hcju/proofwikiqa huggingface dataset:
By the definition of an isometry, $f$ is bijective. By Continuous Mapping is Continuous on Induced Topological Spaces, $f$ is a continuous mapping from $\struct {S_1, \tau_1}$ to $\struct {S_2, \tau_2}$. {{explain|It needs to be established that $f$ is continuous on the original metric spaces in the first place.}} By Inverse of Isometry of Metric Spaces is Isometry, $f^{-1}$ is an isometry. By Continuous Mapping is Continuous on Induced Topological Spaces once more, $f^{-1}$ is a continuous mapping from $\struct {S_2, \tau_2}$ to $\struct {S_1, \tau_1}$. Thus $f$ is a homeomorphism from $\struct {S_1, \tau_1}$ to $\struct {S_2, \tau_2}$. {{qed}} Category:Isometries Category:Homeomorphisms gdeqivac4rd0i7tmb3hgiu4auqz8vfi
=== 1 is equivalent to 1' === By applying Meet is Commutative several times, we have: {{begin-eqn}} {{eqn | l = x \wedge \left({y \vee z}\right)       | r = \left({x \wedge y}\right) \vee \left({x \wedge z}\right) }} {{eqn | ll= \iff       | l = \left({y \vee z}\right) \wedge x       | r = \left({y \wedge x}\right) \vee \left({z \wedge x}\right) }} {{end-eqn}} which (after renaming variables as appropriate) establishes the equivalence. {{qed|lemma}} === 2 is equivalent to 2' === By applying Join is Commutative several times, we have: {{begin-eqn}} {{eqn | l = x \vee \left({y \wedge z}\right)       | r = \left({x \vee y}\right) \wedge \left({x \vee z}\right) }} {{eqn | ll= \iff       | l = \left({y \wedge z}\right) \vee x       | r = \left({y \vee x}\right) \wedge \left({z \vee x}\right) }} {{end-eqn}} which (after renaming variables as appropriate) establishes the equivalence. {{qed|lemma}} === 1 implies 2 === Suppose that $(1)$ holds, and hence $(1')$ as well. {{begin-eqn}} {{eqn | l = \left({x \vee y}\right) \wedge \left({x \vee z}\right)       | r = \left({\left({x \vee y}\right) \wedge x}\right) \vee \left({\left({x \vee y}\right) \wedge z}\right)       | c = by $(1)$ }} {{eqn | r = x \vee \left({\left({x \vee y}\right) \wedge z}\right)       | c = $\wedge$ absorbs $\vee$ }} {{eqn | r = x \vee \left({\left({x \wedge z}\right) \vee \left({y \wedge z}\right)}\right)       | c = by $(1')$ }} {{eqn | r = \left({x \vee \left({x \wedge z}\right)}\right) \vee \left({y \wedge z}\right)       | c = $\vee$ is associative }} {{eqn | r = x \vee \left({y \wedge z}\right)       | c = $\vee$ absorbs $\wedge$ }} {{end-eqn}} {{qed|lemma}} === 2 implies 1 === By inspection, aided by Dual Pairs (Order Theory), we see that $(2)$ is dual to $(1)$. Thus by Global Duality, $(2)$ implies $(1)$ as soon as $(1)$ implies $(2)$. That direction was already established above. {{qed}}
The premises satisfy all elements of the definition of isometry except for bijectivity. As we presume $\phi$ to be surjective we need only show that it is injective. By Distance-Preserving Mapping is Injection of Metric Spaces then $\phi$ is injective. {{qed}} Category:Isometries a6j4t4aomj1hrk6vjl69i65zqgq2xe2
=== Positive Exponent === {{:Existence of Positive Root of Positive Real Number/Positive Exponent|Positive Exponent}} === Negative Exponent === {{:Existence of Positive Root of Positive Real Number/Negative Exponent|Negative Exponent}}
{{BeginTableau|\vdash \left({\left({p \lor q}\right) \implies r}\right) \implies \left({\left({p \implies r}\right) \land \left({q \implies r}\right)}\right)}} {{Assumption|1|\left({p \lor q}\right) \implies r}} {{SequentIntro|2|1|\left({p \implies r}\right) \land \left({q \implies r}\right)|1|Proof by Cases: Formulation 1: Reverse Implication}} {{Implication|3||\left({\left({p \lor q}\right) \implies r}\right) \implies \left({\left({p \implies r}\right) \land \left({q \implies r}\right)}\right)|1|2}} {{EndTableau}} {{qed}} Category:Proof by Cases pcm05869vx9skltwgpdk05y9lobkkvg
{{BeginTableau|\left({\left({p \lor q}\right) \land \left({p \implies r}\right) \land \left({q \implies r}\right)}\right) \implies r}} {{TheoremIntro|1|\left({\left({p \lor r}\right) \land \left({p \implies q}\right) \land \left({r \implies s}\right)}\right) \implies \left({q \lor s}\right)|Constructive Dilemma: Formulation 2}} {{Substitution|2||\left({\left({p \lor q}\right) \land \left({p \implies r}\right) \land \left({q \implies r}\right)}\right) \implies \left({r \lor r}\right)|1|r|q|q|r|s|r}} {{Assumption|3|\left({p \lor q}\right) \land \left({p \implies r}\right) \land \left({q \implies r}\right)}} {{ModusPonens|4|3|r \lor r|2|3}} {{Idempotence|5|3|r|4|Disjunction}} {{Implication|6||\left({\left({p \lor q}\right) \land \left({p \implies r}\right) \land \left({q \implies r}\right)}\right) \implies r|3|5}} {{EndTableau}} {{qed}}
Let $m = -n$. Then $m > 0$. Let $g$ be the real function defined on $\hointr 0 \to$ defined by: :$\map g y = y^m$ Since $x \ge 0$: :$\dfrac 1 x \ge 0$ By Existence of Positive Root of Positive Real Number: Positive Exponent there is a $y > 0$ such that: :$\map g y = \dfrac 1 x$ It follows from the definition of power that: {{begin-eqn}} {{eqn | l = \map f y       | r = \dfrac 1 {\map g y} }} {{eqn | r = \dfrac 1 {1 / x} }} {{eqn | r = x }} {{end-eqn}} {{qed}} Category:Existence of Positive Root of Positive Real Number 2ioe7yjmlsdu080btfmgrb6l8e72o6z
First, suppose that $n$ is even. Then for some $m \in \N_{>0}$: : $n = 2 m = m + m$ Thus since $\circ$ is associative: :$\displaystyle \map {\circ^n} {-x} = \prod_{i \mathop = 1}^m \paren {-x} \circ \paren {-x}$ By Product of Ring Negatives: :$\paren {-x} \circ \paren {-x} = x \circ x = \map {\circ^2} x$ Thus: :$\displaystyle \map {\circ^n} {-x} = \prod_{i \mathop = 1}^m \map {\circ^2} x$ By associativity: :$\map {\circ^n} {-x} = \map {\circ^{2 m} } x = \map {\circ^n} x$ {{qed|lemma}} Now suppose instead that $n$ is odd. If $n = 1$, then: :$\map {\circ^n} {-x} = -x = -\map {\circ^n} x$ Otherwise: {{begin-eqn}} {{eqn | l = \map {\circ^n} {-x}       | r = \paren {-x} \circ \paren {\map {\circ^{n - 1} } {-x} } }} {{eqn | r = \paren {-x} \circ \paren {\map {\circ^{n - 1} } x}       | c = $n-1$ is a strictly positive even integer, so the above case applies }} {{eqn | r = -\paren {x \circ \paren {\map {\circ^{n - 1} } x} }       | c = Product with Ring Negative }} {{eqn | r = -\map {\circ^n} x       | c = Definition of $\map {\circ^n} x$ }} {{end-eqn}} {{qed}} Category:Ring Theory Category:Powers qnscwtcvihfvvopzzazedbfb0gbh64w
=== Finding the constant === First, suppose that $S$ is a subset of $\R^m$. From Closed Real Interval is Compact in Metric Space, it follows that $\closedint a b$ is compact. Then Continuous Image of Compact Space is Compact shows that $\map \gamma {\closedint {x_i} {x_{i + 1} } }$ is compact. From the Heine-Borel Theorem, it follows that $\map \gamma {\closedint {x_i} {x_{i + 1} } }$ is bounded and closed. Suppose that $S \ne \R^m$. As $\R^m \setminus S$ is closed, it follows from Distance between Closed Sets in Euclidean Space that: :$\map d {\map \gamma {\closedint a b }, \R^m \setminus S} > 0$ Put $K = \map d {\map \gamma {\closedint a b }, \R^m \setminus S}$, the distance between $\map \gamma {\closedint a b }$ and $\R^m \setminus S$. If instead $S = \R^m$, then $K$ may be any strictly positive real number. === Finding the subdivision === Let $\epsilon \in \openint 0 K$. From the Heine-Cantor Theorem, it follows that $\gamma$ is uniformly continuous. Then there exists $\delta \in \R_{>0}$ such that: :$\forall y, z \in \closedint a b : \size {y - z} < \delta$ implies $\map d {\map \gamma y, \map \gamma z } < \epsilon$ where $d$ denotes the Euclidean metric. Choose $n \in \N$ such that $\dfrac {b - a} n < \delta$, and put $x_i = a + i \dfrac {b - a} n$ for all $i \in \set {0, 1, \ldots, n}$. Then $\set {x_0, x_1, \ldots, x_n}$ is a normal subdivision of $\closedint a b$. It follows that $\map {B_\epsilon} {\map \gamma {x_i} }$ and $\R^m \setminus S$ are disjoint, as either: :$\epsilon < \map d {\map \gamma {\closedint a b }, \R^m \setminus S}$ or: :$\R \setminus S = \varnothing$ Then:  :$\displaystyle \bigcup_{i \mathop = 0}^n \map {B_\epsilon} {\map \gamma {x_i} } \subseteq S$ Let $x \in \closedint {x_i} {x_{i+1} }$. Then: :$\size {x - x_i} \le \dfrac{b - a} n < \delta$ As the uniorm continuity condition applies, it follows that: :$\map \gamma {\closedint {x_i} {x_{i + 1} } } \subseteq \map {B_\epsilon} {\map \gamma {x_i} }$ {{qed|lemma}} === Complex plane case === Suppose that $S$ is a subset of $\C$. This theorem only uses the properties of $\C$ that depends on the metric. From Complex Plane is Metric Space, it follows that $\C$ and $\R^2$ are homeomorphic. It follows that the proof that applies for $\R^2$ also applies for $\C$. {{qed}}
We apply the Method of Truth Tables to the proposition. As can be seen for all boolean interpretations by inspection, where the truth value under the main connective on the {{LHS}} is $T$, that under the one on the {{RHS}} is also $T$: $\begin{array}{|ccccccc||ccccccc|} \hline (p & \implies & q) & \land & (r & \implies & s) & (p & \land & r) & \implies & (q & \land & s) \\ \hline F & T & F & T & F & T & F & F & F & F & T & F & F & F \\ F & T & F & T & F & T & T & F & F & F & T & F & F & T \\ F & T & F & F & T & F & F & F & F & T & T & F & F & F \\ F & T & F & T & T & T & T & F & F & T & T & F & F & T \\ F & T & T & T & F & T & F & F & F & F & T & T & F & F \\ F & T & T & T & F & T & T & F & F & F & T & T & T & T \\ F & T & T & F & T & F & F & F & F & T & T & T & F & F \\ F & T & T & T & T & T & T & F & F & T & T & T & T & T \\ T & F & F & F & F & T & F & T & F & F & T & F & F & F \\ T & F & F & F & F & T & T & T & F & F & T & F & F & T \\ T & F & F & F & T & F & F & T & T & T & F & F & F & F \\ T & F & F & F & T & T & T & T & T & T & F & F & F & T \\ T & T & T & T & F & T & F & T & F & F & T & T & F & F \\ T & T & T & T & F & T & T & T & F & F & T & T & T & T \\ T & T & T & F & T & F & F & T & T & T & F & T & F & F \\ T & T & T & T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ Hence the result. Note that the two formulas are not equivalent, as the relevant columns do not match exactly. {{qed}}
{{BeginTableau|\vdash \left({\left({p \implies q}\right) \land \left({r \implies s}\right)}\right) \implies \left({\left({p \land r}\right) \implies \left({q \land s}\right)}\right)}} {{Assumption|1|\left({\left({p \implies q}\right) \land \left({r \implies s}\right)}\right)}} {{SequentIntro|2|1|\left({p \land r}\right) \implies \left({q \land s}\right)|1|Praeclarum Theorema: Formulation 1}} {{Implication|3|1|\left({\left({p \implies q}\right) \land \left({r \implies s}\right)}\right) \implies \left({\left({p \land r}\right) \implies \left({q \land s}\right)}\right)|1|2}} {{EndTableau}} {{qed}}
Let $b \in R$. By Directed Set has Strict Successors iff Unbounded Above: :$\exists c \in R: b < c$ :$\exists d \in R: 1 < d$ By the definition of a directed set: :$\exists e \in R: d \le e, c \le e$ $\struct {R, +, \circ, \le}$ is an ordered ring, so $\le$ is by definition a transitive relation. Hence by transitivity: :$b < e$ and: :$1 < e$ By Strictly Positive Power of Strictly Positive Element Greater than One Succeeds Element: :$e \le \map f e$ Thus by transitivity: :$b < \map f e$ {{qed}} Category:Ring Theory 413wwed1qe7xtz89l3kxirtjmke6ll7
We apply the Method of Truth Tables to the proposition. As can be seen for all boolean interpretations by inspection, where the truth value under the main connective on the {{LHS}} is $T$, that under the one on the {{RHS}} is also $T$: : $\begin{array}{|ccccccc||ccccccc|} \hline (p & \implies & q) & \land & (r & \implies & s) & (p & \lor & r) & \implies & (q & \lor & s) \\ \hline F & T & F & T & F & T & F & F & F & F & T & F & F & F \\ F & T & F & T & F & T & T & F & F & F & T & F & T & T \\ F & T & F & F & T & F & F & F & T & T & F & F & F & F \\ F & T & F & T & T & T & T & F & T & T & T & F & T & T \\ F & T & T & T & F & T & F & F & F & F & T & T & T & F \\ F & T & T & T & F & T & T & F & F & F & T & T & T & T \\ F & T & T & F & T & F & F & F & T & T & T & T & T & F \\ F & T & T & T & T & T & T & F & T & T & T & T & T & T \\ T & F & F & F & F & T & F & T & T & F & F & F & F & F \\ T & F & F & F & F & T & T & T & T & F & T & F & T & T \\ T & F & F & F & T & F & F & T & T & T & F & F & F & F \\ T & F & F & F & T & T & T & T & T & T & T & F & T & T \\ T & T & T & T & F & T & F & T & T & F & T & T & T & F \\ T & T & T & T & F & T & T & T & T & F & T & T & T & T \\ T & T & T & F & T & F & F & T & T & T & T & T & T & F \\ T & T & T & T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ Hence the result. Note that the two formulas are not equivalent, as the relevant columns do not match exactly. {{qed}}
{{BeginTableau  |\vdash \paren {\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s} } \implies \paren {q \lor s} }} {{Assumption    |1|\paren {\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s} } }} {{Simplification|2|1|\paren {p \implies q} \land \paren {r \implies s}|1|2|Cutting corners: should use Associativity first}} {{SequentIntro  |3|1|\paren {p \lor r} \implies \paren {q \lor s}|2|Constructive Dilemma: Formulation 1}} {{Simplification|4|1|p \lor r|1|1}} {{ModusPonens   |5|1|q \lor s|3|4}} {{Implication   |6| |\paren {\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s} } \implies \paren {q \lor s}|1|5}} {{EndTableau}} {{qed}}
From the Power Rule for Derivatives: :$\map {D_x} {x^n} = n x^{n - 1}$ As $n$ is odd, $n - 1$ is even. Thus by Even Power is Non-Negative: :$\map {D_x} {x^n} \ge 0$ for all $x$. From Derivative of Monotone Function, it follows that $f_n$ is increasing over the whole of $\R$. The only place where $\map {D_x} {x^n} = 0$ is at $x = 0$. Everywhere else, $f_n$ is strictly increasing. By Sign of Odd Power: :$\map {f_n} x < 0 = \map {f_n} 0$ when $x < 0$ and :$\map {f_n} 0 = 0 < \map {f_n} x$ when $0 < x$. Thus $f_n$ is strictly increasing on $\R$. {{qed}} Category:Real Analysis Category:Real Numbers Category:Odd Power Function is Strictly Increasing 3ztw3ty74w9697jql4779mtobh2zy2y
As a counterexample, let $\mathbf A = \mathbf I_n$ be the unit matrix of order $n > 1$. Let $\mathbf B$ be any invertible matrix over $R$ of order $n$ that is different from the unit matrix. Then $\mathbf A \equiv \mathbf B$, as: :$\mathbf I_n^{-1} \mathbf A \mathbf B = \mathbf I_n^{-1} \mathbf I_n \mathbf B = \mathbf B$. If $\mathbf P$ is an invertible square matrix of order $n$, then: :$\mathbf P^{-1} \mathbf A \mathbf P = \mathbf P^{-1} \mathbf P = \mathbf I_n \ne \mathbf B$ Hence, $\mathbf A$ is not similar to $\mathbf B$. {{qed}}
{{proofread}} Let $x, y \in R$ such that $0 < x < y$. By Power Function is Strictly Increasing on Positive Elements: :$\map f x < \map f y$ Suppose that $x < y < 0$. By Properties of Ordered Ring: :$0 < -y < -x$ By Power Function is Strictly Increasing on Positive Elements (applied to $-y$ and $-x$): :$0 < \map f {-y} < \map f {-x}$ By Power of Ring Negative: :$\map f {-x} = -\map f x$ :$\map f {-y} = -\map f y$ Thus: :$0 < -\map f y < -\map f x$ By Properties of Ordered Ring: :$\map f x < \map f y$ By Sign of Odd Power: :$\map f x < 0 = \map f 0$ when $x < 0$ :$\map f 0 = 0 < \map f x$ when $0 < x$ Thus we have shown that $f$ is strictly increasing on the positive elements and the negative elements, and across zero. {{qed}} Category:Totally Ordered Rings Category:Odd Power Function is Strictly Increasing eh5d5yt5qq4k9gyuhnvli3v2uu8bfkm
Denote the Tychonoff topology on $X$ as $\tau$, and the box topology on $X$ as $\tau'$. Suppose that $U \in \tau'$. Then there exists an index set $I$ such that: :$\displaystyle U = \bigcup_{i \mathop \in I} \struct {U_{i, 1} \times U_{i, 2} \times \cdots \times U_{i, n} }$ where $U_{i, k} \in \tau_k$ for all $i \in I, k \in \set {1, \ldots, n}$. When $\pr_k: X \to X_k$ denotes the $k$th projection on $X$, we have: :$\pr_k^{-1} \sqbrk {U_{i, k} } = X_1 \times X_2 \times \cdots \times X_{k - 1} \times U_{i, k} \times X_{k + 1} \times \cdots \times X_n$ It follows that: :$\displaystyle U = \bigcup_{i \mathop \in I} \struct {U_{i, 1} \times U_{i, 2} \times \cdots \times U_{i, n} } = \bigcup_{i \mathop \in I} \bigcap_{k \mathop = 1}^{n_i} \pr_k^{-1} \sqbrk {U_{i, k} }$ By definition of Tychonoff topology, $U \in \tau$, so $\tau' \subseteq \tau$. {{qed|lemma}} Suppose that $U \in \tau$. Then there exists an index set $I$, so $U$ can be expressed as: $U = \displaystyle \bigcup_{i \mathop \in I} \bigcap_{l \mathop = 1}^{m_i} \pr_{j_l}^{-1} \sqbrk {U_{i, l} }$ where $m_i \in \N$, $j_1, j_2, \ldots, j_{m_i} \in \set {1, \ldots, n}$, and $U_{i, l} \in \tau_{j_l}$ for all $i \in I, l \in \set {1, \ldots, m_i}$. Then: {{begin-eqn}} {{eqn | l = U       | r = \bigcup_{i \mathop \in I} \bigcap_{l \mathop = 1}^{m_i} \pr_{j_l}^{-1} \sqbrk {U_{i, l} } }} {{eqn | r = \bigcup_{i \mathop \in I} \bigcap_{k \mathop = 1}^n \bigcap_{l: k \mathop = j_l} \pr_k^{-1} \sqbrk {U_{i, l} } }} {{eqn | r = \bigcup_{i \mathop \in I} \bigcap_{k \mathop = 1}^n \struct {X_1 \times X_2 \times \cdots \times X_{k - 1} \times \bigcap_{l: k \mathop = j_l} U_{i, l} \times X_{k + 1} \times \cdots \times X_n} }} {{end-eqn}} As $\displaystyle \bigcap_{l: k \mathop = j_l} U_{i, l} \in \tau_k$, it follows that: :$\displaystyle \struct {X_1 \times X_2 \times \cdots \times X_{k - 1} \times \bigcap_{l: k \mathop = j_l} U_{i, l} \times X_{k + 1} \times \cdots \times X_n} \in \tau'$ By definition of topology, it follows that $U \in \tau'$, so $\tau \subseteq \tau'$. {{qed}}
Let $1_R$ be the unity of $R$. Let $b \in R$. We will show that $b$ is not an upper bound of the image of $f$. Since $R$ is totally ordered and unbounded above, there is an element $c \in R$ such that $b < c$ and $1_R < c$. By Strictly Positive Power of Strictly Positive Element Greater than One Succeeds Element, $c \le \map f c$. Thus by transitivity of ordering, $b < \map f c$. {{qed}} Category:Totally Ordered Rings Category:Rings with Unity ourpo7kby1306f7ch5iz4zo25u4c3vb
If $n = 1$, then $f$ is the identity function. By the Archimedean Principle, the real numbers are unbounded above. Thus by definition of the identity function: $f$ is unbounded above. {{qed|lemma}} Let $n \ge 2$. {{AimForCont}} that $f$ is bounded above by $b \in \R$. {{WLOG}} suppose that $b > 0$. Then by the definition of an upper bound: :$\forall x \in \R: x^n \le b$ Let $x > b$. Then: :$\dfrac {x^n - b} {x - b} \le 0$ By the Mean Value Theorem, there exists a point $p$ between $b$ and $x$ such that: :$\map {f'} p = \dfrac {x^n - b} {x - b}$ By Derivative of Power: :$\map {f'} p = n p^{n - 1}$ Therefore $n p^{n - 1} \le 0$. By Power of Strictly Positive Real Number is Positive and Positive Real Numbers Closed under Multiplication it follows that: :$p > 0 \implies n p^{n - 1} > 0$ But this is impossible because it was previously established that $p \ge b > 0$. From this contradiction it follows that there can be no such $b$. Hence the result. {{qed}}
By Properties of Ordered Ring $(6)$: :$z \circ x < z \circ y$ :$z \circ y < w \circ y$ Then by transitivity of $\circ$: :$z \circ x < w \circ y$ Also by Properties of Ordered Ring $(6)$: :$z \circ 0 < z \circ x$ Hence by Ring Product with Zero: :$0 < z \circ x$ {{qed}} Category:Ordered Rings Category:Inequalities e0clz004vppggt5grtn0ls0bmjjxvfu
Proof by induction: For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$0 < \map {\circ^n} x < \map {\circ^n} y$ === Basis for the Induction === $\map P 1$ is the case: :$0 < \map {\circ^1} x < \map {\circ^1} y$ which is just: :$0 < x < y$ Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown 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 the induction hypothesis: :$0 < \map {\circ^k} x < \map {\circ^k} y$ from which it is to be shown that: :$0 < \map {\circ^{k + 1}} x < \map {\circ^{k + 1}} y$ === Induction Step === This is the induction step: We have: :$0 < x < y$ :$0 < \map {\circ^k} x < \map {\circ^k} y$ By Ring Product preserves Inequalities on Positive Elements: :$0 < \map {\circ^k} x \circ x < \map {\circ^k} y \circ y$ Hence: :$0 < \map {\circ^{k + 1}} x < \map {\circ^{k + 1}} y$ So $\map P k \implies \map P {k + 1}$ and thus it follows by the Principle of Mathematical Induction that: :$\forall n \in \N_{> 0}: 0 < \map {\circ^n} x < \map {\circ^n} y$ {{qed}}
Proof of associativity by natural deduction is just too tedious to be considered. We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective match for all boolean interpretations. $\begin{array}{|ccccc||ccccc|} \hline p & \iff & (q & \iff & r) & (p & \iff & q) & \iff & r \\ \hline F & F & F & T & F & F & T & F & F & F \\ F & T & F & F & T & F & T & F & T & T \\ F & T & T & F & F & F & F & T & T & F \\ F & F & T & T & T & F & F & T & F & T \\ T & T & F & T & F & T & F & F & T & F \\ T & F & F & F & T & T & F & F & F & T \\ T & F & T & F & F & T & T & T & F & F \\ T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
{{BeginTableau|\vdash \paren {p \iff \paren {q \iff r} } \iff \paren {\paren {p \iff q} \iff r} }} {{Assumption|1|p \iff \paren {q \iff r} }} {{SequentIntro|2|1|\paren {p \iff q} \iff r|1|Biconditional is Associative: Formulation 1}} {{Implication|3||\paren {p \iff \paren {q \iff r} } \implies \paren {\paren {p \iff q} \iff r}|1|2}} {{Assumption|4|\paren {p \iff q} \iff r}} {{SequentIntro|5|4|p \iff \paren {q \iff r}|4|Biconditional is Associative: Formulation 1}} {{Implication|6||\paren {\paren {p \iff q} \iff r} \implies \paren {p \iff \paren {q \iff r} }|4|5}} {{BiconditionalIntro|7||\paren {p \iff \paren {q \iff r} } \iff \paren {\paren {p \iff q} \iff r}|3|6}} {{EndTableau|qed}}
Let $v \in V$ and $\epsilon \in \R_{>0}$. Denote the open $\epsilon$-ball of $v$ as $B_\epsilon \left({v}\right)$. Let $x, y \in B_\epsilon \left({v}\right)$. Then $x + t \left({y - x}\right)$ lies on line segment joining $x$ and $y$ for all $t \in \left[{0 \,.\,.\, 1}\right]$. The distance between $x + t \left({y - x}\right)$ and $v$ is: {{begin-eqn}} {{eqn |l= \left\Vert{x + t \left({y - x}\right) - v}\right\Vert       |r= \left\Vert{\left({1 - t}\right) \left({x - v}\right) + t \left({y - v}\right) }\right\Vert }} {{eqn |o= \le       |r= \left\Vert{\left({1 - t}\right) \left({x - v}\right) }\right\Vert + \left\Vert{t \left({y - v}\right) }\right\Vert       |c= by triangle inequality }} {{eqn |r= \left({1 - t}\right) \left\Vert{x - v}\right\Vert + t \left\Vert{y-v}\right\Vert       |c= by norm axiom $(N2)$ }} {{eqn |o= <       |r= \left({1 - t}\right) \epsilon + t \epsilon       |c= as $x, y \in B_\epsilon \left({v}\right)$ }} {{eqn |r= \epsilon }} {{end-eqn}} Hence, $x + t \left({y - x}\right) \in B_\epsilon \left({v}\right)$. By definition of convex set, $B_\epsilon \left({v}\right)$ is a convex set. {{qed}} Category:Vector Spaces Category:Open Balls 8lwxjlnq0kh1w3z7tvrti5n8wiaigx4
=== Necessary Condition === Let $M$ be a metric space. From Metric Space is Hausdorff, $M$ is a Hausdorff space. From: :$T_2$ (Hausdorff) Space is $T_1$ Space and: :$T_1$ Space is $T_0$ (Kolmogorov) Space it follows that $M$ is a Kolmogorov space. {{qed|lemma}} === Sufficient Condition === Suppose now that $M$ is a Kolmogorov space. Let $a, b \in S$ such that $a \ne b$. Then $\exists U \in \tau$ such that: :$\paren {a \in U \land b \notin U} \lor \paren {b \in U \land a \notin U}$ {{WLOG}}, suppose that $a \in U$ and $b \notin U$. Then by the definition of the induced topology: :$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} a \subset U$ where $\map {B_\epsilon} a$ denotes the open $\epsilon$-ball of $a$ in $M$. Since $b \notin U$: :$\map d {a, b} \ge \epsilon$ Since $\epsilon > 0$: :$\map d {a, b} > 0$ Since this holds for any pair of distinct points, $M$ is a metric space. {{qed}} Category:Metric Spaces Category:Pseudometric Spaces Category:T0 Spaces rwglezgrcr8npj1mxld4jz80qzfdi6b