal Domain/Corollary 1
== Corollary to Polynomial Forms over Field form Principal Ideal Domain == Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$. Let $X$ be transcendental over $F$. Let $F \sqbrk X$ be the ring of polynomials in $X$ over $F$.  Let $f$ be an irreducible element of $F \sqbrk X$. Then $F \sqbrk X / \ideal f$ is a field, where $\ideal f$ denotes the ideal generated by $f$.  == Proof == It follows from Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal that $\ideal f$ is maximal for irreducible $f$. Therefore by Maximal Ideal iff Quotient Ring is Field, $F \sqbrk X / \ideal f$ is a field.  == Sources == * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Polynomial Forms over Field form Unique Factorization Domain|next = Field of Quotients of Ring of Polynomial Forms on Reals that yields Complex Numbers}}: $\S 65.3$ Some properties of $F \sqbrk X$, where $F$ is a field
No Arithmetic Sequence of 4 Primes with Common Difference 2
== Definition == There exist no $n \in \Z_{>0}$ such that $n, n + 2, n + 4, n + 6$ are all prime. === Corollary === {{:No Arithmetic Sequence of 4 Primes with Common Difference 2/Corollary}} == Proof ==  $S$ is a set of $4$ prime numbers of the form $n, n + 2, n + 4, n + 6$. $S$ must contain as a subset a prime triplet. From Prime Triplet is Unique, the only one of these is $\set {3, 5, 7}$. The only sets of the form $\set {n, n + 2, n + 4, n + 6}$ containing $\set {3, 5, 7}$ are: :$(1): \quad \set {1, 3, 5, 7}$: as $1$ is by convention not a prime, then this is not $S$. :$(2): \quad \set {3, 5, 7, 9}$: as $9 = 3 \times 3$ is not a prime, then this is not $S$. There are no more possible $\set {n, n + 2, n + 4, n + 6}$ all prime. Hence, by Proof by Contradiction, $S$ does not exist.
Cardinal Number Less than Ordinal/Corollary
== Corollary to Cardinal Number Less than Ordinal ==  Let $x$ be an ordinal. Let $\left|{x}\right|$ denote the cardinal number of $x$. Then: :$\left|{x}\right| \le x$  == Proof == By Set Equivalence is Equivalence Relation: : $x \sim x$ By Cardinal Number Less than Ordinal: : $\left|{x}\right| \le x$  == Sources == * {{BookReference|Introduction to Axiomatic Set Theory|1971|Gaisi Takeuti|author2=Wilson M. Zaring}}: $\S 10.13$
Set of Words Generates Group/Corollary
== Corollary to Set of Words Generates Group == Let $G$ be a group. Let $T \subseteq G$. Let $\map W T$ be the set of words of $T$.  If $T$ is closed under taking inverses, then $\map W T$ is a subgroup of $G$.  == Proof == This follows directly from Set of Words Generates Group and the fact that $T$ has the same properties as $\hat S$ in that result.
Subset Product within Semigroup is Associative/Corollary
== Corollary to Subset Product within Semigroup is Associative ==  Let $\struct {S, \circ}$ be a semigroup. Then: {{begin-eqn}} {{eqn | l = x \paren {y S}       | r = \paren {x y} S }} {{eqn | l = x \paren {S y}       | r = \paren {x S} y }} {{eqn | l = \paren {S x} y       | r = S \paren {x y} }} {{end-eqn}}  == Proof == From the definition of Subset Product with Singleton: {{begin-eqn}} {{eqn | l = x \paren {y S}       | r = \set x \paren {\set y S} }} {{eqn | l = x \paren {S y}       | r = \set x \paren {S \set y} }} {{eqn | l = \paren {S x} y       | r = \paren {S \set x} \set y }} {{end-eqn}} The result then follows directly from Subset Product within Semigroup is Associative.  == Also see == * Subset Product within Commutative Structure is Commutative == Sources == * {{BookReference|Sets and Groups|1965|J.A. Green|prev = Definition:Subset Product with Singleton|next = Definition:Right Coset}}: $\S 6.1$. The quotient sets of a subgroup
Partition Topology is T4
== Corollary to Partition Topology is T5 == Let $S$ be a set and let $\PP$ be a partition on $S$ which is not the (trivial) partition of singletons. Let $T = \struct {S, \tau}$ be the partition space whose basis is $\PP$. Then:  :$T$ is a $T_4$ space.  == Proof == We have that the Partition Topology is $T_5$. We also have that a $T_5$ Space is $T_4$ Space. The result follows.  == Sources == * {{BookReference|Counterexamples in Topology|1978|Lynn Arthur Steen|author2 = J. Arthur Seebach, Jr.|ed = 2nd|edpage = Second Edition|prev = Partition Topology is T3 1/2|next = Partition Topology is T5}}: Part $\text {II}$: Counterexamples: $5$. Partition Topology: $2$
Pointwise Addition on Real-Valued Functions is Commutative
== Definition == Let $S$ be a set.  Let $f, g: S \to \R$ be real-valued functions. Let $f + g: S \to \R$ denote the pointwise sum of $f$ and $g$. Then: :$f + g = g + f$  That is, pointwise addition of real-valued functions is commutative. == Proof == {{begin-eqn}} {{eqn | lo= \forall x \in S:       | l = \map {\paren {f + g} } x       | r = \map f x + \map g x       | c = {{Defof|Pointwise Addition of Real-Valued Functions}} }} {{eqn | r = \map g x + \map f x       | c = Real Addition is Commutative }} {{eqn | r = \map {\paren {g + f} } x       | c = {{Defof|Pointwise Addition of Real-Valued Functions}} }} {{end-eqn}}
Compact Subspace of Hausdorff Space is Closed/Corollary
== Corollary to Compact Subspace of Hausdorff Space is Closed ==  A finite subspace of a Hausdorff space is closed.  == Proof == Follows directly from: : Finite Topological Space is Compact and: : Compact Subspace of Hausdorff Space is Closed.
Brahmagupta-Fibonacci Identity/Extension
== Extension to Brahmagupta-Fibonacci Identity ==  Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n$ be integers. Then: :$\displaystyle \prod_{j \mathop = 1}^n \paren { {a_j}^2 + {b_j}^2} = c^2 + d^2$ where $c, d \in \Z$. That is: the product of any number of sums of two squares is also a sum of two squares.  === General Version === {{:Brahmagupta-Fibonacci Identity/Extension/General}} == Proof 1 == {{:Brahmagupta-Fibonacci Identity/Extension/Proof 1}} == Proof 2 == {{:Brahmagupta-Fibonacci Identity/Extension/Proof 2}} == Proof 3 == {{:Brahmagupta-Fibonacci Identity/Extension/Proof 3}}
Sum of Squared Deviations from Mean/Corollary 1
== Theorem  == Let $x_1, x_2, \ldots, x_n$ be real data about some quantitative variable. Let $\overline x$ be the arithmetic mean of the above data. Then:   :$\displaystyle \sum_{i \mathop = 1}^n \paren {x_i - \overline x}^2 = \sum_{i \mathop = 1}^n x_i^2 - n \overline x^2$   == Proof == {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^n \paren {x_i - \overline x}^2       | r = \sum_{i \mathop = 1}^n \paren {x_i^2 - \overline x^2}       | c = Sum of Squared Deviations from Mean }} {{eqn | r = \sum_{i \mathop = 1}^n x_i^2 - \sum_{i \mathop = 1}^n \overline x^2       | c = Summation is Linear }} {{eqn | r = \sum_{i \mathop = 1}^n x_i^2 - n \overline x^2       | c = Sum of Identical Terms }} {{end-eqn}}
Limit of Sequence to Zero Distance Point/Corollary 1
== Corollary to Limit of Sequence to Zero Distance Point == Let $S$ be a non-empty subset of $\R$. Let the distance $\map d {\xi, S} = 0$ for some $\xi \in \R$.  If $S$ is bounded above, then there exists a sequence $\sequence {x_n}$ in $S$ such that $\displaystyle \lim_{n \mathop \to \infty} x_n = \sup S$.  == Proof == Let  $\xi = \sup S$. Then from Distance from Subset of Real Numbers: :$\map d {\xi, S} = 0$ The result then follows directly from Limit of Sequence to Zero Distance Point. Note that the terms of this sequence do not necessarily have to be distinct.  == Sources == * {{BookReference|Mathematical Analysis: A Straightforward Approach|1977|K.G. Binmore|prev = Limit of Sequence to Zero Distance Point|next = Limit of Sequence to Zero Distance Point/Corollary 2}}: $\S 4$: Convergent Sequences: Exercise $\S 4.29 \ (6)$
Derivative of Function of Constant Multiple/Corollary
== Corollary of Derivative of Function of Constant Multiple == Let $f$ be a real function which is differentiable on $\R$.  Let $a, b \in \R$ be constants. Then: :$\map {D_x} {\map f {a x + b} } = a \, \map {D_{a x + b} } {\map f {a x + b} }$  == Proof == First it is shown that $\map {D_x} {a x + b} = a$: {{begin-eqn}} {{eqn | n = 1       | l = \map {D_x} {a x + b}       | r = \map {D_x} {a x} + \map {D_x} b       | c = Sum Rule for Derivatives }} {{eqn | r = a + 0       | c = Derivative of Function of Constant Multiple and Derivative of Constant }} {{eqn | r = a       | c =  }} {{end-eqn}} Next: {{begin-eqn}} {{eqn | l = \map {D_x} {\map f {a x + b} }       | r = \map {D_x} {a x + b} \, \map {D_{a x + b} } {\map f {a x + b} }       | c = Chain Rule for Derivatives }} {{eqn | r = a \, \map {D_{a x + b} } {\map f {a x + b} }       | c = from $(1)$ }} {{end-eqn}}
Open Real Interval is Open Set/Corollary
== Corollary to Open Real Interval is Open Set ==  Let $\R$ be the real number line considered as an Euclidean space. Let $A := \openint a \infty \subset \R$ be an open interval of $\R$. Let $B := \openint {-\infty} b \subset \R$ be an open interval of $\R$. Then both $A$ and $B$ are open sets of $\R$.  == Proof == From Open Real Interval is Open Set we have that for any $c \in \openint a b$ there exists an open $\epsilon$-ball of $c$ lying wholly within $\openint a b$. When either of $a \to -\infty$ or $b \to \infty$ the result still holds. The result follows by definition of open set.
Negative of Absolute Value/Corollary 1
== Corollary to Negative of Absolute Value == Let $x, y \in \R$ be a real numbers. Let $\size x$ be the absolute value of $x$. Then:  :$\size x < y \iff -y < x < y$  == Proof == === Necessary Condition === Let $\size x < y$. Then from Negative of Absolute Value: :$x \le \size x$ and: :$\size x \ge -x$ So $x < y$ and $-x < y$, and so $x > -y$ from Ordering of Inverses in Ordered Monoid. It follows that $-y < x < y$. {{qed|lemma}} === Sufficient Condition === Let $-y < x < y$. Then $x < y$ and $-x < y$. For all $x$: :$\size x = x$ or: :$\size x = -x$ Thus it follows that $\size x < y$.  == Sources == * {{BookReference|Mathematical Analysis: A Straightforward Approach|1977|K.G. Binmore|prev = Reverse Triangle Inequality/Real and Complex Fields/Examples/6 - (-1)|next = Reverse Triangle Inequality/Real and Complex Fields/Proof 2}}: $\S 1$: Real Numbers: Exercise $\S 1.20 \ (1)$
Continuous Image of Connected Space is Connected/Corollary 1
== Corollary to Continuous Image of Connected Space is Connected ==  Connectedness is a topological property.  == Proof == Follows directly from Continuous Image of Connected Space is Connected and the definition of topological property.  == Sources == * {{BookReference|Introduction to Metric and Topological Spaces|1975|W.A. Sutherland|prev = Continuous Image of Connected Space is Connected/Proof 1|next = Continuous Image of Connected Space is Connected/Corollary 2}}: $6.2$: Connectedness: Corollary $6.2.11$
Continuous Image of Connected Space is Connected/Corollary 2
== Corollary to Continuous Image of Connected Space is Connected ==  Let $T$ be a connected topological space. Let $f: T \to \R$ be a continuous real-valued mapping. Then $f \left({T}\right)$ is a real interval.  == Proof == From Continuous Image of Connected Space is Connected, the continuous image of the connected space $T$ is connected. The result follows from Subset of Real Numbers is Interval iff Connected.  == Sources == * {{BookReference|Introduction to Metric and Topological Spaces|1975|W.A. Sutherland|prev = Continuous Image of Connected Space is Connected/Corollary 1|next = Continuous Image of Connected Space is Connected/Corollary 3}}: $6.2$: Connectedness: Corollary $6.2.13$
Image of Countable Set under Mapping is Countable
Image of Countable Set under Mapping is Countable 0 20635 336022 189918 2018-01-12T08:25:39Z Barto 3079 stable link to image of mapping wikitext text/x-wiki == Definition == Let $S$ be a countable set. Let $T$ be a set. Let $f: S \to T$ be a mapping. Then the image of $f$ is countable. == Proof == Let $A$ be the image of $f$. Let $g: S \to A$ be the restriction of $f$ to the cartesian product $S \times A$. By Surjection by Restriction of Codomain, $g$ is a surjection. By Surjection from Natural Numbers iff Countable, there exists a surjection $\phi: \N \to S$. Since the composition of surjections is a surjection, $g \circ \phi: \N \to A$ is a surjection. Hence, by Surjection from Natural Numbers iff Countable, it follows that $A$ is countable.
Infinite Sequence in Countably Compact Space has Accumulation Point
== Corollary to Countably Infinite Set in Countably Compact Space has Omega-Accumulation Point == Let $T = \struct {S, \tau}$ be a countably compact topological space.  Let $\sequence {x_n}_{n \mathop \in \N}$ be an infinite sequence in $S$. Then $\sequence {x_n}$ has an accumulation point in $T$.  == Proof == Let $A \subseteq S$ be the range of $\sequence {x_n}$: :$A = \set {x_n: n \in \N}$ If $A$ is finite, then consider the equality: :$\displaystyle \N = \bigcup_{y \mathop \in A} \set {n \in \N: x_n = y}$ Therefore, there exists a $y \in A$ such that $\set {n \in \N: x_n = y}$ is an infinite set. Hence, $y$ is an accumulation point of $\sequence {x_n}$. Otherwise, $A$ is countably infinite. Then $A$ has an $\omega$-accumulation point in $T$. It follows that $\sequence {x_n}$ has an accumulation point in $T$. {{explain|Link to a result that states that if $T$ has an $\omega$-accumulation point, then $\sequence {x_n}$ has an accumulation point}}  == Sources == * {{BookReference|Counterexamples in Topology|1978|Lynn Arthur Steen|author2 = J. Arthur Seebach, Jr.|ed = 2nd|edpage = Second Edition|prev = Definition:Countable Finite Intersection Axiom|next = Countably Compact Space satisfies Countable Finite Intersection Axiom}}: Part $\text I$: Basic Definitions: Section $3$: Compactness: Global Compactness Properties
L'Hôpital's Rule/Corollary 1
== Corollary to L'Hôpital's Rule == Let $f$ and $g$ be real functions which are continuous on the closed interval $\closedint a b$ and differentiable on the open interval $\openint a b$. Suppose that $\forall x \in \openint a b: \map {g^{\prime} } x \ne 0$.  Suppose that $\exists c \in \openint a b: \map f c = \map g c = 0$. Then: :$\displaystyle \lim_{x \mathop \to c} \frac {\map f x} {\map g x} = \lim_{x \mathop \to c} \frac {\map {f^{\prime} } x} {\map {g^{\prime} } x}$ provided that the second limit exists.  == Proof == This follows directly from the definition of limit. If $\displaystyle \lim_{x \mathop \to c} \frac {\map {f^{\prime} } x} {\map {g^{\prime} } x}$ exists, it follows that: :$\displaystyle \lim_{x \mathop \to c} \frac {\map {f^{\prime} } x} {\map {g^{\prime} } x} = \lim_{x \to c^+} \frac {\map {f^{\prime} } x} {\map {g^{\prime} } x}$ That is, if there exists such a limit, it is also a limit from the right.  {{Namedfor|Guillaume de l'Hôpital|cat = L'Hôpital}} == Historical Note == {{:L'Hôpital's Rule/Historical Note}}