hcju/proofwikiqa huggingface dataset:
From the definition of complex numbers, we define the following: {{begin-eqn}} {{eqn | l = z_1       | o = :=       | r = \tuple {x_1, y_1} }} {{eqn | l = z_2       | o = :=       | r = \tuple {x_2, y_2} }} {{eqn | l = z_3       | o = :=       | r = \tuple {x_3, y_3} }} {{end-eqn}} where $x_1, x_2, x_3, y_1, y_2, y_3 \in \R$. Thus: {{begin-eqn}} {{eqn | l = z_1 \paren {z_2 + z_3}       | r = \tuple {x_1, y_1} \paren {\tuple {x_2, y_2} + \tuple {x_3, y_3} }       | c = {{Defof|Complex Number|index = 2}} }} {{eqn | r = \tuple {x_1, y_1} \tuple {x_2 + x_3, y_2 + y_3}       | c = {{Defof|Complex Number/Definition 2/Addition|Complex Addition}} }} {{eqn | r = \tuple {x_1 \paren {x_2 + x_3} - y_1 \paren {y_2 + y_3}, x_1 \paren {y_2 + y_3} + y_1 \paren {x_2 + x_3} }       | c = {{Defof|Complex Number/Definition 2/Multiplication|Complex Multiplication}} }} {{eqn | r = \tuple {x_1 x_2 + x_1 x_3 - y_1 y_2 - y_1 y_3, x_1 y_2 + x_1 y_3 + y_1 x_2 + y_1 x_3}       | c = Real Multiplication Distributes over Addition }} {{eqn | r = \tuple {\paren {x_1 x_2 - y_1 y_2}\ + \paren {x_1 x_3 - y_1 y_3}, \paren {x_1 y_2 + y_1 x_2} + \paren {x_1 y_3 + y_1 x_3} }       | c = Real Addition is Commutative }} {{eqn | r = \tuple {x_1 x_2 - y_1 y_2, x_1 y_2 + y_1 x_2} + \tuple {x_1 x_3 - y_1 y_3, x_1 y_3 + y_1 x_3}       | c = {{Defof|Complex Number/Definition 2/Addition|Complex Addition}} }} {{eqn | r = \tuple {x_1, y_1} \tuple {x_2, y_2} + \tuple {x_1, y_1} \tuple {x_3, y_3}       | c = {{Defof|Complex Number/Definition 2/Multiplication|Complex Multiplication}} }} {{eqn | r = z_1 z_2 + z_1 z_3       | c = {{Defof|Complex Number|index = 2}} }} {{end-eqn}} The result $\paren {z_2 + z_3} z_1 = z_2 z_1 + z_3 z_1$ follows directly from the above, and the fact that Complex Multiplication is Commutative. {{qed}}
Let $\overline z$ denote the complex conjugate of $z$.  Then: {{begin-eqn}} {{eqn | l = \cmod {z_1 z_2}       | r = \sqrt {\paren {z_1 z_2} \overline {\paren {z_1 z_2} } }       | c = Modulus in Terms of Conjugate }} {{eqn | r = \sqrt {z_1 \overline {z_1} z_2 \overline {z_2} }       | c = Product of Complex Conjugates, Complex Multiplication is Commutative }} {{eqn | r = \sqrt {z_1 \overline {z_1} } \sqrt {z_2 \overline {z_2} }       | c = Power of Product }} {{eqn | r = \cmod {z_1} \cmod {z_2} }} {{end-eqn}} {{qed}}
Let: : $z_1 = r_1 \left({\cos \theta_1 + i \sin \theta_1}\right)$ : $z_2 = r_2 \left({\cos \theta_2 + i \sin \theta_2}\right)$ Then: {{begin-eqn}} {{eqn | l = \left\vert{z_1 z_2}\right\vert       | r = \left\vert{r_1 \left({\cos \theta_1 + i \sin \theta_1}\right) r_2 \left({\cos \theta_2 + i \sin \theta_2}\right)}\right\vert       | c = {{Defof|Polar Form of Complex Number}} }} {{eqn | r = \left\vert{r_1 r_2 \left({\cos \left({\theta_1 + \theta_2}\right) + i \sin \left({\theta_1 + \theta_2}\right)}\right)}\right\vert       | c = Product of Complex Numbers in Polar Form }} {{eqn | r = r_1 r_2       | c = {{Defof|Polar Form of Complex Number}} }} {{eqn | r = \left\vert{z_1}\right\vert \left\vert{z_2}\right\vert       | c = {{Defof|Polar Form of Complex Number}} }} {{end-eqn}} {{qed}}
Let $z = x + i y$. Then: {{begin-eqn}} {{eqn | l = z       | r = \overline z       | c =  }} {{eqn | ll= \leadsto       | l = x + i y       | r = x - i y       | c = {{Defof|Complex Conjugate}} }} {{eqn | ll= \leadsto       | l = +y       | r = -y       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = 0       | c =  }} {{end-eqn}} Hence by definition, $z$ is wholly real. {{qed|lemma}} Now suppose $z$ is wholly real. Then: {{begin-eqn}} {{eqn | l = z       | r = x + 0 i       | c =  }} {{eqn | r = x       | c =  }} {{eqn | r = x - 0 i       | c =  }} {{eqn | r = \overline z       | c =  }} {{end-eqn}} {{qed}}
{{AimForCont}} there exists a relation $\preceq$ on $\C$ which is ordering compatible with the ring structure of $\C$. That is: :$(1): \quad z \ne 0 \implies 0 \prec z \lor z \prec 0$, but not both :$(2): \quad 0 \prec z_1, z_2 \implies 0 \prec z_1 z_2 \land 0 \prec z_1 + z_2$ By Totally Ordered Ring Zero Precedes Element or its Inverse, $(1)$ can be replaced with: :$(1'): \quad z \ne 0 \implies 0 \prec z \lor 0 \prec -z$, but not both. As $i \ne 0$, it follows that: :$0 \prec i$ or $0 \prec -i$ Suppose $0 \prec i$. Then: {{begin-eqn}} {{eqn | l = 0       | o = \prec       | r = i \times i       | c = from $(2)$ }} {{eqn | r = -1       | c = {{Defof|Complex Number|index = 1}} }} {{end-eqn}} Otherwise, suppose $0 \prec \paren {-i}$. Then: {{begin-eqn}} {{eqn | l = 0       | o = \prec       | r = \paren {-i} \times \paren {-i}       | c = from $(2)$ }} {{eqn | r = -1       | c = {{Defof|Complex Number|index = 1}} }} {{end-eqn}} Thus by Proof by Cases: :$0 \prec -1$ Thus it follows that: {{begin-eqn}} {{eqn | l = 0       | o = \prec       | r = \paren {-1} \times \paren {-1}       | c = from $(2)$ }} {{eqn | r = 1       | c =  }} {{end-eqn}} Thus both: :$0 \prec -1$ and: :$0 \prec 1$ This contradicts hypothesis $(1')$: :$(1): \quad z \ne 0 \implies 0 \prec z \lor 0 \prec -z$, but not both Hence, by Proof by Contradiction, there can be no such ordering. {{qed}}
{{AimForCont}} such a total ordering $\preceq$ exists. By the definition of a total ordering, $\preceq$ is connected. That is: :$0 \preceq i \lor i \preceq 0$ Using Proof by Cases, we will prove that: :$0 \preceq -1$ ;Case 1 Assume that $0 \preceq i$. By definition of an ordering compatible with the ring structure of $\C$: :$\forall x, y \in \C: \left({0 \preceq x \land 0 \preceq y}\right) \implies 0 \preceq x \times y$ Substituting $x = i$ and $y = i$ gives: :$0 \preceq i \times i$ Simplifying: :$0 \preceq -1$ which is the result required. ;Case 2 Assume that $i \preceq 0$. By definition of compatibility with addition: :$\forall x, y, z \in \C: x \preceq y \implies \paren {x + z} \preceq \paren {y + z}$ Substituting $x = i$, $y = 0$, $z = -i$ gives: :$i + \paren {-i} \preceq 0 + \paren {-i}$ Simplifying: :$0 \preceq -i$ By definition of an ordering compatible with the ring structure of $\C$: :$\forall x, y \in \C: \paren {0 \preceq x \land 0 \preceq y} \implies 0 \preceq x \times y$ Substituting $x = -i$ and $y = -i$ gives: :$0 \preceq \paren {-i} \times \paren {-i}$ Simplifying: :$0 \preceq -1$ This has been demonstrated to follow from both cases, and so by Proof by Cases: :$0 \preceq -1$ {{qed|lemma}} By definition of an ordering compatible with the ring structure of $\C$: :$\forall x, y \in \C: \paren {0 \preceq x \land 0 \preceq y} \implies 0 \preceq x \times y$ Substituting $x = -1$ and $y = -1$: :$0 \preceq \paren {-1} \times \paren {-1}$ Simplifying: :$0 \preceq 1$ By definition of compatibility with addition: :$\forall x, y, z \in \C: x \preceq y \implies \paren {x + z} \preceq \paren {y + z}$ Substituting $x = 0$, $y = 1$, $z = -1$ gives: :$0 + \paren {-1} \preceq 1 + \paren {-1}$ Simplifying: :$-1 \preceq 0$ From the definition of ordering: :$\forall a, b \in \C: \paren {a \preceq b \land b \preceq a} \implies a = b$ Substituting $a = -1$ and $b = 0$ gives: :$-1 = 0$ which is a contradiction. Hence, from Proof by Contradiction, there can be no such ordering. {{qed}}
From Complex Numbers form Integral Domain, $\struct {\C, +, \times}$ is an integral domain. {{AimForCont}} that $\struct {\C, +, \times}$ can be ordered. Thus, by definition, it possesses a (strict) positivity property $P$. Then from Strict Positivity Property induces Total Ordering, let $\le$ be the total ordering induced by $P$. From Unity of Ordered Integral Domain is Strictly Positive: :$1$ is strictly positive. Thus by strict positivity, axiom $3$: :$-1$ is not strictly positive. Consider the element $i \in \C$. By definition of strict positivity, axiom $3$, either: :$i$ is strictly positive or: :$-i$ is strictly positive. Suppose $i$ is strictly positive. Then by Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive: :$i^2 = -1$ is strictly positive. Similarly, suppose $-i$ is strictly positive. Then by Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive: :$\paren {-i}^2 = -1$ is strictly positive. In both cases we have that $-1$ is strictly positive. But it has already been established that $-1$ is not strictly positive. Hence, by Proof by Contradiction, there can be no such ordering. {{qed}}
=== Proof for Complex Analysis === Suppose $L' \ne L$ is another limit of $f \left({z}\right)$ at $z_0$. Let us take $\epsilon = \dfrac {\left\vert{L - L'}\right\vert} 2$. Then we can find $\delta_1 > 0, \delta_2 > 0$ such that: * $z \in S, 0 < \left\vert{z - z_0}\right\vert < \delta_1 \implies \left\vert{f \left({z}\right) - L}\right\vert < \epsilon$ * $z \in S, 0 < \left\vert{z - z_0}\right\vert < \delta_2 \implies \left\vert{f \left({z}\right) - L'}\right\vert < \epsilon$ Because $z_0$ is a limit point: : $\exists z^* \in S: 0 < \left\vert{z - z_0}\right\vert < \min \left\{{\delta_1, \delta_2}\right\}$ Then: {{begin-eqn}} {{eqn | l=\left\vert{L - L'}\right\vert       | r=\left\vert{L - f \left({z^*}\right) + f \left({z^*}\right) - L'}\right\vert       | c= }} {{eqn | o=\le       | r=\left\vert{L - f \left({z^*}\right)}\right\vert + \left\vert{f \left({z^*}\right) - L'}\right\vert       | c=Triangle Inequality }} {{eqn | o=<       | r=\epsilon + \epsilon       | c= }} {{eqn | r=2 \epsilon       | c= }} {{end-eqn}} This contradicts the choice we made of $\epsilon$. {{qed}} {{wtd|Expand the context to include the more general result for metric spaces etc.}} Category:Complex Analysis 711tzlnlrulnvtl4yuph2knofq1ko50
Let $z_1 = x_1 + i y_1, z_2 = x_2 + i y_2$. From the definition of modulus: :$\size {z_1 - z_2} = \sqrt {\paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2}$ This is the euclidean metric on the real number plane This is shown in Euclidean Metric on Real Vector Space is Metric to be a metric. Thus the complex plane is a 2-dimensional Euclidean space. {{qed}}
=== Proof of $\text M 1$ === {{begin-eqn}} {{eqn | l = \map d {x, x}       | r = \size {x - x}       | c = Definition of $d$ }} {{eqn | r = 0       | c = {{Defof|Absolute Value}} }} {{end-eqn}} So axiom $\text M 1$ holds for $d$. {{qed|lemma}} === Proof of $\text M 2$ === {{begin-eqn}} {{eqn | l = \map d {x, y} + \map d {y, z}       | r = \size {x - y} + \size {y - z}       | c = Definition of $d$ }} {{eqn | o = \ge       | r = \size {\paren {x - y} + \paren {y - z} }       | c = Triangle Inequality for Real Numbers }} {{eqn | r = \size {x - z}       | c =  }} {{eqn | r = \map d {x, z}       | c = Definition of $d$ }} {{end-eqn}} So axiom $\text M 2$ holds for $d$. {{qed|lemma}} === Proof of $\text M 3$ === {{begin-eqn}} {{eqn | l = \map d {x, y}       | r = \size {x - y}       | c = Definition of $d$ }} {{eqn | r = \size {y - x}       | c = {{Defof|Absolute Value}} }} {{eqn | r = \map d {y, x}       | c = Definition of $d$ }} {{end-eqn}} So axiom $\text M 3$ holds for $d$. {{qed|lemma}} === Proof of $\text M 4$ === {{begin-eqn}} {{eqn | l = x       | o = \ne       | r = y       | c =  }} {{eqn | ll= \leadsto       | l = \size {x - y}       | o = >       | r = 0       | c = {{Defof|Absolute Value}} }} {{eqn | ll= \leadsto       | l = \map d {x, y}       | o = >       | r = 0       | c = Definition of $d$ }} {{end-eqn}} So axiom $\text M 4$ holds for $d$. {{qed}}
=== Proof of $M1$ === {{begin-eqn}} {{eqn | l = d_p \left({x, x}\right)       | r = \left({\sum_{i \mathop = 1}^n \left \vert {x_i - x_i} \right \vert^p}\right)^{\frac 1 p}       | c = Definition of $d_p$ }} {{eqn | r = \left({\sum_{i \mathop = 1}^n 0^p}\right)^{\frac 1 p}       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} So axiom $M1$ holds for $d_p$. {{qed|lemma}} === Proof of $M2$ === It is required to be shown: :$d \left({x, y}\right) + d \left({y, z}\right) \ge d \left({x, z}\right)$ for all $x, y, z \in \R^n$. Let: :$(1): \quad z = \left({z_1, z_2, \ldots, z_n}\right)$ :$(2): \quad$ all summations be over $i = 1, 2, \ldots, n$ :$(3): \quad \left\vert{x_i - y_i}\right\vert = r_i$ :$(4): \quad \left\vert{y_i - z_i}\right\vert = s_i$. Thus we need to show that: :$\displaystyle \left({\sum \left\vert{x_i - y_i}\right\vert^p}\right)^{\frac 1 p} + \left({\sum \left\vert{y_i - z_i}\right\vert^p}\right)^{\frac 1 p} \ge \left({\sum \left\vert{x_i - z_i}\right\vert^p}\right)^{\frac 1 p}$ We have: {{begin-eqn}} {{eqn | l = d_p \left({x, y}\right) + d_p \left({y, z}\right)       | r = \left({\sum \left\vert{x_i - y_i}\right\vert^p}\right)^{\frac 1 p} + \left({\sum \left\vert{y_i - z_i}\right\vert^p}\right)^{\frac 1 p}       | c = Definition of $d_p$ }} {{eqn | r = \left({\sum r_i^p}\right)^{\frac 1 p} + \left({\sum s_i^p}\right)^{\frac 1 p}       | c =  }} {{eqn | o = \ge       | r = \left({\sum \left({r_i + s_i}\right)^p}\right)^{\frac 1 p}       | c = Minkowski's Inequality for Sums }} {{eqn | o = \ge       | r = \left({\sum \left({\left\vert{x_i - y_i}\right\vert + \left\vert{y_i - z_i}\right\vert}\right)^p}\right)^{\frac 1 p}       | c = Definition of $r_i$ and $s_i$ }} {{eqn | r = \left({\sum \left\vert{x_i - z_i}\right\vert^p}\right)^{\frac 1 p}       | c = Triangle Inequality for Real Numbers }} {{eqn | r = d_p \left({x, z}\right)       | c = Definition of $d_p$ }} {{end-eqn}} So axiom $M2$ holds for $d_p$. {{qed|lemma}} === Proof of $M3$ === {{begin-eqn}} {{eqn | l = d_p \left({x, y}\right)       | r = \left({\sum \left\vert{x_i - y_i}\right\vert^p}\right)^{\frac 1 p}       | c = Definition of $d_p$ }} {{eqn | r = \left({\sum \left\vert{y_i - x_i}\right\vert^p}\right)^{\frac 1 p}       | c =  }} {{eqn | r = d_p \left({y, x}\right)       | c = Definition of $d_p$ }} {{end-eqn}} So axiom $M3$ holds for $d_p$. {{qed|lemma}} === Proof of $M4$ === {{begin-eqn}} {{eqn | l = x       | o = \ne       | r = y       | c =  }} {{eqn | ll= \implies       | l = \exists k \in \left[{1 \,.\,.\, n}\right]: x_k       | o = \ne       | r = y_k       | c =  }} {{eqn | ll= \implies       | l = x_k - y_k       | o = >       | r = 0       | c = as $d_k$ fulfils axiom $M4$ }} {{eqn | ll= \implies       | l = \left({\sum \left\vert{x_k - y_k}\right\vert^p}\right)^{\frac 1 p}       | o = >       | r = 0       | c =  }} {{eqn | ll= \implies       | l = d_p \left({x, y}\right)       | o = >       | r = 0       | c = Definition of $d_p$ }} {{end-eqn}} So axiom $M4$ holds for $d_p$. {{qed}}
Follows directly from P-Product Metric is Metric, where in this case $p = 1$. {{qed}}
It is sufficient to prove this for two subsets, as the general result follows by induction. Suppose $S_1$ and $S_2$ are bounded subsets of $M = \left({A, d}\right)$. Let $a_1, a_2 \in A$. Let $K_1, K_2 \in \R$ such that: : $(1): \quad \forall x \in S_1: d \left({x, a_1}\right) \le K_1$ : $(2): \quad \forall x \in S_2: d \left({x, a_2}\right) \le K_2$ {{WLOG}}, let $a = a_1$ and $K = \max \left\{{K_1, K_2 + d \left({a_1, a_2}\right)}\right\}$. Then $\forall x \in S_1 \cup S_2$: Either: : $x \in S_1$ and so $d \left({x, a}\right) \le K_1 \le K$ or: : $x \in S_2$ and so $d \left({x, a}\right) = d \left({x, a_1}\right) \le d \left({x, a_2}\right) + d \left({a_2, a_1}\right) \le K_2 + d \left({a_2, a_1}\right) \le K$ Hence the result. {{qed}}
=== Proof of $M1$ === {{begin-eqn}} {{eqn | l = d_p \left({x, x}\right)       | r = \left({\sum_{i \mathop = 1}^n \left({d_{i'} \left({x_i, x_i}\right)}\right)^p}\right)^{\frac 1 p}       | c = Definition of $d_p$ }} {{eqn | r = \left({\sum_{i \mathop = 1}^n 0^p}\right)^{\frac 1 p}       | c = as $d_{i'}$ fulfils axiom $M1$ }} {{eqn | r = 0       | c =  }} {{end-eqn}} So axiom $M1$ holds for $d_p$. {{qed|lemma}} === Proof of $M2$ === Let: :$(1): \quad z = \left({z_1, z_2, \ldots, z_n}\right)$ :$(2): \quad$ all summations be over $i = 1, 2, \ldots, n$ :$(3): \quad d_{i'} \left({x_i, y_i}\right) = r_i$ :$(4): \quad d_{i'} \left({y_i, z_i}\right) = s_i$. Thus we need to show that: :$\displaystyle \left({\sum \left({d_{i'} \left({x_i, y_i}\right)}\right)^p}\right)^{\frac 1 p} + \left({\sum \left({d_{i'} \left({y_i, z_i}\right)}\right)^p}\right)^{\frac 1 p} \ge \left({\sum \left({d_{i'} \left({x_i, z_i}\right)}\right)^p}\right)^{\frac 1 p}$ We have: {{begin-eqn}} {{eqn | l = d_p \left({x, y}\right) + d_p \left({y, z}\right)       | r = \left({\sum \left({d_{i'} \left({x_i, y_i}\right)}\right)^p}\right)^{\frac 1 p} + \left({\sum \left({d_{i'} \left({y_i, z_i}\right)}\right)^p}\right)^{\frac 1 p}       | c = Definition of $d_p$ }} {{eqn | r = \left({\sum r_i^p}\right)^{\frac 1 p} + \left({\sum s_i^p}\right)^{\frac 1 p}       | c =  }} {{eqn | o = \ge       | r = \left({\sum \left({r_i + s_i}\right)^p}\right)^{\frac 1 p}       | c = Minkowski's Inequality for Sums }} {{eqn | r = \left({\sum \left({d_{i'} \left({x_i, y_i}\right) + d_{i'} \left({y_i, z_i}\right)}\right)^p}\right)^{\frac 1 p}       | c = Definition of $r_i$ and $s_i$ }} {{eqn | o = \ge       | r = \left({\sum \left({d_{i'} \left({x_i, z_i}\right)}\right)^p}\right)^{\frac 1 p}       | c = as $d_{i'}$ fulfils axiom $M2$ }} {{eqn | r = d_p \left({x, z}\right)       | c = Definition of $d_p$ }} {{end-eqn}} So axiom $M2$ holds for $d_p$. {{qed|lemma}} === Proof of $M3$ === {{begin-eqn}} {{eqn | l = d_p \left({x, y}\right)       | r = \left({\sum \left({d_{i'} \left({x_i, y_i}\right)}\right)^p}\right)^{\frac 1 p}       | c = Definition of $d_p$ }} {{eqn | r = \left({\sum \left({d_{i'} \left({y_i, x_i}\right)}\right)^p}\right)^{\frac 1 p}       | c = as $d_i$ fulfils axiom $M3$ }} {{eqn | r = d_p \left({y, x}\right)       | c = Definition of $d_p$ }} {{end-eqn}} So axiom $M3$ holds for $d_p$. {{qed|lemma}} === Proof of $M4$ === {{begin-eqn}} {{eqn | l = x       | o = \ne       | r = y       | c =  }} {{eqn | ll= \implies       | l = \exists k \in \left[{1 \,.\,.\, n}\right]: x_k       | o = \ne       | r = y_k       | c =  }} {{eqn | ll= \implies       | l = d_k \left({x_k, y_k}\right)       | o = >       | r = 0       | c = as $d_k$ fulfils axiom $M4$ }} {{eqn | ll= \implies       | l = \left({\sum \left({d_{i'} \left({y_i, x_i}\right)}\right)^p}\right)^{\frac 1 p}       | o = >       | r = 0       | c =  }} {{eqn | ll= \implies       | l = d_p \left({x, y}\right)       | o = >       | r = 0       | c = Definition of $d_p$ }} {{end-eqn}} So axiom $M4$ holds for $d_p$. {{qed}}
=== Lemma === {{:Euler Triangle Formula/Lemma}} {{qed|lemma}} :400px Let the incenter of $\triangle ABC$ be $I$. Let the circumcenter of $\triangle ABC$ be $O$. Let $OI$ be produced to the circumcircle at $G$ and $J$. Let $F$ be the point where the incircle of $\triangle ABC$ meets $BC$. We are given that: :the distance between the incenter and the circumcenter is $d$ :the inradius is $\rho$ :the circumradius is $R$. Thus: :$OI = d$ :$OG = OJ = R$ Therefore: :$IJ = R + d$ :$GI = R - d$ By the Intersecting Chord Theorem: : $GI \cdot IJ = IP \cdot CI$ By the lemma: :$IP = PB$ and so: :$GI \cdot IJ = PB \cdot CI$ Now using the Extension of Law of Sines in $\triangle CPB$: :$\dfrac {PB} {\sin \left({\angle PCB}\right)} = 2 R$ and so: :$GI \cdot IJ = 2 R \sin \left({\angle PCB}\right) \cdot CI$ By the $4$th of Euclid's common notions: :$\angle PCB = \angle ICF$ and so: :$(1): \quad GI \cdot IJ = 2 R \sin \left({\angle ICF}\right) \cdot CI$ We have that: :$IF = \rho$ and by Radius at Right Angle to Tangent: :$\angle IFC$ is a right angle. By the definition of sine: :$\sin \left({\angle ICF}\right) = \dfrac {\rho} {CI}$ and so: :$\sin \left({\angle ICF}\right) \cdot CI = \rho$ Substituting in $(1)$: {{begin-eqn}} {{eqn | l = GI \cdot IJ       | r = 2 R \rho       | c =  }} {{eqn | ll= \implies       | l = \left({R + d}\right) \left({R - d}\right)       | r = 2 R \rho       | c =  }} {{eqn | ll= \implies       | l = R^2 - d^2       | r = 2 R \rho       | c = Difference of Two Squares }} {{eqn | ll= \implies       | l = d^2       | r = R^2 - 2 R \rho       | c =  }} {{eqn | r = R \left({R - 2 \rho}\right)       | c =  }} {{end-eqn}} {{qed}} {{namedfor|Leonhard Paul Euler|cat = Euler}} Category:Triangles 3epb0l67d8pfil90dh4r9ab1iotbqma
Let $\delta = \epsilon - d \left({x, y}\right)$. From the definition of open ball, this is strictly positive, since $y \in B_\epsilon \left({x}\right)$. If $z \in B_\delta \left({y}\right)$, then $d \left({y, z}\right) < \delta$. So: : $d \left({x, z}\right) \le d \left({x, y}\right) + d \left({y, z}\right) < d \left({x, y}\right) + \delta = \epsilon$ Thus $z \in B_\epsilon \left({x}\right)$. So $B_\delta \left({y}\right) \subseteq B_\epsilon \left({x}\right)$. {{qed}} This diagram illustrates the proof in $M = \left({\R^2, d_2}\right)$: :300px In $\R^2$, we can draw a disc radius $\delta$ whose center is $y$ and which lies entirely within the larger disc whose center is $x$ with radius $\epsilon$.
We have the results: : Empty Set is Open in Metric Space : Metric Space is Open in Itself {{qed}}
=== Definition by Points implies Definition by Open Sets === Suppose that $f$ is continuous at every point $x \in A_1$. Let $U \subseteq M_2$ be open in $M_2$. Let $x \in f^{-1} \left({U}\right)$. Since $U$ is open in $M_2$: : $\exists \epsilon \in \R_{>0}: B_\epsilon \left({f \left({x}\right); d_2}\right) \subseteq U$ where $B_\epsilon \left({f \left({x}\right); d_2}\right)$ denotes the open $\epsilon$-ball of $f \left({x}\right)$ in $M_2$. By the definition of continuity at a point: : $\exists \delta \in \R_{>0}: f \left({B_\delta \left({x}\right); d_1}\right) \subseteq B_\epsilon \left({f \left({x}\right); d_2}\right)$ So: : $f \left({B_\delta \left({x}\right)}\right) \subseteq U$ and so: : $B_\delta \left({x}\right) \subseteq f^{-1} \left({U}\right)$ Thus $f^{-1} \left({U}\right)$ is open in $M_1$. {{qed|lemma}} === Definition by Open Sets implies Definition by Points === Suppose $f$ is defined to be continuous in the sense that: :for every $U \subseteq A_2$ which is open in $M_2$, $f^{-1} \left({U}\right)$ is open in $M_1$. Let $x \in A_1$. Then by Open Ball of Point Inside Open Ball: : $\exists \epsilon \in \R_{>0}: B_\epsilon \left({f \left({x}\right); d_2}\right)$ is open in $M_2$ So by hypothesis, $f^{-1} \left({B_\epsilon \left({f \left({x}\right); d_2}\right)}\right)$ is open in $M_1$. As $f \left({x}\right) \in B_\epsilon \left({f \left({x}\right); d_2}\right)$, it follows that: : $x \in f^{-1} \left({B_\epsilon \left({f \left({x}\right); d_2}\right)}\right)$ So by hypothesis: : $\exists \delta \in \R_{>0}: B_\delta \left({x; d_1}\right) \subseteq f^{-1} \left({B_\epsilon \left({f \left({x}\right); d_2}\right)}\right)$ Then: : $f \left({B_\delta \left({x}\right)}\right) \subseteq B_\epsilon \left({f \left({x}\right)}\right)$ Thus by the $\epsilon$-Ball definition, $f$ is continuous at $x$. {{qed}}
Let $\ds x \in \bigcap_{i \mathop = 1}^n U_i$. By the definition of intersection, for each $i \in \closedint 1 n$, we have $x \in U_i$. Thus, since $U_i$ is open in $M$: :$\exists \epsilon_i > 0: \map {B_{\epsilon_i} } x \subseteq U_i$ where $\map {B_{\epsilon_i} } x$ is the open $\epsilon_i$-ball of $x$. Let $\displaystyle \epsilon = \min_{i \mathop = 1}^n \set {\epsilon_i}$. Then, by Open Ball contains Smaller Open Ball: :$\map {B_\epsilon} x \subseteq \map {B_{\epsilon_i} } x$ for all $i \in \closedint 1 n$. So: :$\ds \map {B_\epsilon} x \subseteq \bigcap_{i \mathop = 1}^n U_i$ The result follows. {{qed}}
Let $I$ be any indexing set. Let $U_i$ be open in $M$ for all $i \in I$. Let $\ds x \in \bigcup_{i \mathop \in I} U_i$. Then by definition of set union, $x \in U_k$ for some $k \in I$. Since $U_k$ is open in $M$: :$\ds \exists \epsilon > 0: \map {B_\epsilon} x \subseteq U_k$ where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$ in $M$. By Set is Subset of Union: :$\ds U_k \subseteq \bigcup_{i \mathop \in I} U_i$ Hence by Subset Relation is Transitive: :$\ds \map {B_\epsilon} x \subseteq \bigcup_{i \mathop \in I} U_i$ and the result follows by definition of open set in metric space. {{qed}}