hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = \map {\frac \d {\d x} } {x^{x^x} }       | r = \map {\frac \d {\d x} } {e^{x^x \ln x} } }} {{eqn | r = e^{x^x \ln x} \map {\frac \d {\d x} } {x^x \ln x}       | c = Chain Rule for Derivatives }} {{eqn | r = x^{x^x} \paren {x^x \map {\frac \d {\d x} } {\ln x} + \map {\frac \d {\d x} } {x^x} \ln x}       | c = Product Rule for Derivatives }} {{eqn | r = x^{x^x} \paren {x^x \frac 1 x + x^x \ln x \paren {\ln x + 1} }       | c = Derivative of Natural Logarithm Function, Derivative of $x^x$ }} {{eqn | r = x^{x^x + x} \paren {\frac 1 x + \ln x \paren {\ln x + 1} } }} {{end-eqn}} For $\ln x \le 0$ and $\ln x \ge -1$, the above is positive. For $-1 < \ln x < 0$, we have $e^{-1} < x < 1$. {{begin-eqn}} {{eqn | l = x^{x^x + x} \paren {\frac 1 x + \ln x \paren {\ln x + 1} }       | o = >       | r = x^{x^x + x} \paren {\frac 1 1 + \paren {-1} \paren {0 + 1} } }} {{eqn | r = 0}} {{end-eqn}} so we have $\dfrac \d {\d x} x^{x^x} > 0$ for every $x > 0$. We also have: :$\displaystyle \lim_{x \mathop \to 0^+} x^{x^x} = \lim_{x \mathop \to 0^+} x^{\lim_{x \mathop \to 0^+} x^x} = 0^1 = 0$ Thus $x^{x^x}: \R_{>0} \to \R_{>0}$ is bijective. For each $y > 0$, we can find some $x > 0$ such that $y = x^{x^x}$. The result above is therefore equivalent to: :$a^{a^a}$ is irrational if $a$ is rational and not a natural number. By Rational Number as Power of Number with Itself, as long as $a$ is not a natural number, $a^a$ is irrational. Therefore if $a$ is not a natural number, by Gelfond-Schneider Theorem, $a^{a^a}$ is transcendental, so it is also irrational. This proves the result. {{qed}}
$\dfrac \d {\d x} x^x = \dfrac \d {\d x} e^{x \ln x} = e^{x \ln x} \paren {\ln x + 1}$ So we have $\dfrac \d {\d x} x^x > 0$ for every $x > \dfrac 1 e$. Thus $x^x: \openint {\dfrac 1 e} {+\infty} \to \openint {\paren {\dfrac 1 e}^{\frac 1 e} } {+\infty}$ is bijective. For each $y \in \openint {\paren {\dfrac 1 e}^{\frac 1 e} } {+\infty}$, we can find some $x \in \openint {\dfrac 1 e} {+\infty}$ such that $y = x^x$. The result above is therefore equivalent to: :$a^a$ is irrational if $a$ is rational and not a natural number. Suppose $a^a$ is rational for some rational number $a > \dfrac 1 e$. We express $a^a$ and $a$ in canonical form:  :$\paren {\dfrac m n}^{\frac m n} = \dfrac b c$ where $m, n, b, c \in \N_{> 0}$ and $\map \gcd {b, c} = \map \gcd {m, n} = 1$. This gives $\dfrac {m^m} {n^m} = \dfrac {b^n} {c^n}$. By Canonical Form of Rational Number is Unique: :$n^m = c^n$ If $n = 1$ then $a = \dfrac m n$ is a natural number, and $a^a$ is rational. {{AimForCont}} $n \ge 1$. Then there is a prime such that $p \divides n$. This prime must also divide $c$. Write $n = p^i k$, $c = p^j l$, where $i, j, k, l \in \N_{> 0}$ and $\map \gcd {p, k} = \map \gcd {p, l} = 1$. Then $n^m = p^{i m} k^m = p^{j n} l^n = c^n$. Since $\map \gcd {p, k^m} = \map \gcd {p, l^n} = 1$, we must have $p^{i m} = p^{j n}$. Thus $i m = j n = j k p^i$. Since $\map \gcd {m, n} = 1$, $p \nmid m$. So we have $p^i \divides i$. By Absolute Value of Integer is not less than Divisors, $p^i \le i$. However by Bernoulli's Inequality: :$p^i \ge 2^i \ge 1 + i > i$ This is a contradiction. Therefore if $a$ is not a natural number, $a^a$ is irrational. This proves the result. {{qed}}
Let $\displaystyle z = \int_X f \rd \mu \in \C$. By Complex Multiplication as Geometrical Transformation, there is a complex number $\alpha$ with $\cmod \alpha = 1$ such that: :$\alpha z = \cmod z \in \R$ Let $u = \map \Re {\alpha f}$, where $\Re$ denotes the real part of a complex number. By Modulus Larger than Real Part, we have that: :$u \le \cmod {\alpha f} = \cmod f$ Thus we get the inequality: {{begin-eqn}} {{eqn | l = \cmod {\int_X f \rd \mu}       | r = \alpha \int_X f \rd \mu       | c = }} {{eqn | r = \int_X \alpha f \rd \mu       | c = Integral of Integrable Function is Homogeneous }} {{eqn | r = \int_X u \rd \mu       | c = }} {{eqn | o = \le       | r = \int_X \cmod f  \rd \mu       | c = Integral of Integrable Function is Monotone }} {{end-eqn}} {{qed}}
Let $N$ be the set defined by: :$N = \set {x \in X: \map f x \ne \map g x}$ By hypothesis, $N$ is a $\mu$-null set. If $N = \O$, then $f = g$, trivially implying the result. If $N \ne \O$, then by Set with Relative Complement forms Partition: :$X = N \cup \paren {X \setminus N}$ Now: {{begin-eqn}} {{eqn | l = \int f \rd \mu       | r = \int f \chi_X \rd \mu       | c = Characteristic Function of Universe }} {{eqn | r = \int f \chi_{N \cup \paren {X \setminus N} } \rd \mu }} {{eqn | r = \int \map f {\chi_N + \chi_{X \setminus N} } \rd \mu       | c = Characteristic Function of Union }} {{eqn | r = \int f \chi_N \rd \mu + \int f \chi_{X \setminus N} \rd \mu       | c = Integral of Integrable Function is Additive }} {{eqn | r = 0 + \int f \chi_{X \setminus N} \rd \mu       | c = Integral of Integrable Function over Null Set }} {{eqn | r = \int g \chi_{X \setminus N} \rd \mu       | c = Definition of $N$ }} {{eqn | r = \int g \chi_N \rd \mu + \int g \chi_{X \setminus N} \rd \mu       | c = Integral of Integrable Function over Null Set }} {{eqn | r = \int \map g {\chi_N + \chi_{X \setminus N} } \rd \mu       | c = Integral of Integrable Function is Additive }} {{eqn | r = \int g \chi_X \rd \mu       | c = Characteristic Function of Union }} {{eqn | r = \int g \rd \mu       | c = Characteristic Function of Universe }} {{end-eqn}} {{Disambiguate|Characteristic Function of Union|This variant is not posted yet}} which establishes the result. {{qed}}
Let $X \subseteq S_i$ for all $i \in I$. Then from Set is Subset of Intersection of Supersets: General Result: :$\ds X \subseteq \bigcap_{i \mathop \in I} S_i$ {{qed|lemma}} Now suppose that $\ds X \subseteq \bigcap_{i \mathop \in I} S_i$. From Intersection is Subset: Family of Sets we have: :$\forall i \in I: \bigcap_{j \mathop \in I} S_j \subseteq S_i$ So from Subset Relation is Transitive, it follows that: :$\ds \forall i \in I: X \subseteq \bigcap_{j \mathop \in I} S_j \subseteq S_i$ So it follows that $\forall i \in I: X \subseteq S_i$. So: :$\ds X \subseteq \bigcap_{i \mathop \in I} S_i \implies \paren {\forall i \in I: X \subseteq S_i}$ {{qed|lemma}} Hence: :$\ds \paren {\forall i \in I: X \subseteq S_i} \iff X \subseteq \bigcap_{i \mathop \in I} S_i$ {{Qed}}
By Surjection from Natural Numbers iff Countable, there exists a surjection $f: \N \to S$. It follows that: :$S = \displaystyle \bigcup_{n \mathop \in \N} \left\{{f \left({n}\right)}\right\}$ As Lebesgue Measure is Diffuse, it holds that: :$\forall n \in \N: \lambda \left({\left\{{f \left({n}\right)}\right\}}\right) = 0$ Thus, by Null Sets Closed under Countable Union, it follows that: :$\lambda \left({S}\right) = 0$ {{qed}}
=== Necessary Condition === From Union of Subsets is Subset/Family of Sets we have that: :$\ds \paren {\forall i \in I: S_i \subseteq X} \implies \bigcup_{i \mathop \in I} S_i \subseteq X$ {{qed|lemma}} === Sufficient Condition === Now suppose that $\ds \bigcup_{i \mathop \in I} S_i \subseteq X$. Consider any $i \in I$ and take any $x \in S_i$. From Set is Subset of Union: Family of Sets we have that: :$\ds S_i \subseteq \bigcup_{i \mathop \in I} S_i$ Thus: :$\ds x \in \bigcup_{i \mathop \in I} S_i$ But: :$\ds \bigcup_{i \mathop \in I} S_i \subseteq X$ So it follows that $S_i \subseteq X$. So: :$\ds \bigcup_{i \mathop \in I} S_i \subseteq X \implies \paren {\forall i \in I: S_i \subseteq X}$ {{qed|lemma}} Hence: :$\ds \paren {\forall i \in I: S_i \subseteq X} \iff \bigcup_{i \mathop \in I} S_i \subseteq X$ {{qed}}
{{MissingLinks}} Let $\lambda_0 \in U$ be arbitrary. Let $\sequence {\lambda_n}_{n \mathop \ge 1}$ be a sequence in $U$ which converges to $\lambda_0$. Define the sequence of functions $f_n: X \to \R$, for $n = 0$ and $n \ge 1$ by $\map {f_n} x = \map f {\lambda_n, x}$. By hypothesis $(1)$, for each $n \ge 1$, the function $f_n$ is $\mu$-integrable. By hypothesis $(2)$, and by Sequential Continuity is Equivalent to Continuity in Metric Space, we have: :$\displaystyle \map f x = \lim_{n \mathop \to \infty} \map {f_n} x$ for $\mu$-almost all $x \in X$. By hypothesis $(3)$, we have: :$\size {\map {f_n} x} \le \map g x$ for all $x \in X$ and all $n \ge 1$. Therefore by Lebesgue's Dominated Convergence Theorem: :$\displaystyle \lim_{n \mathop \to \infty} \int f_n \rd \mu = \int f_0 \rd \mu$ That is: :$\displaystyle \lim_{n \mathop \to \infty} \int \map f {\lambda_n, x} \rd \mu = \int \map f {\lambda_0, x} \rd \mu$ or in terms of $h$: :$\displaystyle \lim_{n \mathop \to \infty} \map h {\lambda_n} = \map h {\lambda_0}$ Since the sequence $\sequence {\lambda_n}_{n \mathop \ge 1}$ was arbitrary, by Sequential Continuity is Equivalent to Continuity in Metric Space this shows that $h$ is continuous at $\lambda_0 \in U$. But $\lambda_0 \in U$ was arbitrary, so $h$ is continuous in $U$. {{qed}}
Let $0 < \alpha < a \le x \le y \le b < \beta$. Let $0 < \delta < \Delta$. Then: :$(1): \quad \displaystyle \left|{\int_\delta^\Delta t^{x - 1} e^{-t} \ \mathrm d t - \int_\delta^\Delta t^{y - 1} e^{-t} }\right| \le \int_\delta^\Delta \left({t^{x - 1} - t^{y - 1}}\right)e^{-t} \ \mathrm d t$ {{explain|why?}} From the Mean Value Theorem: :$(2): \quad \dfrac {t^{x - 1} - t^{y - 1} }{x - y} = \left({\ln t}\right) t^{\xi - 1}$ for some $\xi \in \R$ such that $x \le \xi \le y$. Let $r \in \R_{>0}$. Then: : $t^{-r} \ln t \to 0$ as $t \to \infty$ : $t^r \ln t \to 0$ as $t \to 0^+$ From $(2)$ it follows that: :$\exists H \in \R: \left|{t^{x - 1} - t^{y - 1} }\right| \le H \left({t^{\alpha - 1} + t^{\beta - 1} }\right) \left|{x - y}\right|$ From $(1)$ it follows that: :$\left|{\Gamma \left({x}\right) - \Gamma \left({y}\right) }\right| \le H \left|{x - y}\right| \left({\Gamma \left({\alpha}\right) - \Gamma \left({\beta}\right) }\right)$ The result follows from the Squeeze Theorem. {{qed}} {{explain|Quite a few steps in the above need to be filled in and/or justified with links to results}}
By definition, the Gamma function $\Gamma: \R_{> 0} \to \R$ is defined as: :$\displaystyle \Gamma \left({z}\right) = \int_0^{\infty} t^{z - 1} e^{-t} \rd t$ :$\forall z > 0: \Gamma \left({z}\right) > 0$, as an integral of a strictly positive function in $t$. {{explain|A separate page is needed for the above statement}}   The function is smooth according to Gamma Function is Smooth on Positive Reals, and :$\displaystyle \forall k \in \N: \Gamma^\left({k}\right) \left({z}\right) = \int_0^{\infty} \ln \left({t}\right)^k t^{z - 1} e^{-t} \, \mathrm d t$ {{explain|Prove the above}} Let $f \left({z}\right) := \ln \left({\Gamma \left({z}\right) }\right)$. :$f$ is smooth because $\Gamma$ is smooth and positive. {{explain|A link to why this follows}} Then: : $f' \left({z}\right) = \dfrac {\Gamma' \left({z}\right)} {\Gamma \left({z}\right)}$ :$f^{\left({2}\right)} \left({z}\right) = \dfrac {\Gamma^{\left(2\right)} \left({z}\right) \Gamma \left({z}\right) - \Gamma' \left({z}\right)^2} {\Gamma \left({z}\right)^2} > 0$ {{explain|Invoke the result that this comes from: Derivative of Quotient or whatever it is}} The numerator is positive due to the Cauchy-Bunyakovsky-Schwarz Inequality applied to the scalar products: :$\displaystyle \left \langle {g, h} \right \rangle = \int_0^\infty g \left({t}\right) h \left({t}\right) t^{z - 1} e^{-t} \rd t \quad \forall z \gt 0$ applied to $g = \ln$ and $h = 1$. :$\forall z \in \R_{>0}: f^{\left({2}\right)} \left({z}\right) \gt 0 \implies$ $f$ is convex. {{qed}}
The strategy is to show that: :$\map \ln {\map \Gamma {\dfrac x 2 + \dfrac y 2} } \le \dfrac 1 2 \map \ln {\map \Gamma x} + \dfrac 1 2 \map \ln {\map \Gamma y}$ Let $0 < \delta < \Delta$. Then: {{begin-eqn}} {{eqn | l = \paren {\int_\delta^\Delta t^{\paren {x + y - 2} / 2} e^{-t} \rd t}^2       | r = \paren {\int_\delta^\Delta \paren {t^{\paren {x - 1} / 2} e^{-t/2} } \paren {t^{\paren {y - 1} / 2} e^{-t/2} } \rd t}^2       | c =  }} {{eqn | n = 1       | o = \le       | r = \paren {\int_\delta^\Delta t^{x - 1} e^{-t} \rd t} \paren {\int_\delta^\Delta t^{y - 1} e^{-t} \rd t}       | c = Cauchy-Bunyakovsky-Schwarz Inequality for Definite Integrals }} {{end-eqn}} Letting $\delta \to 0$ and $\Delta \to \infty$, $(1)$ becomes equivalent to: :$\paren {\map \Gamma {\dfrac {x + y} 2} }^2 \le \paren {\map \Gamma x} \paren {\map \Gamma y}$ {{begin-eqn}} {{eqn | l = \paren {\map \Gamma {\dfrac {x + y} 2} }^2       | o = \le       | r = \paren {\map \Gamma x} \paren {\map \Gamma y}       | c =  }} {{eqn | ll= \leadsto       | l = \map \ln {\paren {\map \Gamma {\dfrac {x + y} 2} }^2}       | o = \le       | r = \map \ln {\paren {\map \Gamma x} \paren {\map \Gamma y} }       | c =  }} {{eqn | ll= \leadsto       | l = 2 \map \ln {\map \Gamma {\dfrac {x + y} 2} }       | o = \le       | r = \map \ln {\map \Gamma x} + \map \ln {\map \Gamma y}       | c =  }} {{eqn | ll= \leadsto       | l = \map \ln {\map \Gamma {\dfrac x 2 + \dfrac y 2} }       | o = \le       | r = \frac 1 2 \map \ln {\map \Gamma x} + \frac 1 2 \map \ln {\map \Gamma y}       | c =  }} {{end-eqn}} The result follows by definition of convex function. {{qed}}
The strategy is to use the Euler Form of the Gamma function and directly calculate the second derivative of $\ln \map \Gamma z$. {{begin-eqn}} {{eqn | l = \map \Gamma z       | r = \lim_{m \mathop \to \infty} \frac {m^z m!} {z \paren {z + 1} \paren {z + 2} \dotsm \paren {z + m} }       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {\d^2 \ln \map \Gamma z} {\d z^2}        | r = \dfrac {\d^2} {\d z^2} \map \ln {\lim_{m \mathop \to \infty} \frac {m^z m!} {z \paren {z + 1} \paren {z + 2} \dotsm \paren {z + m} } }       | c =  }} {{eqn | r = \lim_{m \mathop \to \infty} \dfrac {\d^2} {\d z^2} \map \ln {\frac {m^z m!} {z \paren {z + 1} \paren {z + 2} \dotsm \paren {z + m} } }       | c =  }} {{end-eqn}} The limit interchange is justified because Gamma Function is Smooth on Positive Reals. {{begin-eqn}} {{eqn | l = \lim_{m \mathop \to \infty} \dfrac {\d^2} {\d z^2} \paren {z \map \ln m + \map \ln {m!} - \sum_{n \mathop = 0}^m \map \ln {z + n} }       | r = \lim_{m \mathop \to \infty} \sum_{n \mathop = 0}^m \dfrac 1 {\paren {z + n}^2}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {\d^2 \ln \map \Gamma z} {\d z^2}       | r = \sum_{n \mathop = 0}^\infty \dfrac 1 {\paren {z + n}^2} > 0       | c =  }} {{end-eqn}} Logarithmic convexity then follows from Second Derivative of Strictly Convex Real Function is Strictly Positive. {{qed}}
Let $x \in X$. Let: :$a_x := \dfrac {\size {\map f x} } {\norm f_p}$ and: :$b_x := \dfrac {\size {\map g x} } {\norm g_q}$ Applying Young's Inequality for Products to $a_x$ and $b_x$: :$\dfrac {\size {\map f x \map g x} } {\norm f_p \cdot \norm g_q} \le \dfrac {\size {\map f x}^p} {p \norm f_p^p} + \dfrac {\size {\map g x}^q} { q \norm g_q^q}$ By Integral of Positive Measurable Function is Monotone, integrating both sides of this inequality over x yields: :$\displaystyle \dfrac {\int size {\map f x \map g x} \map \mu {\d x} } {\norm f_p \cdot \norm g_q} \le \frac {\norm f_p^p} {p \norm f_p^p} + \frac {\norm g_p^q} {q \norm g_q^q} = \frac 1 p + \frac 1 q = 1$ so: :$\displaystyle \int \size {\map f x \map g x} \map \mu {\d x} \le \norm f_p \cdot \norm g_q$ If we have equality, then: :$\displaystyle \int \frac {\size {\map f x}^p} {p \norm f_p^p} + \frac {\size {\map g x}^q} {q \norm g_q^q} - \frac {\size {\map f x \map g x} } {\norm f_p \cdot \norm g_q} \map \mu {\d x} = 0$ As: :$\dfrac {\size {\map f x}^p} {p \norm f_p^p} + \dfrac {\size {\map g x}^q} {q \norm g_q^q} - \dfrac {\size {\map f x \map g x} } {\norm f_p \cdot \norm g_q} \ge 0$ it follows from Integrable Function Zero A.E. iff Absolute Value has Zero Integral that: :$\dfrac {\size {\map g x}^p} {p \norm f_p^p} + \dfrac {\size {\map g x}^q} {q \norm g_q^q} = \dfrac {\size {\map f x \map g x} } {\norm f_p \cdot \norm g_q}$ a.e. By Young's Inequality for Products, we have equality {{iff}} $b_x = {a_x}^{p - 1}$. Raising both sides to the $q$th power gives: :$\dfrac {\size {\map g x}^p} {\norm f_p^p} = \dfrac {\size {\map g x}^q} {\norm g_q^q}$ as $\paren {p - 1} q = p$.  {{qed}} {{Namedfor|Otto Ludwig Hölder|cat = Hölder}}
From the definition of an equivalence class: :$\eqclass x \RR = \set {y \in S: \tuple {x, y} \in \RR}$ Let $y \in S$ such that $y \in \eqclass x \RR$. Then by definition $\tuple {x, y} \in \RR$. Similarly, let $\tuple {x, y} \in \RR$. Again by definition, $y \in \eqclass x \RR$. {{qed}}
From Linearly Ordered Space is Completely Normal, $\struct {S, \tau}$ is a completely normal space. The result follows from Completely Normal Space is Normal Space. {{qed}} Category:Order Topology Category:Normal Spaces rnfql1287pv66psszvqlpe6476dvbaw
From the definition: :For all sets $X$ and all indexed families $\family {f_i}_{i \mathop \in I}$ of mappings $f_i: X \to S_i$ there exists a unique mapping $h: X \to P$ such that: ::$\forall i \in I: \phi_i \circ h = f_i$ Let: :$i \in I$ :$X = S_i$ :$f_i = I_{S_i}$ where $I_{S_i}$ is the identity mapping on $S_i$. Then we have: :$\phi_i \circ h = I_{S_i}$ From Identity Mapping is Surjection: :$I_{S_i}$ is a surjection. From Surjection if Composite is Surjection: :$\phi_i$ is a surjection. $i \in I$ is arbitrary so the argument applies whatever its value. Hence the result. {{qed}}
Let $i \in I$. Consider any set $X$ and any indexed family of mappings $\family {f_i: X \to S_i}_{i \mathop \in I}$. Define $h: X \to \displaystyle \prod_{j \mathop \in I} \family {S_j}$ by: :$\forall x \in X: \map h x = \family {\map {f_i} x}_{i \mathop \in I}$ Then for all $x \in X$ and $i \in I$ we have: :$\map {\paren {\pr_i \mathop \circ h} } x = \map {\pr_i} {\family {\map {f_i} x}_{i \mathop \in I} } = \map {f_i} x$ So: :$\pr_i \mathop \circ h = f_i$ Thus $h$ is shown to exist. Suppose there exists a mapping $k: X \to \displaystyle \prod_{j \mathop \in I} \family {S_j}$ such that: :$\forall i \in I: \pr_1 \mathop \circ k = f_i$ Let $x \in X$ and let: :$\map k x = \family {s_j}_{j \mathop \in I}$ for some $\family {s_j}_{j \mathop \in I} \in \displaystyle \prod_{j \mathop \in I} \family {S_j}$. Then: :$\map {f_i} x = \map {\paren {\pr_i \mathop \circ k} } x = \map {\pr_i} {\family {s_j}_{j \mathop \in I} } = s_i$ and so: :$\map k x = \family {s_j}_{j \mathop \in I} = \family {\map {f_i} x}_{i \mathop \in I} = \map h x$ and so $k = h$. So $h$ is unique. Hence the result. {{qed}}
Follows directly from Hölder's Inequality with $p = q = 2$. {{qed}}
This follows from Image of Intersection under Relation: General Result, and the fact that $\mathcal R^{-1}$ is itself a relation, and therefore obeys the same rules. {{qed}} Category:Relation Theory Category:Set Intersection trz34aptagvw7bpr8mhw0fk2hh5i4cy
{{begin-eqn}} {{eqn | l = t       | o = \in       | r = \RR \sqbrk {\bigcup_{i \mathop \in I} S_i} }} {{eqn | ll= \leadstoandfrom       | l = \exists s \in \bigcup_{i \mathop \in I} S_i: t       | o = \in       | r = \map \RR s       | c = Image of Subset under Relation equals Union of Images of Elements }} {{eqn | ll= \leadstoandfrom       | l = \exists i \in I: \exists s \in S_i: t       | o = \in       | r = \map \RR s       | c = {{Defof|Union of Family}} }} {{eqn | ll= \leadstoandfrom       | l = \exists i \in I: t       | o = \in       | r = \RR \sqbrk {S_i}       | c = {{Defof|Image of Subset under Relation}} }} {{eqn | ll= \leadstoandfrom       | l = t       | o = \in       | r = \bigcup_{i \mathop \in I} \RR \sqbrk {S_i}       | c = {{Defof|Union of Family}} }} {{end-eqn}} {{Qed}}
{{begin-eqn}} {{eqn | l = \forall i \in I: \bigcap_{j \mathop \in I} S_j       | o = \subseteq       | r = S_i       | c = Intersection is Subset: Family of Sets }} {{eqn | ll= \leadsto       | l = \forall i \in I: \RR \sqbrk {\bigcap_{j \mathop \in I} S_j}       | o = \subseteq       | r = \RR \sqbrk {S_i}       | c = Image of Subset is Subset of Image }} {{eqn | ll= \leadsto       | l = \RR \sqbrk {\bigcap_{i \mathop \in I} S_i}       | o = \subseteq       | r = \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}       | c = Intersection is Largest Subset: Family of Sets }} {{end-eqn}} {{qed}}
We have that $\RR^{-1}$ is itself a relation. The result follows from Image of Union under Relation: Family of Sets. {{qed}} Category:Preimage of Union under Relation Category:Indexed Families lpebiysmpj0xboy29sx9jg903d9m8nq
This follows from Image of Intersection under Relation: Family of Sets, and the fact that $\RR^{-1}$ is itself a relation, and therefore obeys the same rules. {{qed}} Category:Relation Theory Category:Set Intersection ay5y7m74ej87s8gd6b8vmfa30p39edj
An injection is a type of one-to-one relation, and therefore also a one-to-many relation. Therefore Image of Intersection under One-to-Many Relation: Family of Sets applies: :$\displaystyle \mathcal R \sqbrk {\bigcap_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} \mathcal R \sqbrk {S_i}$ {{iff}} $\mathcal R$ is a one-to-many relation. We have that $f$ is a mapping and therefore a many-to-one relation. So $f$ is a one-to-many relation {{iff}} $f$ is also an injection. It follows that: :$\displaystyle f \sqbrk {\bigcap_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} f \sqbrk {S_i}$ {{iff}} $f$ is an injection. {{qed}}
As $f$, being a mapping, is also a relation, we can apply Image of Intersection under Relation: Family of Sets: :$\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i} \subseteq \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}$ {{qed}}
As $f$, being a mapping, is also a relation, we can apply Image of Union under Relation: Family of Sets: :$\ds \RR \sqbrk {\bigcup_{i \mathop \in I} S_i} = \bigcup_{i \mathop \in I} \RR \sqbrk {S_i}$ {{qed}}
=== Sufficient Condition === Suppose: :$\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}$  where $\family {S_i}_{i \mathop \in I}$ is ''any'' family of subsets of $S$. Then by definition of $\family {S_i}_{i \mathop \in I}$: :$\forall i, j \in I: \RR \sqbrk {S_i \cap S_j} = \RR \sqbrk {S_i} \cap \RR \sqbrk {S_j}$ and the sufficient condition applies for Image of Intersection under One-to-Many Relation. So $\RR$ is one-to-many. {{qed|lemma}} === Necessary Condition === Suppose $\RR$ is one-to-many. From Image of Intersection under Relation: Family of Sets, we already have: :$\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i} \subseteq \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}$ so we just need to show: :$\ds \bigcap_{i \mathop \in I} \RR \sqbrk {S_i} \subseteq \RR \sqbrk {\bigcap_{i \mathop \in I} S_i}$ Let: :$\ds t \in \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}$ Then: {{begin-eqn}} {{eqn | l = t       | o = \in       | r = \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}       | c =  }} {{eqn | ll= \leadsto       | l = \forall i \in I: t       | o = \in       | r = \RR \sqbrk {S_i}       | c = {{Defof|Set Intersection}} }} {{eqn | ll= \leadsto       | l = \forall i \in I: \exists x \in S_i: \tuple {x, t}       | o = \in       | r = \RR       | c = {{Defof|Relation}} }} {{eqn | ll= \leadsto       | l = x       | o = \in       | r = \RR \sqbrk {\bigcap_{i \mathop \in I} S_i}       | c = $\RR$ is one-to-many }} {{eqn | ll= \leadsto       | l = \map \RR x       | o = \subseteq       | r = \RR \sqbrk {\bigcap_{i \mathop \in I} S_i}       | c = Image of Element is Subset }} {{eqn | ll= \leadsto       | l = \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}       | o = \subseteq       | r = \RR \sqbrk {\bigcap_{i \mathop \in I} S_i}       | c = {{Defof|Subset}} }} {{end-eqn}} So if $\RR$ is one-to-many, it follows that: :$\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}$  {{qed|lemma}} Putting the results together: $\RR$ is one-to-many {{iff}}: :$\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}$ where $\family {S_i}_{i \mathop \in I}$ is ''any'' family of subsets of $S$. {{qed}} Category:Relation Theory Category:Set Intersection lui5nuntmhxua0i42wst8iq47deu0ld