hcju/proofwikiqa huggingface dataset:
From the definition, the real numbers are the set of all equivalence classes $\eqclass {\sequence {x_n} } {}$ of Cauchy sequences of rational numbers. Let $x = \eqclass {\sequence {x_n} } {}, y = \eqclass {\sequence {y_n} } {}, z = \eqclass {\sequence {z_n} } {}$, where $\eqclass {\sequence {x_n} } {}$, $\eqclass {\sequence {y_n} } {}$ and $\eqclass {\sequence {z_n} } {}$ are such equivalence classes. From the definition of real addition, $x + y$ is defined as $\eqclass {\sequence {x_n} } {} + \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n + y_n} } {}$. Thus we have: {{begin-eqn}} {{eqn | l = x + \paren {y + z}       | r = \eqclass {\sequence {x_n} } {} + \paren {\eqclass {\sequence {y_n} } {} + \eqclass {\sequence {z_n} } {} }       | c =  }} {{eqn | r = \eqclass {\sequence {x_n} } {} + \eqclass {\sequence {y_n + z_n} } {}       | c =  }} {{eqn | r = \eqclass {\sequence {x_n + \paren {y_n + z_n} } } {}       | c =  }} {{eqn | r = \eqclass {\sequence {\paren {x_n + y_n} + z_n} } {}       | c = Rational Addition is Associative }} {{eqn | r = \eqclass {\sequence {x_n + y_n} } {} + \eqclass {\sequence {z_n} } {}       | c =  }} {{eqn | r = \paren {\eqclass {\sequence {x_n} } {} + \eqclass {\sequence {y_n} } {} } + \eqclass {\sequence {z_n} } {}       | c =  }} {{eqn | r = \paren {x + y} + z       | c =  }} {{end-eqn}} {{qed}}
From the definition, the real numbers are the set of all equivalence classes $\eqclass {\sequence {x_n} } {}$ of Cauchy sequences of rational numbers. Let $x = \eqclass {\sequence {x_n} } {}, y = \eqclass {\sequence {y_n} } {}, z = \eqclass {\sequence {z_n} } {}$, where $\eqclass {\sequence {x_n} } {}$, $\eqclass {\sequence {y_n} } {}$ and $\eqclass {\sequence {z_n} } {}$ are such equivalence classes. From the definition of real multiplication, $x \times y$ is defined as $\eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n \times y_n} } {}$. Thus we have: {{begin-eqn}} {{eqn | l = x \times \paren {y \times z}       | r = \eqclass {\sequence {x_n} } {} \times \paren {\eqclass {\sequence {y_n} } {} \times \eqclass {\sequence {z_n} } {} }       | c =  }} {{eqn | r = \eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n \times z_n} } {}       | c =  }} {{eqn | r = \eqclass {\sequence {x_n \times \paren {y_n \times z_n} } } {}       | c =  }} {{eqn | r = \eqclass {\sequence {\paren {x_n \times y_n} \times z_n} } {}       | c = Rational Multiplication is Associative }} {{eqn | r = \eqclass {\sequence {x_n \times y_n} } {} \times \eqclass {\sequence {z_n} } {}       | c =  }} {{eqn | r = \paren {\eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n} } {} } \times \eqclass {\sequence {z_n} } {}       | c =  }} {{eqn | r = \paren {x \times y} \times z       | c =  }} {{end-eqn}} {{qed}}
From the definition, the real numbers are the set of all equivalence classes $\eqclass {\sequence {x_n} } {}$ of Cauchy sequences of rational numbers. Let $x = \eqclass {\sequence {x_n} } {}, y = \eqclass {\sequence {y_n} } {}$, where $\eqclass {\sequence {x_n} } {}$ and $\eqclass {\sequence {y_n} } {}$ are such equivalence classes. Thus: {{begin-eqn}} {{eqn | l = x + y       | r = \eqclass {\sequence {x_n} } {} + \eqclass {\sequence {y_n} } {}       | c = {{Defof|Real Number|subdef = Cauchy Sequences}} }} {{eqn | r = \eqclass {\sequence {x_n + y_n} } {}       | c = {{Defof|Real Addition}} }} {{eqn | r = \eqclass {\sequence {y_n + x_n} } {}       | c = Rational Addition is Commutative }} {{eqn | r = \eqclass {\sequence {y_n} } {} + \eqclass {\sequence {x_n} } {}       | c = {{Defof|Real Addition}} }} {{eqn | r = y + x       | c = {{Defof|Real Number|subdef = Cauchy Sequences}} }} {{end-eqn}} {{qed}}
From the definition, the real numbers are the set of all equivalence classes $\eqclass {\sequence {x_n} } {}$ of Cauchy sequences of rational numbers. Let $x = \eqclass {\sequence {x_n} } {}, y = \eqclass {\sequence {y_n} } {}$, where $\eqclass {\sequence {x_n} } {}$ and $\eqclass {\sequence {y_n} } {}$ are such equivalence classes. From the definition of real multiplication, $x \times y$ is defined as $\eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n \times y_n} } {}$. Thus we have: {{begin-eqn}} {{eqn | l = x \times y       | r = \eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n} } {}       | c =  }} {{eqn | r = \eqclass {\sequence {x_n \times y_n} } {}       | c =  }} {{eqn | r = \eqclass {\sequence {y_n \times n_n} } {}       | c = Rational Multiplication is Commutative }} {{eqn | r = \eqclass {\sequence {y_n} } {} \times \eqclass {\sequence {x_n} } {}       | c =  }} {{eqn | r = y \times x       | c =  }} {{end-eqn}} {{qed}}
Let $\left\{{X_k}\right\}_{k \mathop = 1}^n$ be a decomposition of $A$ and $C$, so that there are isometries $\left\{{\phi_k}\right\}_{k \mathop = 1}^n$ and $\left\{{\psi_k}\right\}_{k \mathop = 1}^n$ such that: $\displaystyle A = \bigcup_{k \mathop = 1}^n \phi_k \left({X_k}\right)$ $\displaystyle C = \bigcup_{k \mathop = 1}^n \psi_k \left({X_k}\right)$ Let $Y_k = \psi_k^{-1} \left({ B \cap \psi_k \left({X_k}\right)}\right)$. Then: : $\displaystyle \bigcup_{k \mathop = 1}^n \psi_k \left({Y_k}\right) = B \cap \bigcup_{k \mathop = 1}^n \psi_k \left({X_k}\right) = B \cap C = B$ Let $Z_k = \phi^{-1} \left({ A \cap \phi_k \left({X_k}\right)}\right)$. Then: : $\displaystyle \bigcup_{k \mathop = 1}^n \phi_k \left({Z_k}\right) = A \cap \bigcup_{k \mathop = 1}^n \phi_k \left({X_k}\right) = A \cap C = A$ {{finish}} Category:Topology bwgmj0y0xe77s29ciiqljand7maxwif
From the definition of absolute value: :$\size {x_1 - x_2} = \sqrt {\paren {x_1 - x_2}^2}$ This is the same as the Euclidean metric. This is shown in Euclidean Metric on Real Vector Space is Metric to be a metric. From Rational Numbers form Vector Space, it follows that the set of all rational numbers is a 1-dimensional Euclidean space. {{qed}}
From the definition, a vector space is a unitary module whose scalar ring is a field. From Rational Numbers form Field, we have that $\Q$ is a field. So the $\Q$-module $\Q^n$ fits the description. {{qed}} Category:Linear Algebra Category:Rational Numbers n74hux90kj7e88gtvcb4nzycbi1nbre
The proof proceeds by the Principle of Mathematical Induction on the number of vertices. For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$G_n$ can be assigned a proper vertex $k$-coloring such that $k \le 5$. === Basis for the Induction === $\map P r$ is trivially true for $1 \le r \le 5$, as there are no more than $5$ vertices to be colored. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P r$ is true, where $r \ge 5$, then it logically follows that $\map P {r + 1}$ is true. So this is our induction hypothesis: :$G_r$ can be assigned a proper vertex $k$-coloring such that $k \le 5$. Then we need to show: :$G_{r + 1}$ can be assigned a proper vertex $k$-coloring such that $k \le 5$. === Induction Step === This is our induction step: According to the Minimum Degree Bound for Simple Planar Graph, $G_{r + 1}$ has at least one vertex with at most $5$ edges. :Let this vertex be labeled $x$.   Remove vertex $x$ from $G_{r + 1}$ to create another graph, $G'_r$. By the induction hypothesis, $G'_r$ is five-colorable. Suppose all five colors were not connected to $x$. Then we can give $x$ the missing color and thus five-color $G_{r + 1}$. Suppose all five colors are connected to $x$. Then examine the five vertices $x$ was adjacent to. Call them $y_1, y_2, y_3, y_4$ and $y_5$ in clockwise order around $x$. Let $y_1, y_2, y_3, y_4$ and $y_5$ be colored respectively by colors $c_1, c_2, c_3, c_4$ and $c_5$. Let us denote $H_{i,j}$ a subgraph of $G'_r$ induced by the vertices colored with $c_i$ and $c_j$.  Consider $H_{1, 3}$.  Suppose there exists no path between $y_1$ and $y_3$ in $H_{1,3}$.  Thus, $H_{1, 3}$ is disconnected into two components. We can, then, interchange the colors $c_1$ and $c_3$ in the component that is connected to $y_1$. Thus $x$ is no longer adjacent to a vertex of color $c_1$, so $x$ can be colored $c_1$. Suppose there exists a path between $y_1$ and $y_3$ in $H_{1,3}$. Including the vertex $x$ in this path we get a circuit $C$. Since we indexed the vertices $y_1, y_2, y_3, y_4$ and $y_5$ clockwise, exactly one of the vertices $y_2$ and $y_4$ is inside $C$. Hence, $y_2$ and $y_4$ are in different connected components of $H_{2, 4}$ Then  we can switch colors $c_2$ and $c_4$ in the component of $H_{2, 4}$ that is connected to $y_2$. Now $x$ is no longer adjacent to a vertex of color $c_2$, so we can color it $c_2$. :300px This graph illustrates the case in which the path from $y_1$ to $y_3$ can be completed. :$\text{Blue} = c_1, \text{Yellow} = c_2, \text{Red} = c_3, \text{Green} = c_4, \text{Turquoise} = c_5$. The dotted lines represent edges and vertices that might exist, as this is simply a fairly minimal example graph that matches the conditions. So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :For all $n \in \N_{>0}$, $G_n$ can be assigned a proper vertex $k$-coloring such that $k \le 5$.
This proof works directly from the definition of the cosine function: {{begin-eqn}} {{eqn | l = \cos x       | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}       | c = {{Defof|Real Cosine Function}} }} {{eqn | r = \paren {-1}^0 \cdot \frac {x^{2 \cdot 0} } {\paren {2 \cdot 0}!} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} }} {{eqn | r = 1 + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}       | c = {{Defof|Factorial|Zero Factorial}} and {{Defof|Power (Algebra)|Zeroth Power}} }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \lim_{x \mathop \to 0} \frac {\map \cos x - 1} x       | r = \lim_{x \mathop \to 0} \frac {1 + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} - 1} x }} {{eqn | r = \lim_{x \mathop \to 0} \frac {\sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} } x }} {{eqn | r = \lim_{x \mathop \to 0} \frac {\sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n - 1} } {\paren {2 n - 1}!} } 1       | c = Power Series is Differentiable on Interval of Convergence and L'Hôpital's Rule }} {{eqn | r = \lim_{x \mathop \to 0} \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n - 1} } {\paren {2 n - 1}!} }} {{end-eqn}} Now let: :$\map {f_n} x = \paren {-1}^n \dfrac {x^{2 n - 1} } {\paren {2 n - 1}!}$ Then for every $n \in \N_{> 0}$, and for all $x \in \closedint {\dfrac 1 2} {\dfrac 1 2}$: {{begin-eqn}} {{eqn | l = \map {f_n} x       | o = \le       | r = \size {\paren {-1}^n \frac {x^{2 n - 1} } {\paren {2 n - 1}!} }       | rr= \; = \frac { {\size x}^{2 n - 1} } {\paren {2 n - 1}!}       | c = Absolute Value Function is Completely Multiplicative }} {{eqn | o = \le       | r = \frac 1 {2^{2 n - 1} \paren {2 n - 1}!}       | c = Power Function is Strictly Increasing over Positive Reals }} {{eqn | o = \le       | r = \frac 1 {2^{2 n - 1} }       | c = because the factorial is strictly increasing }} {{eqn | o = \le       | r = \frac 1 {2^n}       | c = because $n \ge 1 \iff 2 n - 1 \ge n$ }} {{end-eqn}} But from Sum of Infinite Geometric Sequence: :$\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {2^n} = 2 < \infty$ By the Weierstrass M-Test, $\displaystyle \sum_{n \mathop = 1}^\infty \map {f_n} x$ converges uniformly to some function $f$ on $\closedint {\dfrac 1 2} {\dfrac 1 2}$. But from Real Polynomial Function is Continuous, and the Uniform Limit Theorem $f$ is continuous on $\closedint {\dfrac 1 2} {\dfrac 1 2}$. So: :$\displaystyle \lim_{x \mathop \to 0} \map f x = \map f 0 = \sum_{n \mathop = 1}^\infty \paren {-1} \frac {0^{2 n - 1} } {\paren {2 n - 1}!} = 0$ {{qed}}
This proof assumes the truth of the Derivative of Cosine Function: From Cosine of Zero is One: :$\cos 0 = 1$ From Derivative of Cosine Function: :$D_x \left({\cos x}\right) = - \sin x$ and by Derivative of Constant: :$D_x \left({-1}\right) = 0$ So by Sum Rule for Derivatives: :$D_x \left({\cos x - 1}\right) = - \sin x$ By Sine of Zero is Zero: :$\sin 0 = 0$ From Derivative of Identity Function: :$D_x \left({x}\right) = 1$ Thus L'Hôpital's Rule applies and so: : $\displaystyle \lim_{x \mathop \to 0} \frac {\cos x - 1} x = \lim_{x \mathop \to 0} \frac {-\sin x} 1 = \frac {-0} 1 = 0$ {{qed}}
{{begin-eqn}} {{eqn | l = \lim_{x \mathop \to 0} \ \frac {\cos x - 1} x       | r = \lim_{x \mathop \to 0} \ \frac {\left({\cos x - 1}\right) \left({\cos x + 1}\right)} {x \left({\cos x + 1}\right)} }} {{eqn | r = \lim_{x \mathop \to 0} \ \frac {\cos^2 x - 1} {x \left({\cos x + 1}\right)} }} {{eqn | r = \lim_{x \mathop \to 0} \ \frac {-\sin^2 x} {x  \left({\cos x + 1}\right)}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \left({\lim_{x \mathop \to 0} \ \frac {\sin x} x}\right) \left({\lim_{x \mathop \to 0} \ \frac {-\sin x} {\cos x + 1} }\right)       | c = Product Rule for Limits of Functions }} {{eqn | r = 1 \times \left({\lim_{x \mathop \to 0} \ \frac{-\sin x} {\cos x + 1} }\right)       | c = Limit of Sine of X over X }} {{eqn | r = \frac {\lim_{x \mathop \to 0} \left({- \sin x}\right)} {\lim_{x \mathop \to 0} \left({\cos x + 1}\right)}       | c = Quotient Rule for Limits of Functions }} {{eqn | r = \frac 0 2 }} {{eqn | r = 0 }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \frac {\cos x - 1} x        | r = \frac {\cos x - \cos 0} x       | c = Cosine of Zero is One }} {{eqn | o = \to       | r = \left.{\dfrac {\mathrm d} {\mathrm dx} \cos x}\right \vert_{x \mathop = 0}       | c = as $x \to 0$, from definition of derivative at a point }} {{eqn | r = \sin x \vert_{x \mathop = 0}       | c = Derivative of Cosine Function }} {{eqn | r = 0       | c = Sine of Zero is Zero }} {{end-eqn}} {{qed}}
From the definition, the real numbers are the set of all equivalence classes $\eqclass {\sequence {x_n} } {}$ of Cauchy sequences of rational numbers. Let $x = \eqclass {\sequence {x_n} } {}, y = \eqclass {\sequence {y_n} } {}, z = \eqclass {\sequence {z_n} } {}$, where $\eqclass {\sequence {x_n} } {}$, $\eqclass {\sequence {y_n} } {}$ and $\eqclass {\sequence {z_n} } {}$ are such equivalence classes. From the definition of real multiplication, $x \times y$ is defined as: :$\eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n \times y_n} } {}$ From the definition of real addition, $x + y$ is defined as: :$\eqclass {\sequence {x_n} } {} + \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n + y_n} } {}$ Thus: {{begin-eqn}} {{eqn | l = x \times \paren {y + z}       | r = \eqclass {\sequence {x_n} } {} \times \paren {\eqclass {\sequence {y_n} } {} + \eqclass {\sequence {z_n} } {} }       | c =  }} {{eqn | r = \eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {x_n + y_n} } {}       | c =  }} {{eqn | r = \eqclass {\sequence {x_n \times \paren {y_n + z_n} } } {}       | c =  }} {{eqn | r = \eqclass {\sequence {\paren {x_n \times y_n} + \paren {x_n \times z_n} } } {}       | c = Rational Multiplication Distributes over Addition }} {{eqn | r = \eqclass {\sequence {x_n \times y_n} } {} + \eqclass {\sequence {x_n \times z_n} } {}       | c =  }} {{eqn | r = \paren {\eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n} } {} } + \paren {\eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {z_n} } {} }       | c =  }} {{eqn | r = \paren {x \times y} + \paren {x \times z}       | c =  }} {{end-eqn}} By Real Addition is Commutative and Real Multiplication is Commutative, it follows that: : $\paren {y + z} \times x = \paren {y \times x} + \paren {z \times x}$ {{qed}}
{{:Euclid:Proposition/II/1}} :400px Let $A$ and $BC$ be two straight lines. Let $BC$ be cut at random at points $D$ and $E$. Then the rectangle contained by $A$ and $BC$ is equal to the sum of the rectangles contained by $A$ and $BD$, by $A$ and $DE$, and by $A$ and $EC$, as follows: Construct $BF$ perpendicular to $BC$. Construct $BG$ on $BF$ equal to $A$. Construct $GH$ through $G$ parallel to $BC$. Construct $DK, EL, CH$ through $D, E, C$ parallel to $BG$. Then we have that: : $\Box BCHG = \Box BDKG + \Box DELK + \Box ECHL$ Now $\Box BCHG$ is the rectangle contained by $A$ and $BC$, because it is contained by $BC$ and $BG$, and $BG = A$. Similarly, from Opposite Sides and Angles of Parallelogram are Equal: : $\Box BDKG$ is the rectangle contained by $A$ and $BD$, because it is contained by $BD$ and $BG = A$ : $\Box DEKL$ is the rectangle contained by $A$ and $DE$, because it is contained by $DE$ and $DK$, and $DK = A$ : $\Box ECHL$ is the rectangle contained by $A$ and $EC$, because it is contained by $EC$ and $EL$, and $EL = A$. Hence the result. {{qed}}
For clarity the Cayley table of $Q$ is presented below: {{:Quaternion Group/Cayley Table}} By definition $Q$ is Hamiltonian {{iff}}: :$Q$ is non-abelian and: :every subgroup of $Q$ is normal. $Q$ is non-abelian as demonstrated by the counter-example: :$a b \ne b a$ From Subgroups of Quaternion Group: {{:Subgroups of Quaternion Group}} From Trivial Subgroup and Group Itself are Normal: :$Q$ and $\set e$ are normal subgroups of $Q$. From Center of Quaternion Group, $\gen {a^2} = \set {e, a^2}$ is the center of $Q$. From Center of Group is Normal Subgroup, $\set {e, a^2}$ is normal in $Q$. The remaining subgroups of $Q$ are of order $4$, and so have index $2$. From Subgroup of Index 2 is Normal it follows that all of these order $4$ subgroups of $Q$ are normal. That accounts for all subgroups of $Q$. Hence the result. {{qed}}
Let $x \in X$. From Point in Finite Metric Space is Isolated, $x$ is an isolated point. The result follows by definition of isolated point: :$x$ is an isolated point {{iff}} $x$ is not a limit point. {{qed}}
As $X$ is finite, its elements can be placed in one-to-one correspondence with the elements of $\N^*_n$ for some $n \in \N$. So let $X = \left\{{x_1, x_2, \ldots, x_n}\right\}$. Now let $\epsilon := \min \left\{{\forall i, j \in \N^*_n: i \ne j: d \left({x_i, x_j}\right)}\right\}$. That is, $\epsilon$ is the minimum distance between any two elements of $X$. We have that $\epsilon > 0$ from the definition of metric. Let $N_\epsilon \left({x_k}\right)$ be the open $\epsilon$-ball of any $x_k \in X$. By the method of construction of $\epsilon$, it is clear that $N_\epsilon \left({x_k}\right) = \left\{{x_k}\right\}$. Hence $x_k$ is an isolated point. {{qed}}
A Metric Space is Hausdorff. Therefore Point in Finite Hausdorff Space is Isolated can be applied. Hence the result. {{qed}}
As $X$ is finite, its elements can be placed in one-to-one correspondence with the elements of $\left\{{1, 2, \ldots, n}\right\}$ for some $n \in \N$. So let $X = \left\{{x_1, x_2, \ldots, x_n}\right\}$. From the definition of Hausdorff space: : $\forall x_i, x_j \in X: x_i \ne x_j: \exists U, V \in \tau: x_i \in U, x_j \in V: U \cap V = \varnothing$ Let $x_k \in X$. Then by definition: :$\forall x_i \in X, x_i \ne x_k: \exists U_i, V \in \tau: x_k \in U_i, x_i \in V: U_i \cap V = \varnothing$ Note that $U_i$ may be different for each $X_i$. That is: :$\forall x_i \in X: \exists U_i \in \tau: x_k \in U_i, x_i \notin U_i$ Now consider: :$\displaystyle U = \bigcap_{i \mathop \ne k} U_i$ We have by definition of a topology that $U \in \tau$, as the intersection $U$ is of a finite number of sets $U_i$. We have that: : $\forall i: x_k \in U_i$ : $\forall i: x_i \notin U_i$ So: : $U \cap X = \left\{{x_k}\right\}$ and so $U$ is an open set of $T$ which contains no points of $X$ other than $x_k$. So by definition, $x_k$ is an isolated point of $X$. Hence the result. {{qed}}
Let $T = \left({S, \tau}\right)$ be a $T_2$ (Hausdorff) space. Let $X \subseteq T$ be finite. From Separation Properties Preserved in Subspace, it follows that $\left({X, \tau_X}\right)$ is also a $T_2$ (Hausdorff) space. From $T_2$ Space is $T_1$ Space it follows that $\left({X, \tau_X}\right)$ is a $T_1$ (Fréchet) space. From Finite $T_1$ Space is Discrete, it follows that $\left({X, \tau_X}\right)$ is a discrete space. The result follows from Topological Space is Discrete iff All Points are Isolated. {{qed}}
By Finite Subset of Metric Space has no Limit Points, $S$ is infinite (or it has no limit points). We may assume that $x_n$ never equals $x$. Otherwise we delete all instances of $x$ from $\sequence {x_n}$ and create a new sequence $x_m$ which is a subsequence of $x_n$ which ''does'' never equal $x$. Then $S \setminus \set x$ is still infinite, and it still has $x$ as a limit point. So, since $x$ is a limit point of $S$, there is an integer $\map n 1$, say, such that $x_{\map n 1} \in \map {B_1} x$, where $\map {B_1} x$ is the open $1$-ball of $x$. Suppose we choose the integers $\map n 1 < \map n 2 < \cdots < \map n k$ so that $x_{\map n i} \in \map {B_{1/i} } x$ for $i = 1, 2, \ldots, k$. Now we put $\epsilon = \min \set {\dfrac 1 {k+1}, \map d {x_1, x}, \map d {x_2, x}, \map d {x_{\map n k}, x} }$. Since $x_n \ne x$ for any $n$, it follows that $\epsilon > 0$. Since $x$ is a limit point of $S$, there exists an integer $\map n {k+1}$ such that $x_{\map n {k+1} } \in \map {B_\epsilon} x$. Now $\epsilon$ has been chosen so as to force $\map n {k+1} > \map n k$, since $\map d {x_i, x} \ge \epsilon$ for all $i \le \map n k$. Also, note that $x_{\map n {k+1} } \in \map {B_{\frac 1 {k+1}} } x$. This completes the inductive step in constructing a subsequence. This converges to $x$, because $\forall k_0 \in \N_{>0}$ and $\forall k \ge k_0$ we have $\map d {x_{\map n k}, x} < \dfrac 1 k \ge \dfrac 1 {k_0}$. Hence the result. {{qed}}
First we note that by Integer Less One divides Power Less One: :$\paren {m - 1} \divides \paren {m^n - 1}$ where $\divides$ denotes divisibility. Thus $m^n - 1$ is composite for all $m \in \Z: m > 2$. Let $m = 2$, and consider $2^n - 1$. Suppose $n$ is composite. Then $n = r s$ where $r, s \in \Z_{> 1}$. Then by the corollary to Integer Less One divides Power Less One: :$\paren {2^r - 1} \divides \paren {2^{r s} - 1}$ Thus if $n$ is composite, then so is $2^n - 1$. So $2^n - 1$ can be prime only when $n$ is prime. {{qed}}
{{AimForCont}} $\mathcal U$ is an open cover of $M$ for which no Lebesgue number exists. Then for any $n \in \N$, there exists some $x_n \in M$ such that $\map {B_{1 / n} } {x_n} \subseteq U$ is false for each $U \in \mathcal U$. (Otherwise $1/n$ would be a Lebesgue number for $\mathcal U$.) As $M$ is sequentially compact, $\sequence {x_n}$ has a subsequence, say $\sequence {x_{\map n r} }$ which converges to some $x \in M$. Since $\mathcal U$ covers $M$, $x \in U_0$ for some $U_0 \in \mathcal U$. Since $U_0$ is open, $\exists m \in \N: \map {B_{2 / m} } x \subseteq U_0$. Now $\map {B_{1 / m} } x$ contains $x_{\map n r}$ for all $r \ge R$, say. Choose $r \ge R$ such that $\map n r \ge m$ and write $s = \map n r$. Then $\map {B_{1 / s} } {x_s} \subseteq \map {B_{2 / m} } x$ since: {{begin-eqn}} {{eqn | l = \map d {x_s, y}       | o = <       | r = \frac 1 s       | c =  }} {{eqn | ll= \leadsto       | l = \map d {x, y}       | o = \le       | r = \map d {x, x_s} + \map d {x_s, y}       | c =  }} {{eqn | o = <       | r = \frac 1 m + \frac 1 s       | c =  }} {{eqn | o = \le       | r = \frac 2 m       | c =  }} {{end-eqn}} So $\map {B_{1 / s} } {x_s} \subseteq U_0$ which contradicts the choice of $x_s$. So, by Proof by Contradiction, there has to be a Lebesgue number for $\mathcal U$. {{qed}} {{Namedfor|Henri Léon Lebesgue|cat = Lebesgue}}