hcju/proofwikiqa huggingface dataset:
From the $1$st logarithm form: :$\displaystyle \int \frac {\d x} {a^2 - x^2} = \begin {cases} \dfrac 1 {2 a} \map \ln {\dfrac {a + x} {a - x} } + C & : \size x < a\\ & \\  \dfrac 1 {2 a} \map \ln {\dfrac {x + a} {x - a} } + C & : \size x > a \\ & \\ \text {undefined} & : \size x = a \end {cases}$ From Primitive of Reciprocal of a squared minus x squared: Logarithm Form: Lemma: :$\map \ln {\dfrac {a + x} {a - x} }$ is defined {{iff}} $\size x < a$ :$\map \ln {\dfrac {x + a} {x - a} }$ is defined {{iff}} $\size x > a$ Let $\size x < a$. Then $\map \ln {\dfrac {a + x} {a - x} }$ is defined. We have that: {{begin-eqn}} {{eqn | l = \dfrac {a + x} {a - x}       | o = >       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {a + x} {a - x}       | r = \size {\dfrac {a + x} {a - x} }       | c =  }} {{end-eqn}} So the result holds for $\size x < a$. Let $\size x > a$. Then $\map \ln {\dfrac {x + a} {x - a} }$ is defined. We have: We have that: {{begin-eqn}} {{eqn | l = \dfrac {x + a} {x - a}       | r = -\dfrac {a + x} {a - x}       | c =  }} {{eqn | o = <       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {x + a} {x - a}       | r = \size {\dfrac {a + x} {a - x} }       | c =  }} {{end-eqn}} The result follows. {{qed}}
Let $\size x > a$. Let: {{begin-eqn}} {{eqn | l = u       | r = \coth^{-1} {\frac x a}       | c = {{Defof|Real Inverse Hyperbolic Cotangent}}, which is defined where $\size {\dfrac x a} > 1$ }} {{eqn | ll= \leadsto       | l = x       | r = a \coth u       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d x} {\d u}       | r = -a \csch^2 u       | c = Derivative of Hyperbolic Cotangent }} {{eqn | ll= \leadsto       | l = \int \frac 1 {x^2 - a^2} \rd x       | r = \int \frac {-a \csch^2 u} {a^2 \coth^2 u - a^2} \rd u       | c = Integration by Substitution }} {{eqn | r = -\frac a {a^2} \int \frac {\csch^2 u} {\coth^2 u - 1} \rd u       | c = Primitive of Constant Multiple of Function }} {{eqn | r = -\frac 1 a \int \frac {\csch^2 u} {\csch^2 u} \rd u       | c = Difference of Squares of Hyperbolic Cotangent and Cosecant }} {{eqn | r = -\frac 1 a \int \rd u }} {{eqn | r = -\frac 1 a u + C       | c = Integral of Constant }} {{eqn | r = -\frac 1 a \coth^{-1} {\frac x a} + C       | c = Definition of $u$ }} {{end-eqn}}
=== Case where $\size x < a$ === {{:Primitive of Reciprocal of x squared minus a squared/Logarithm Form 1/size of x less than a}} === Case where $\size x > a$ === {{:Primitive of Reciprocal of x squared minus a squared/Logarithm Form 1/size of x greater than a}}
Let $\size x < a$. Then: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^2 - a^2}       | r = -\frac 1 a \tanh^{-1} {\frac x a} + C       | c = Primitive of $\dfrac 1 {x^2 - a^2}$: $\tanh^{-1}$ form }} {{eqn | r = -\frac 1 a \paren {\dfrac 1 2 \map \ln {\dfrac {a + x} {a - x} } } + C       | c = $\tanh^{-1} \dfrac x a$ in Logarithm Form }} {{eqn | r = -\dfrac 1 {2 a} \map \ln {\dfrac {a + x} {a - x} } + C       | c = simplifying }} {{eqn | r = \dfrac 1 {2 a} \map \ln {\dfrac {a - x} {a + x} } + C       | c = Logarithm of Reciprocal }} {{end-eqn}}
Let $\size x > a$. Then: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^2 - a^2}       | r = -\frac 1 a \coth^{-1} {\frac x a} + C       | c = Primitive of $\dfrac 1 {x^2 - a^2}$: $\coth^{-1}$ form }} {{eqn | r = -\frac 1 a \paren {\dfrac 1 2 \map \ln {\dfrac {x + a} {x - a} } } + C       | c = $\coth^{-1} \dfrac x a$  in Logarithm Form }} {{eqn | r = -\dfrac 1 {2 a} \map \ln {\frac {x + a} {x - a} } + C       | c = simplifying }} {{eqn | r = \dfrac 1 {2 a} \map \ln {\frac {x - a} {x + a} } + C       | c = Reciprocal of Logarithm }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = A       | o = \in       | r = \powerset S       | c =  }} {{eqn | ll= \leadsto       | l = A       | o = \subseteq       | r = S       | c = {{Defof|Power Set}} }} {{eqn | ll= \leadsto       | l = S \setminus A       | o = \subseteq       | r = S       | c = Set Difference is Subset }} {{eqn | ll= \leadsto       | l = \relcomp S A       | o = \subseteq       | r = S       | c = {{Defof|Relative Complement}} }} {{eqn | ll= \leadsto       | l = \relcomp S A       | o = \in       | r = \powerset S       | c = {{Defof|Power Set}} }} {{end-eqn}} {{qed}}
From Symmetric Difference with Intersection forms Ring: :$\struct {\powerset S, *, \cap}$ is a commutative ring with unity. From Intersection is Idempotent, $\cap$ is an idempotent operation on $S$. Hence the result by definition of Boolean ring. {{qed}}
By definition of parallel lines, the contemporary definition is for a straight line to be declared parallel to itself. Hence for a straight line $l$: :$l \parallel l$ Thus $\parallel$ is seen to be reflexive.
Let $l_1 \parallel l_2$. By definition of parallel lines, $l_1$ does not meet $l_2$ when produced indefinitely. Hence $l_2$ similarly does not meet $l_1$ when produced indefinitely. That is: :$l_2 \parallel l_1$ Thus $\parallel$ is seen to be symmetric.
From Parallelism is Transitive Relation: :$l_1 \parallel l_2$ and $l_2 \parallel l_3$ implies $l_1 \parallel l_3$. Thus $\parallel$ is seen to be transitive.
=== Positive definiteness === Let $x \in \CC \closedint a b$. Then $\forall t \in \closedint 0 1 : \size {\map x t} \ge 0 $. Hence: :$\displaystyle \int_a^b \size {\map x t} \rd t = \norm x_1 \ge 0$. Suppose $\forall t \in \closedint a b : \map x t = 0$. Then $\norm x_1 = 0$. Therefore: :$\paren {x = 0} \implies \paren {\norm x_1 = 0}$ Let $x \in \CC \closedint a b : \norm x_1 = 0$. Suppose: :$\forall t \in \openint a b : \map x t = 0$. By assumption of continuity of $x$: :$\forall t \in \closedint a b : \map x t = 0$. {{AimForCont}}: :$\exists t_0 \in \openint a b : \map x {t_0} \ne 0$. By assumption, $x$ is continuous at $t_0$. :$\forall \epsilon \in \R_{> 0} : \exists \delta \in \R_{> 0} : \size {t - t_0} < \delta \implies \size {\map x t - \map x {t_0} } < \epsilon$ Furthermore: :$\exists \delta \in \R_{> 0} : \paren{ a < t_0 - \delta} \land \paren {t_0 + \delta < b}$ Let $\displaystyle \epsilon = \frac {\size {\map x {t_0}}} 2$. We have that: {{begin-eqn}} {{eqn | l = \size {\map x t}       | r = \size {\map x t + \map x {t_0} - \map x {t_0} } }} {{eqn | o = \ge       | r = \size {\map x {t_0} } - \size {\map x t - \map x {t_0} }       | c = Reverse triangle inequality }} {{eqn | o = >       | r = \size {\map x {t_0} } - \frac {\size {\map x {t_0} } } 2 }} {{eqn | r = \frac {\size {\map x {t_0} } } 2  }} {{eqn | o = >       | r = 0       | c = $\map x {t_0} \ne 0$ }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = 0       | r = \norm x_1 }} {{eqn | r = \int_a^b \size {\map x t} \rd t }} {{eqn | o = \ge       | r = \int_{t_0 - \delta}^{t_0 + \delta} \size {\map x t} \rd t       | c = $t_0 + \delta < b$, $a < t_0 - \delta$ }} {{eqn | o = \ge       | r = \int_{t_0 - \delta}^{t_0 + \delta} \frac {\size {\map x {t_0} } } 2 \rd t }} {{eqn | r = 2 \delta \frac {\size {\map x {t_0} } } 2 }} {{eqn | r = \delta \size {\map x {t_0} } }} {{eqn | o = >       | r = 0 }} {{end-eqn}} Hence, we reached a contradiction. Therefore: :$\paren {\norm x_1 = 0} \implies \paren {x = 0}$ === Positive homogeneity === Let $x \in \CC \closedint a b$, $\alpha \in \R$. Then: {{begin-eqn}} {{eqn | l = \size {\alpha x}_1       | r = \int_a^b \size {\alpha \map x t} \rd t }} {{eqn | r =\size \alpha \int_a^b \size {\map x t} \rd t }} {{eqn | r =\size \alpha \norm x_1 }} {{end-eqn}} === Triangle inequality === Let $x, y \in \CC \closedint a b$ {{begin-eqn}} {{eqn | l = \norm {x + y}_1       | r = \int_a^b \size {\map {\paren {x + y} } t} \rd t }} {{eqn | r = \int_a^b \size {\map x t + \map y t} \rd t       | c = {{defof|Pointwise Addition of Real-Valued Functions}} }} {{eqn | o = \le       | r = \int_a^b \paren {\size x + \size y} \rd t       | c = Triangle Inequality for Real Numbers }} {{eqn | r = \int_a^b \size x \rd t + \int_a^b \size y \rd t }} {{eqn | r = \norm x_1 + \norm y_1 }} {{end-eqn}} {{qed}}
Let $l_1 \perp l_2$. By definition of perpendicular lines, $l_1$ meets $l_2$ at a right angle. Hence $l_2$ similarly meets $l_1$ at a right angle. That is: :$l_2 \perp l_1$ Thus $\parallel$ is seen to be symmetric.
By definition of perpendicular lines, for $l_1$ to be perpendicular to itself would mean it would have to meet itself in a right angle. This it does not do. So $l_1 \not \perp l_1$. Thus $\perp$ is seen to be antireflexive.
Let $U, V \in \mathscr I$ such that: :$\card V < \card U$ === Lemma 1 === {{:Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom/Sufficient Condition/Lemma/Lemma 1}}{{qed|lemma}} === Lemma 2 === {{:Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom/Sufficient Condition/Lemma/Lemma 2}}{{qed|lemma}} From Subset of Set Difference iff Disjoint Set: :$U \setminus V \subseteq B_1 \setminus B_2$ We have: {{begin-eqn}} {{eqn | l = \card{B_2}        | r = \card{B_1}       | c = Leigh.Samphier/Sandbox/Matroid Base Axiom Implies Sets Have Same Cardinality }} {{eqn | ll = \leadsto        | l = \card{B_2 \cap B_1} + \card{V \setminus B_1} + \card{\paren{B_2 \setminus B_1} \setminus V}        | r = \card{B_1 \cap B_2} + \card{B_1 \setminus B_2} }} {{eqn | ll = \leadsto        | l = \card{V \setminus B_1} + \card{\paren{B_2 \setminus B_1} \setminus V}        | r = \card{B_1 \setminus B_2}       | c = Subtract $\card{B_1 \cap B_2}$ from both sides of equation }} {{eqn | o = \ge        | r = \card{U \setminus V}       | c = Cardinality of Subset of Finite Set }} {{eqn | o = >        | r = \card{V \setminus U} }} {{eqn | o = \ge        | r = \card{V \setminus B_1}       | c = As $U \subseteq B_1$ }} {{eqn | ll = \leadsto        | l = \card{\paren{B_2 \setminus B_1} \setminus V}        | o = >       | r = 0       | c = Subtract $\card{V \setminus B_1}$ from both sides of equation }} {{end-eqn}} From the contrapositive statement of Cardinality of Empty Set: :$\paren{B_2 \setminus B_1} \setminus V \ne \O$ Hence: $\exists y \in \paren{B_2 \setminus B_1} \setminus V$ From $(\text B 1)$: :$\exists z \in B_1 \setminus B_2 : \paren{B_2 \setminus \set y} \cup \set z \in \mathscr B$ Let: :$B_3 = \paren{B_2 \setminus \set y} \cup \set z$ We have: :{{begin-eqn}} {{eqn | l = V       | o = \subseteq       | r = B_2 \setminus \set y       | c = As $y \notin V$ and $V \subseteq B_2$ }} {{eqn | o = \subseteq       | r = \paren{B_2 \setminus \set y} \cup \set z       | c = Set is Subset of Union }} {{eqn | r = B_3 }} {{end-eqn}} === Lemma 3 === {{:Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom/Sufficient Condition/Lemma/Lemma 3}}{{qed|lemma}} Hence: :$\card{B_1 \cap B_3} > \card{B_1 \cap B_2}$ {{qed}} Category:Matroid Satisfies Base Axiom 2j3iv9ails9r00l223vn9augij2zms8
Let $l_1 \perp l_2$ and $l_2 \perp l_3$. Then $l_1$ and $l_3$ are parallel, and not perpendicular. Thus $\perp$ is seen to be antitransitive.
We have: {{begin-eqn}} {{eqn | l = \card {B_1}       | r = \card{ \paren{B_1 \cap B_2} \cup \paren{B_1 \setminus B_2} }       | c = Set Difference Union Intersection }} {{eqn | r = \card{B_1 \cap B_2} + \card{B_1 \setminus B_2}       | c = Set Difference and Intersection are Disjoint and Corollary to Cardinality of Set Union }} {{end-eqn}} and {{begin-eqn}} {{eqn | l = \card {B_2}       | r = \card{ \paren{B_2 \cap B_1} \cup \paren{B_2 \setminus B_1} }       | c = Set Difference Union Intersection }} {{eqn | r = \card{B_2 \cap B_1} + \card{B_2 \setminus B_1}       | c = Set Difference and Intersection are Disjoint and Corollary to Cardinality of Set Union }} {{eqn | r = \card{B_2 \cap B_1} + \card{\paren{\paren{B_2 \setminus B_1} \cap V} \cup \paren{\paren{B_2 \setminus B_1} \setminus V} }       | c = Set Difference Union Intersection }} {{eqn | r = \card{B_2 \cap B_1} + \card{\paren{B_2 \setminus B_1} \cap V} + \card{\paren{B_2 \setminus B_1} \setminus V}       | c = Set Difference and Intersection are Disjoint and Corollary to Cardinality of Set Union }} {{eqn | r = \card{B_2 \cap B_1} + \card{V \setminus B_1} + \card{\paren{B_2 \setminus B_1} \setminus V}       | c = Intersection with Set Difference is Set Difference with Intersection }} {{end-eqn}} {{qed}} Category:Matroid Satisfies Base Axiom ntzh0sbv53wpqpq4txhx6u57p808oq0
We have: {{begin-eqn}} {{eqn | l = B_3 \cap B_1       | r = \paren{\paren{B_2 \setminus \set y} \cup \set z} \cap B_1 }} {{eqn | r = \paren{\paren{B_2 \setminus \set y} \cap B_1 } \cup \paren{ \set z \cap B_1}       | c = Intersection Distributes over Union }} {{eqn | r = \paren{\paren{B_2 \setminus \set y} \cap B_1 } \cup \set z       | c = As $z \in B_1$ }} {{eqn | r = \paren{\paren{B_2 \cap B_1} \setminus \paren{\set y \cap B_1} } \cup \set z       | c = Set Intersection Distributes over Set Difference }} {{eqn | r = \paren{\paren{B_2 \cap B_1} \setminus \O } \cup \set z       | c = As $y \notin B_1$ }} {{eqn | r = \paren{B_2 \cap B_1} \cup \set z       | c = Set Difference with Empty Set is Self }} {{eqn | ll = \leadsto       | l = \card{B_3 \cap B_1}       | r = \card{\paren{B_2 \cap B_1} \cup \set z} }} {{eqn | r = \card{\paren{B_2 \cap B_1} } +  \card {\set z}       | c = As $z \notin B_2$ }} {{eqn | r = \card{\paren{B_2 \cap B_1} } +  1       | c = Cardinality of Singleton }} {{end-eqn}} {{qed}} Category:Matroid Satisfies Base Axiom nvp65ssgy60e8jk5mdmozb5psrgp2hg
Suppose that Formulation 2 holds. That is, the Cartesian product of a non-empty family of non-empty sets is non-empty. Let $\CC$ be a non-empty set of non-empty sets. $\CC$ may be converted into an indexed set by using $\CC$ itself as the indexing set and using the identity mapping on $\CC$ to do the indexing. Then the Cartesian product of all the sets in $\CC$ has at least one element. An element of such a Cartesian product is a mapping, that is a family whose domain is the indexing set which in this context is $\CC$. The value of this mapping at each index is an element of the set which bears that index. So we have a mapping $f$ whose domain is $\CC$ such that: :$A \in \CC \implies \map f A \in A$ Now let $\CC$ be the set of all non-empty subsets of some set $X$. Then the assertion means that there exists a mapping $f$ whose domain is $\powerset X \setminus \set \O$ such that: :$A \in \powerset X \setminus \set \O \implies \map f A \in A$ That is, Formulation 1 holds.
Let $\SS$ be the set: :$\SS = \set {s: \O \notin s \land \forall t, u \in s: t = u \lor t \cap u = \O}$ Let $c$ be a choice function on $\SS$ and consider the image set $c \sqbrk \SS$: :$c \sqbrk \SS = \set {\map c s: \O \notin s \land \forall t, u \in s: t = u \lor t \cap u = \O}$ By the definition of choice function: :$\map c s \in s$ By construction of $\SS$, for any $s \in \SS$: :$s \cap c \sqbrk \SS = \set {\map c s}$
Let $\BB$ be a non-empty indexed family of non-empty sets indexed by $\II$. Consider sets of the following form: :$\CC = \set {\tuple {B_i, x}: i \in \II, B_i \in \BB, x \in B_i}$ That is, it is the set of ordered pairs of which the first coordinate is a set $B_i \in \BB$ and the second coordinate is an element of $B_i$. This is, by construction, a subset of the cartesian product: :$\CC \subseteq \displaystyle \BB \times \bigcup_{i \mathop \in \II} B_i$ By hypothesis all sets $B$ are non-empty. Thus there exists at least one $\tuple {B_i, x}$ element in $\CC$ for each $B_i \in \CC$. By Equality of Ordered Pairs, if $B_j \ne B_k$ in $\BB$, then $\tuple {B_i, x} \ne \tuple {B_j, x}$ for all such pairs in $\CC$. So any two sets $\set {\tuple {B_j, x}: x \in B_j}$ and $\set {\tuple {B_k, x} :x \in B_k}$ are disjoint, by the inequality of their first ccoordinates. Then $\CC$ is an indexed family of disjoint non-empty sets in $\BB \times \bigcup_{i \mathop \in \II} B_i$. Hence $\CC$ satisfies the hypotheses of the third formulation of the Axiom of Choice. Let $c$ be a set satisfying: :$\forall r \in \CC: c \cap r = \set t$ All elements of $c$ are ordered pairs: :the first coordinate of which is a set in $B_i \in \BB$ and: :the second coordinate of which is an element $x \in B_i$. Viewed as a relation, each set in $\BB$ bears $c$ to an element in that set. Every set in $\BB$ bears $c$ to exactly one element inside that set. So $c$ is in fact a mapping and satisfies the criteria of a choice function as expounded in formulation $(1)$.
=== Positive definiteness === Let $x \in \map {\CC^k} I$. Then: {{begin-eqn}} {{eqn | l = \norm x_{\map {C^k} I}       | r = \sum_{i \mathop = 0}^k \norm {x^{\paren i} }_\infty }} {{eqn | o = \ge       | r = \sum_{i \mathop = 0}^k 0       | c = Supremum Norm is Norm, Norm Axiom $\paren {N1}:$ Positive Definiteness }} {{eqn | r = 0 }} {{end-eqn}} Suppose $\norm x_{\map {C^k} I} = 0$. We have that the sum of non-negatives is zero if every element is zero. Hence: :$\forall i \in \N : 0 \le i \le k : \norm {x^{\paren i}}_\infty = 0$ Namely, $\norm x_\infty = 0$. By Supremum Norm is Norm and Norm Axiom $\paren {N1}:$ Positive Definiteness: :$\forall t \in I : \map x t = 0$. === Positive homogeneity === Let $x \in \map {\CC^k} I$, $\alpha \in \R$. Then: {{begin-eqn}} {{eqn | l = \norm {\alpha x}_{\map {C^k} I}       | r = \sum_{i \mathop = 0}^k \norm {\paren {\alpha x}^{\paren i} }_\infty       | c = {{defof|C^k Norm}} }} {{eqn | r = \sum_{i \mathop = 0}^k \norm {\alpha x^{\paren i} }_\infty       | c = {{defof|Pointwise Scalar Multiplication of Real-Valued Functions}} }} {{eqn | r = \size \alpha \sum_{i \mathop = 0}^k \norm {x^{\paren i} }_\infty       | c = Supremum norm on continuous real-valued functions is a norm: positive homogeneity }} {{eqn | r = \size \alpha \norm {x}_{\map {C^k} I}       | c = {{defof|C^k Norm}} }} {{end-eqn}} === Triangle inequality === Let $x, y \in \map {\CC^k} I$ {{begin-eqn}} {{eqn | l = \norm {x + y}_{\map {C^k} I}       | r = \sum_{i \mathop = 0}^k \norm {\paren {x + y}^{\paren i} }_\infty       | c = {{defof|Pointwise Addition of Real-Valued Functions}} }} {{eqn | r = \sum_{i \mathop = 0}^k \norm {x^{\paren i} + y^{\paren i} }_\infty       | c = {{defof|Pointwise Addition of Real-Valued Functions}} }} {{eqn | o = \le       | r = \sum_{i \mathop = 0}^k \norm {x^{\paren i} }_\infty + \sum_{i \mathop = 0}^k \norm {y^{\paren i} }_\infty       | c = Triangle Inequality for Real Numbers }} {{eqn | r = \norm x_{\map {C^k} I} + \norm y_{\map {C^k} I}       | c = {{defof|C^k Norm}} }} {{end-eqn}} {{qed}}
By Matrices over Field form Vector Space: :$K^{n \times n}$ forms a vector space under usual matrix addition and scalar multiplication. By Dimension of Vector Space of Matrices: :$K^{n \times n}$ has dimension $n^2$.  Consider the collection of vectors:  :$I, A, A^2, \ldots, A^{n^2}$ Since this is a collection of $n^2 + 1$ vectors, and $K^{n \times n}$ has dimension $n^2$, we have by Size of Linearly Independent Subset is at Most Size of Finite Generator: :$I, A, A^2, \ldots, A^{n^2}$ are linearly dependent. That is, there exists $\alpha_0, \alpha_1, \ldots, \alpha_{n^2} \in K$ not all zero such that:  :$\displaystyle \sum_{i \mathop = 0}^{n^2} \alpha_i A^i = 0$ That is, the polynomial:  :$\displaystyle \sum_{i \mathop = 0}^{n^2} \alpha_i X^i \in K \sqbrk X$ has $\map P A = 0$, and degree at most $n^2$.  Let:  :$\displaystyle S = \set {P \in K \sqbrk X \setminus \set 0 \mid \map P A = 0}$  $S$ is certainly non-empty since we have found such an element in the computation above.  Now consider the set:  :$\displaystyle D = \set {\deg P \mid P \in S}$ Since $D$ is a subset of the natural numbers, it contains a least element $N$ by the Well-Ordering Principle. Since the polynomial we constructed has degree at most $n^2$, we have $N \le n^2$. Let $Q \in S$ be of degree $N$. Let $a_N$ be the coefficient of the $X^N$ term in $Q$.  Then $\mu = \dfrac 1 {a_N} Q$ is a monic polynomial of minimum degree with $\map \mu A = 0$.  So $\mu$ is a minimal polynomial for $A$.  {{qed}} Category:Linear Algebra Category:Minimal Polynomials tb8lxiwzhe3qho6xa228up6rkyunipb
Let $p, q \in S$ such that $p \ne q$. Let $\tau_p$ and $\tau_q$ be the particular point topologies on $S$ by $p$ and $q$ respectively. From Homeomorphic Non-Comparable Particular Point Topologies, $\struct {S, \tau_p}$ and $\struct {S, \tau_q}$ are homeomorphic. But $\struct {S, \tau_p} \ne \struct {S, \tau_q}$, as $\set p \in \tau_p$ but $\set p \notin \tau_q$, and similarly $\set q \in \tau_q$ but $\set q \notin \tau_p$. {{Qed}}
Let $\tau_1$ be the indiscrete topology on $S$. Let $\tau_2$ be the discrete topology on $S$. Then $\struct {S, \tau_1}$ has $2$ elements: $S$ and $\O$. Let $a, b \in S$ such that $a \ne b$. Then $\set a \in \tau_2$ and $\set b \in \tau 2$, as well as $S$ and $\O$. So there cannot be a bijection between $\struct {S, \tau_1}$ and $\struct {S, \tau_2}$. {{qed}}
By definition of indiscrete topological space, the only open sets in $\struct {S, \tau}$ are $S$ and $\O$. From Open and Closed Sets in Topological Space, both $S$ and $\O$ are both closed and open in $\struct {S, \tau}$ are $S$ and $\O$.  Hence the result. {{qed}}
=== Definition by No Clopen Sets implies Definition by Separation === Let $T$ be connected by having no clopen sets. {{AimForCont}} $T$ admits a separation, $A \mid B$ say. Then both $A$ and $B$ are clopen sets of $T$, neither of which is either $S$ or $\O$. From this contradiction it follows that $T$ can admit no separation {{qed|lemma}} === Definition by Separation implies Definition by No Clopen Sets === Let $T$ be connected by admitting no separation. Suppose $\exists H \subseteq S$ which is clopen. Then $\relcomp S H$ is also clopen. Hence $H \mid \relcomp S H$ is a separation of $T$. From this contradiction it follows that $T$ can have no proper subsets which are clopen.
From the contrapositive statement of Superset of Dependent Set is Dependent: :$C \not \subseteq X$ From the contrapositive statement of Set Difference with Superset is Empty Set: :$C \setminus X \ne \O$ From Set Difference over Subset: :$C \setminus X \subseteq \paren{X \cup \set x} \setminus X = \set x$ From Power Set of Singleton: :$C \setminus X = \set x$ From Set Difference is Subset: :$\set x \subseteq C$ From Singleton of Element is Subset: :$x \in C$ {{qed}} Category:Matroid Theory h3wje0fo4f9qmhw0uc6wwo1cj132rz2
We prove the contrapositive statement: :$x \in X \implies X$ is a dependent subset. Let $x \in X$. We have: {{begin-eqn}} {{eqn | l = \paren{C \setminus \set x} \cup \set x       | o = \subseteq       | r =  X       | c = Union of Subsets is Subset }} {{eqn | ll = \leadsto       | l = C \cup \set x       | o = \subseteq       | r =  X       | c = Set Difference Union Second Set is Union }} {{eqn | ll = \leadsto       | l = C       | o = \subseteq       | r =  X       | c = Set is Subset of Union }} {{end-eqn}} From Superset of Dependent Set is Dependent: :$X$ is a dependent subset {{qed}} Category:Matroid Theory ndartfsndj1oorn109xfk0oppyk3883
From Union of Matroid Base with Element of Complement is Dependent: :$B \cup \set x$ is dependent. From Matroid Unique Circuit Property there exists a unique circuit $C$ such that: :$x \in C \subseteq B \cup \set x$ {{qed}} Category:Matroid Theory eh68f7pq174bzyyyftuozku64gonq7v