hcju/proofwikiqa huggingface dataset:
Let $\left({A_i}\right)_{i \mathop \in I}, \left({B_i}\right)_{i \mathop \in I}$ be families of non empty sets. Let $\left({f_i}\right)_{i \mathop \in I}$ be a family of mappings such that: :$\forall i \in I: f_i \in \mathcal F \left({A_i, B_i}\right)$ === Sufficient Condition === Let: :$\displaystyle \bigcup_{i \mathop \in I} f_i \in \mathcal F \left({\bigcup_{i \mathop \in I} A_i, \bigcup_{i \mathop \in I} B_i}\right)$ Let $i, j \in I$ be such that: :$\operatorname{Dom} f_i \cap \operatorname{Dom} f_j \ne \varnothing$ Let $a \in \left({\operatorname{Dom} f_i \cap \operatorname{Dom} f_j}\right)$ Let $\displaystyle b \in \bigcup_{i \mathop \in I} B_i$ be such that: :$\left({a, b}\right) \in f_i$ Aiming for a contradiction, suppose: :$\left({a, b}\right) \notin f_j$ As $a \in \left({\operatorname{Dom} f_i \cap \operatorname{Dom} f_j} \right)$: :$\displaystyle \exists c \in \bigcup_{i \mathop \in I} B_i: \left({a, c}\right) \in f_j$  As $\left({a, b}\right) \in f_i$: :$\displaystyle \left({a, b}\right) \in \bigcup_{i \mathop \in I} f_i$ Thus: :$\displaystyle \left({a, b}\right), \left({a, c}\right) \in \bigcup_{i \mathop \in I} f_i$ such that $b \ne c$ and $\displaystyle \bigcup_{i \mathop \in I} f_i$ is a mapping. This is a contradiction. Thus the supposition that  the fact $\left({a, b}\right) \notin f_j$ was false. So: :$\left({a, b}\right) \in f_j$ {{qed|lemma}} === Necessary Condition === Let: :$\forall i, j \in I: \operatorname{Dom} f_i \cap \operatorname{Dom} f_j \ne \varnothing \implies \left({\forall a \in \left({\operatorname{Dom} f_i \cap \operatorname{Dom} f_j}\right), \left({a, b}\right) \in f_i \implies \left({a, b}\right) \in f_j}\right)$ Let $\displaystyle a \in \bigcup_{i \mathop \in I} A_i$. Hence: :$\exists k \in I: a \in A_k$ Let $k \in I$. Thus: :$a \in \operatorname{Dom} f_k$ Let $l = f_k \left({a}\right)$ It follows that: :$\left({a, l}\right) \in f_k$ and so: :$\displaystyle \left({a, l}\right) \in \bigcup_{i \mathop \in I} f_i$ Aiming for a contradiction, suppose: :$\displaystyle \exists m \in \bigcup_{i \mathop \in I} B_i: \left({\left({a, m}\right) \in \bigcup_{i \mathop \in I} f_i \land m \ne l}\right)$ Let $\displaystyle m \in \bigcup_{i \mathop \in I} B_i$. Let $j \in I$ be such that: :$\left({a, m}\right) \in f_j$ We have: :$a \in \left({\operatorname{Dom} f_k \cap \operatorname{Dom} f_j}\right)$  As $\left({a, l}\right) \in f_k$: :$\left(a,l \right) \in f_j$. Therefore: :$\left({a, m}\right), \left({a, l}\right) \in f_j$ where $f_j \in \mathcal F \left({A_j, B_j}\right)$ and $m \ne l$. This contradicts the definition of mapping. So: :$\displaystyle \nexists m \in \bigcup_{i \mathop \in I} B_i: \left({\left({a, m}\right) \in \bigcup_{i \mathop \in I} f_i \land m \ne l}\right)$ and so: :$\displaystyle \bigcup_{i \mathop \in I} f_i \in \mathcal F \left({\bigcup_{i \mathop \in I} A_i, \bigcup_{i \mathop \in I} B_i}\right)$ {{qed}} Category:Mapping Theory Category:Set Union 4742h2h4m9o7qykrnfpt0mgy6tfbfrx
If $S$ equals $S_j$ for a $j$ in $\set {1, 2, \ldots, n}$, it is trivially true that $\sup S = \sup S_j$. Now assume that $S$ is unequal to $S_i$ for every $i$ in $\left\{{1, 2, \ldots, n}\right\}$. By Supremum of Set of Real Numbers is at least Supremum of Subset, $\sup S \ge \sup S_i$ for every $i$ in $\set{1, 2, \ldots, n}$. There are two alternatives; either: :$\sup S > \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$ or: :$\sup S = \sup S_j$ for at least one $j$ in $\set {1, 2, \ldots, n}$. Suppose that: :$\sup S > \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$ Let $\epsilon = \sup S - \map \max {\sup S_1, \sup S_2, \ldots, \sup S_n}$. We note that $\epsilon > 0$. By Supremum of Subset of Real Numbers is Arbitrarily Close, $S$ has an element $x$ that satisfies: :$x > \sup S - \epsilon$ We have: {{begin-eqn}} {{eqn | o = >       | l = x       | r = \sup S - \epsilon }} {{eqn | r = \sup S - \paren {\sup S - \map \max {\sup S_1, \sup S_2, \ldots, \sup S_n} }       | c = definition of $\epsilon$ }} {{eqn | r = \map \max {\sup S_1, \sup S_2, \ldots, \sup S_n} }} {{end-eqn}} Therefore: :$x > \map \max {\sup S_1, \sup S_2, \ldots, \sup S_n}$ This means that $x > \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$. However, $x$ must be an element of $S_j$ for some $j$ in $\set {1, 2, \ldots, n}$ as $x \in S$ and $S = \bigcup S_i$. Accordingly, it is not true that $\sup S > \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$. We just concluded that the alternative: :$\sup S > \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$ is not true. Therefore, the other alternative: :$\sup S = \sup S_j$ for a $j$ in $\set {1, 2, \ldots, n}$ is true. {{qed}} Category:Real Analysis 2p2kel5xg9ahcfw856r2xlxijud240n
{{begin-eqn}} {{eqn | l = \tan \left({3 \theta}\right)       | r = \frac {\sin \left({3 \theta}\right)} {\cos \left({3 \theta}\right)}       | c = Tangent is Sine divided by Cosine }} {{eqn | r = \frac {3 \sin \theta - 4 \sin^3 \theta} {4 \cos^3 \theta - 3 \cos \theta}       | c = Triple Angle Formula for Sine and Triple Angle Formula for Cosine }} {{eqn | r = \frac {3 \sin \theta - 4 \sin^3 \theta} {4 \cos^3 \theta - 3 \cos \theta}\frac {\cos^3 \theta} {\cos^3 \theta}       | c =  }} {{eqn | r = \frac {\frac {3 \tan \theta}{\cos^2 \theta} - 4 \tan^3 \theta} {4 - \frac {3 \cos \theta}{\cos^3 \theta} }       | c = Tangent is Sine divided by Cosine }} {{eqn | r = \frac {3 \tan \theta \sec^2 \theta - 4 \tan^3 \theta} {4 - 3 \sec^2 \theta}       | c = Secant is Reciprocal of Cosine }} {{eqn | r = \frac {3 \tan \theta \left({1 + \tan^2 \theta}\right) - 4 \tan^3 \theta} {4 - 3  \left({1 + \tan^2 \theta}\right)}       | c = Difference of Squares of Secant and Tangent }} {{eqn | r = \frac {3 \tan \theta + 3 \tan^3 \theta - 4 \tan^3 \theta} {4 - 3 - 3 \tan^2 \theta}       | c = multiplying out }} {{eqn | r = \frac {3 \tan \theta - \tan^3 \theta} {1 - 3 \tan^2 \theta}       | c = gathering terms }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \tan \left({3 \theta}\right)       | r = \dfrac {\tan \theta + \tan \left({2 \theta}\right)} {1 - \tan \theta \tan \left({2 \theta}\right)}       | c = Tangent of Sum }} {{eqn | r = \dfrac {\tan \theta + \dfrac {2\tan \theta} {1 - \tan^2 \theta} } {1 - \tan \theta \dfrac {2 \tan \theta} {1 - \tan^2 \theta} }       | c = Double Angle Formula for Tangent }} {{eqn | r = \dfrac {\tan \theta \left({1 - \tan^2 \theta}\right) + 2 \tan \theta} {\left({1 - \tan^2 \theta}\right) - 2 \tan^2 \theta}       | c = simplifying }} {{eqn | r = \dfrac {\tan \theta - \tan^3 \theta + 2 \tan \theta} {1 - \tan^2 \theta - 2 \tan^2 \theta}       | c = simplifying }} {{eqn | r = \dfrac {3 \tan \theta - \tan^3 \theta} {1 - 3 \tan^2 \theta}       | c = simplifying }} {{end-eqn}} {{qed}}
Let $A$ be a commutative ring with unity. Let $I$ be an ideal of $A$ such that the quotient ring $A / I$ is a field. Let $J$ be an ideal of $A$ such that $I \subsetneq J$. From Ideal is Subring: :$J \subseteq A$ It remains to be proved that that $A \subseteq J$. Let $a \in A$. As $I \subsetneq J$, it follows from definition of proper subset that: :$\exists j \in J: j \notin I$ Consider the coset $j + I \in  A / I$. As $A / I$ is a field: :$\exists C \in A / I: \left({j + I}\right) \circ C = 1 + I$ Let $j' \in A$ be such that $C = j' + I$. Then: {{begin-eqn}} {{eqn | l = \left({j + I}\right) \circ \left({j' + I}\right)       | r = 1 + I       | c =  }} {{eqn | ll= \implies       | l = j \circ j' + I       | r = 1 + I       | c =  }} {{eqn | ll= \implies       | l = a \circ \left({j \circ j'}\right) + I       | r = a + I       | c =  }} {{eqn | ll= \implies       | lo= \exists i, i' \in I:       | l = a \circ \left({j \circ j'}\right) + i       | r = a + i'       | c =  }} {{eqn | ll= \implies       | l = a       | r = a \circ j \circ j' + i - i'       | c =  }} {{eqn | ll= \implies       | l = a \circ j       | o = \in       | r = J       | c = {{Defof|Ideal of Ring}} }} {{eqn | ll= \implies       | l = a \circ j \circ j'       | o = \in       | r = J       | c =  }} {{eqn | ll= \implies       | l = a \cdot j \cdot j' + i - i'       | o = \in       | r = J       | c = {{Defof|Ideal of Ring}} and $I \subsetneq J$ }} {{eqn | ll= \implies       | l = a       | o = \in       | r = J       | c = {{Defof|Ideal of Ring}} and $I \subsetneq J$ }} {{end-eqn}} By definition of subset: :$A \subseteq J$ Thus $A = J$ follows by definition of set equality. {{qed}} {{improve|Might be able to get there quicker by using Ideals of Field on the quotient ring $A / I$.}} Category:Ideal Theory 7q3s92y3hxf8h7dz1s77i9bae1pexhv
Let $\mathcal S_\alpha$ be the area of the segment whose base subtends $\alpha$. Let $\mathcal S_\theta$ be the area of the segment whose base subtends $\theta$. === Case $(1)$: Center included in Area === Let the center $O$ be included in the area. The area between the two chords is given by: :the area of the whole circle minus: :the areas of the segments $\mathcal S_\alpha$ and $\mathcal S_\theta$ . Thus: {{begin-eqn}} {{eqn | l = \mathcal A       | r = \pi r^2 - \mathcal S_\alpha - \mathcal S_\theta       | c = Area of Circle: $\pi r^2$ }} {{eqn | r = \pi r^2 - \frac 1 2 r^2 \left({\theta - \sin \theta}\right) - \frac 1 2 r^2 \left({\alpha - \sin \alpha}\right)       | c = Area of Segment of Circle }} {{eqn | r = r^2 \left({\pi - \frac 1 2 \left({\theta - \sin \theta + \alpha - \sin \alpha}\right)}\right)       | c = rearranging }} {{end-eqn}} {{qed|lemma}} === Case $(2)$: Center not included in Area === Let $\theta \ge \alpha$. The area between the two chords is given by: :the area of the segment $\mathcal S_\theta$ minus: :the area of the segment $\mathcal S_\alpha$. Thus: {{begin-eqn}} {{eqn | l = \mathcal A       | r = \mathcal S_\theta - \mathcal S_\alpha }} {{eqn | r = \frac 1 2 r^2 \left({\theta - \sin \theta}\right) - \frac 1 2 r^2 \left({\alpha - \sin \alpha}\right)       | c = Area of Segment of Circle }} {{eqn | r = \frac {r^2} 2 \left({\theta - \sin \theta - \alpha + \sin \alpha}\right)       | c = rearranging }} {{end-eqn}} {{qed}} Category:Circles tpt20mes3aa6vungexvygtvr7phgzp5
Consider the set $\set {x - y: x, y \in S, x - y \le 0}$. There is a number $x'$ in $S$ as $S$ is non-empty. Therefore, $0 \in \set {x - y: x, y \in S, x - y \le 0}$ as $x = y = x'$ implies that $x - y = 0$, $x, y \in S$, and $x - y \le 0$. Also, $0$ is an upper bound for $\set {x - y: x, y \in S, x - y \le 0}$ by definition. Accordingly: :$\displaystyle \sup_{x, y \mathop \in S, x − y \mathop \le 0} \paren {x - y} = 0$ Consider the set $\left\{{x - y: x, y \in S, x - y \ge 0}\right\}$. There is a number $x'$ in $S$ as $S$ is non-empty. Therefore, $0 \in \left\{{x - y: x, y \in S, x - y \ge 0}\right\}$ as $x = y = x'$ implies that $x - y = 0$, $x, y \in S$, $x - y \ge 0$. Accordingly: :$\displaystyle \sup_{x, y \mathop \in S, x − y \mathop \ge 0} \paren {x - y} \ge 0$ {{improve|I can't immediately think of how it would be done, but it would be good if we could devise a neater and more compact notation that what is used here. All the complicated mathematics is being done in the underscript, which makes it not easy to follow. (Improved Dec. 2016.)}} {{begin-eqn}} {{eqn | l = \sup_{x, y \mathop \in S} \paren {x - y}       | r = \sup_{x, y \mathop \in S, x − y \mathop \ge 0 \text { or } x − y \mathop \le 0} \paren {x - y}       | c = as ($x - y \ge 0$ or $x - y \le 0$) is true }} {{eqn | r = \max \set {\sup_{x, y \mathop \in S, x − y \mathop \ge 0} \paren {x - y}, \sup_{x, y \mathop \in S, x − y \mathop \le 0} \paren {x - y} }       | c = by Supremum of Set Equals Maximum of Suprema of Subsets }} {{eqn | r = \max \set {\sup_{x, y \mathop \in S, x − y \mathop \ge 0} \paren {x - y}, 0}       | c = as $\displaystyle \sup_{x, y \mathop \in S, x − y \mathop \le 0} \paren {x - y} = 0$ }} {{eqn | r = \sup_{x, y \mathop \in S, x − y \mathop \ge 0} \paren {x - y}       | c = as $\displaystyle \sup_{x, y \mathop \in S, x − y \mathop \ge 0} \paren {x - y} \ge 0$ }} {{eqn | r = \sup_{x, y \mathop \in S, x − y \mathop \ge 0} \size {x - y}       | c = as $\size {x − y} = x − y$ since $x − y \ge 0$ }} {{eqn | r = \max \set {\sup_{x, y \mathop \in S, x − y \mathop \ge 0} \size {x - y}, \sup_{x, y \mathop \in S, x − y \mathop \ge 0} \size {x - y} }       | c = as the two arguments of max are equal }} {{eqn | r = \max \set {\sup_{x, y \mathop \in S, x − y \mathop \ge 0} \size {x - y}, \sup_{y, x \mathop \in S, y − x \mathop \ge 0} \size {y - x} }       | c = by renaming variables $x \leftrightarrow y$ }} {{eqn | r = \max \set {\sup_{x, y \mathop \in S, x − y \mathop \ge 0} \size {x - y}, \sup_{x, y \mathop \in S, x − y \mathop \le 0} \size {x - y} }       | c =  }} {{eqn | r = \sup_{x, y \mathop \in S, x − y \mathop \ge 0 \text { or } x − y \mathop \le 0} \size {x - y}       | c = by Supremum of Set Equals Maximum of Suprema of Subsets }} {{eqn | r = \sup_{x, y \mathop \in S} \size {x - y}       | c = as ($x - y \ge 0$ or $x - y \le 0$) is true }} {{end-eqn}} {{qed}} Category:Real Analysis pku05d5dpl8yk89v4yqaxaa6u3fbnjn
Assume first that $A$ and $B$ have suprema. We have: :$x \le \sup A$ for an arbitrary $x$ in $A$ :$y \le \sup B$ for an arbitrary $y$ in $B$ Adding these inequalities, we get: :$x + y \le \sup A + \sup B$ The number $x + y$ is an arbitrary element of $A + B$ as $x$ and $y$ are arbitrary elements of $A$ and $B$ respectively. Therefore, $\sup A + \sup B$ is an upper bound for $A + B$. $A + B$ is non-empty as $A$ and $B$ are non-empty. Accordingly, $A + B$ has a supremum by the Continuum Property. Next, assume that $\sup \paren {A + B}$ has a supremum. We need to prove that $A$ and $B$ have suprema. Let $y$ be a point in $B$. We have: {{begin-eqn}} {{eqn | l = x + y       | o = \le       | r = \sup \paren {A + B}       | c = for every $x$ in $A$ as $x + y$ is a point in $A + B$ }} {{eqn | ll = \iff        | l = x       | o = \le       | r = \sup \paren {A + B} - y }} {{end-eqn}} Therefore, $A$ has an upper bound as $x$ is an arbitrary point in $A$. Also, we have that $A$ is non-empty. Accordingly, $A$ has a supremum by the Continuum Property. A similar argument gives that $B$ has a supremum. So, we have shown that all $\sup A$, $\sup B$, and $\sup \paren {A + B}$ exist. We proceed to show that $\sup \paren {A + B} = \sup A + \sup B$. We have $\sup \paren {A + B} \le \sup A + \sup B$ as $\sup A + \sup B$ is an upper bound for $A + B$. Accordingly, either: :$\sup \paren {A + B} < \sup A + \sup B$ or: :$\sup \paren {A + B} = \sup A + \sup B$. Suppose that: :$\sup \paren {A + B} < \sup A + \sup B$. Let $\epsilon = \sup A + \sup B - \sup \paren {A + B}$. We note that $\epsilon > 0$. Since $\sup A$ is the least upper bound of $A$, there is an element $x$ in $A$ such that: :$x > \sup A - \dfrac \epsilon 2$ by Supremum of Subset of Real Numbers is Arbitrarily Close Since $\sup B$ is the least upper bound of $B$, there is an element $y$ in $B$ such that: :$y > \sup B - \dfrac \epsilon 2$ by Supremum of Subset of Real Numbers is Arbitrarily Close Adding these inequalities, we get: {{begin-eqn}} {{eqn | l = x + y       | o = >       | r = \sup A - \frac \epsilon 2 + \sup B - \frac \epsilon 2 }} {{eqn | ll = \iff       | l = x + y       | o = >       | r = \sup A + \sup B - \epsilon }} {{eqn | ll = \iff       | l = x + y       | o = >       | r = \sup A + \sup B - \paren {\sup A + \sup B - \sup \paren {A + B} }       | c = definition of $\epsilon$ }} {{eqn | ll = \iff       | l = x + y       | o = >       | r = \sup \paren {A + B} }} {{end-eqn}} which is impossible since the number $x + y$ is an element of $A + B$ as $x \in A$ and $y \in B$. We have found that: :$\sup \paren {A + B} < \sup A + \sup B$ is not true. Therefore: :$\sup \paren {A + B} = \sup A + \sup B$ as $\sup \paren {A + B} \le \sup A + \sup B$. {{qed}} Category:Real Analysis 0vhqmkayoxmgy72m9hooj17zbybrm9s
=== Necessary Condition === Let $e$ be a bridge. That is, suppose the edge deletion $G - e$ is disconnected. Let $T$ be an arbitrary spanning tree for $G$. By definition $T$ is a connected subgraph of $G$. If $T$ did not contain $e$, then it would also be a subgraph of $G - e$. This contradicts the fact that $G - e$ is disconnected. Therefore $e$ is in $T$. {{qed|lemma}} === Sufficient Condition === Suppose $T$ is a spanning tree for $G$. Suppose that $T$ does not contain $e$. Then $T$ is a subgraph of the edge deletion $G - e$. Since $T$ is by definition connected, so is $G - e$. Hence $e$ is not a bridge. {{qed}} Category:Graph Theory g7xleezy3dx5oqvng31hcp2096mefwl
=== Necessary Condition === Aiming for a contradiction, suppose $e$ is a bridge of minimum weight that does not belong to some minimum spanning tree $Q$. Let $e$ be added to $Q$ to make $Q'$. Then $e$ forms part of a unique cycle $C$ in $Q$. Thus there exists an edge $f \in C$ such that $w \left({Q}\right) < w \left({Q + e - f}\right)$. This contradicts the minimality of $Q$. {{qed|lemma}} {{handwaving|The above argument (cleaned up considerably from the original, which was appallingly inadequate) needs to be made rigorous.}} === Sufficient Condition === {{proof wanted}} Category:Network Theory 8ouniob19cjt6260sxp6crbsrpbobgu
The equation can be rearranged as: :$\left\vert a \right\vert \delta \left({a t}\right) = \delta \left({t}\right)$ We will check the definition of Dirac delta function in turn. Definition of Dirac delta function: :$\left({1}\right):\delta \left({t}\right) = \begin{cases} +\infty & : t = 0 \\ 0 & : \text{otherwise} \end{cases}$ :$\left({2}\right):\displaystyle \int_{-\infty}^{+\infty} \delta \left({t}\right) \rd t = 1$ $\left({1}\right):$ {{begin-eqn}} {{eqn | l = \left\vert a \right\vert \delta \left({a t}\right)       | r = \begin{cases} \left({\left\vert a \right\vert}\right) \left({+\infty}\right) & : at = 0 \\ \left({\left\vert a \right\vert}\right) 0 & : \text{otherwise} \end{cases}       | c = Definition of Dirac delta function }} {{eqn | ll= \iff       | l = \left\vert a \right\vert \delta \left({a t}\right)       | r = \begin{cases} +\infty & : t = 0 \\ 0 & : \text{otherwise} \end{cases}       | c = simplifying }} {{end-eqn}} $\left({2}\right):$ The proof of this part will be split into two parts, one for positive $a$ and one for negative $a$. For $a>0$: {{begin-eqn}} {{eqn | l = \int_{-\infty}^{+\infty} \left\vert a \right\vert \delta \left({at}\right) \rd t       | r = \int_{-\infty}^{+\infty} \left\vert a \right\vert \delta \left({t}\right) \dfrac {\rd t} a       | c = Substitute $t \mapsto \dfrac t a$ }} {{eqn | r = \dfrac {\left\vert a \right\vert} a \int_{-\infty}^{+\infty} \delta \left({t}\right) \rd t       | c = Simplifying }} {{eqn | r = \dfrac a a \int_{-\infty}^{+\infty} \delta \left({t}\right) \rd t       | c = $a > 0$ }} {{eqn | r = 1       | c = {{Defof|Dirac Delta Function}} }} {{end-eqn}} {{qed|lemma}} For $a<0$: {{begin-eqn}} {{eqn | l = \int_{-\infty}^{+\infty} \left\vert a \right\vert \delta \left({at}\right) \rd t       | r = \int_{+\infty}^{-\infty} \left\vert a \right\vert \delta \left({t}\right) \dfrac {\rd t} a       | c = Substitute $t \mapsto \dfrac t a$ }} {{eqn | r = \dfrac {\left\vert a \right\vert} a \int_{+\infty}^{-\infty} \delta \left({t}\right) \rd t       | c = Simplifying }} {{eqn | r = \dfrac {-\left\vert a \right\vert} a \int_{-\infty}^{+\infty} \delta \left({t}\right) \rd t       | c = Reversal of Limits of Definite Integral }} {{eqn | r = \dfrac a a \int_{-\infty}^{+\infty} \delta \left({t}\right) \rd t       | c = $a < 0$ }} {{eqn | r = 1       | c = {{Defof|Dirac Delta Function}} }} {{end-eqn}} {{qed|lemma}} Therefore, by definition, $\left\vert a \right\vert \delta \left({a t}\right) = \delta \left({t}\right)$. The result follows after rearrangement. {{qed}} {{explain|Sorry, I'm going to have to ask the awkward question: can we link to a proof that: $\int_a^b f \left({x}\right) \rd x {{=}} \int_a^b g \left({x}\right) \rd x \implies f {{=}} g$?}} Category:Dirac Delta Function 8l8omeuh20gft0p93m703ed0jhstc7c
We have: {{begin-eqn}} {{eqn | l = \sup g       | r = \sup \left({f + \left({g - f}\right)}\right) }} {{eqn | r = \sup f + \sup \left({g - f}\right)       | c = Supremum of Sum equals Sum of Suprema }} {{end-eqn}} Supremum of Sum equals Sum of Suprema also gives that $\sup f$ and $\sup \left({g - f}\right)$ exist. We have: {{begin-eqn}} {{eqn | lo= \forall x \in S:       | l = g \left({x}\right)       | o = \ge       | r = f \left({x}\right)       | c =  }} {{eqn | ll= \iff       | lo= \forall x \in S:       | l = g \left({x}\right) - f \left({x}\right)       | o = \ge       | r = 0       | c =  }} {{eqn | ll= \implies       | lo= \forall x \in S:       | l = \sup \left({g - f}\right)       | o = \ge       | r = g \left({x}\right) - f \left({x}\right) \ge 0       | c = as $\sup \left({g - f}\right)$ is an upper bound for $\left\{ {g \left({x}\right) - f \left({x}\right): x \in S}\right\}$ }} {{eqn | ll= \implies       | l = \sup \left({g - f}\right)       | o = \ge       | r = 0 }} {{eqn | ll= \iff       | l = \sup f + \sup \left({g - f}\right)       | o = \ge       | r = \sup f }} {{eqn | ll= \iff       | l = \sup g       | o = \ge       | r = \sup f       | c = as $\sup g = \sup f + \sup \left({g - f}\right)$ }} {{end-eqn}} {{qed}} Category:Real Analysis k0jbjvo8a72k44x24vcm4tqhvagzz1c
=== Sufficient Condition === Let $x \in \operatorname{Fr} A$. Then by Boundary is Intersection of Closure with Closure of Complement: :$x \in \left({\complement_S \left({A}\right)}\right)^-$ and $x \in A^-$ where $A^-$ denotes the closure of $A$. Hence by Condition for Point being in Closure, for every open set $U$ of $T$: :$x \in U \implies A \cap U \ne \varnothing$ and: :$x \in U \implies \complement_S \left({A}\right) \cap U \ne \varnothing$ === Necessary Condition === Let $x$ be such that for every open set $U$ of $T$: :if $x \in U$ :then $A \cap U \ne \varnothing$ and $\complement_S \left({A}\right) \cap U \ne \varnothing$. Then by Condition for Point being in Closure: :$x \in \left({\complement_S \left({A}\right)}\right)^-$ and $x \in A^-$. Hence by Boundary is Intersection of Closure with Closure of Complement: :$x \in \operatorname{Fr} A$ {{qed}}
=== Sufficient Condition === Let $x \in A^-$. {{AimForCont}} there exists an open set $U$ of $T$ such that: :$x \in U$ and $A \cap U = \varnothing$ We have that $U$ is open in $T$. So by definition of closed set, $\complement_S \left({U}\right)$ is closed in $T$. Then: {{begin-eqn}} {{eqn | l = A \cap U       | r = \varnothing       | c =  }} {{eqn | ll= \implies       | l = A       | o = \subseteq       | r = \complement_S \left({U}\right)       | c = Empty Intersection iff Subset of Complement }} {{eqn | ll= \implies       | l = A^-       | o = \subseteq       | r = \complement_S \left({U}\right)       | c = Definition of Set Closure: $\complement_S \left({U}\right)$ is closed }} {{eqn | ll= \implies       | l = A^- \cap U       | r = \varnothing       | c = Empty Intersection iff Subset of Complement }} {{end-eqn}} But we have: :$x \in A^-$ and also: :$x \in U$ and thus by definition of set intersection: :$x \in A^- \cap U$ This contradicts $A^- \cap U = \varnothing$ Hence by Proof by Contradiction the assumption that there exists an open set $U$ of $T$ such that $x \in U$ and $A \cap U = \varnothing$ was false. So for every open set $U$ of $T$: :$x \in U \implies A \cap U \ne \varnothing$ {{qed|lemma}}  === Necessary Condition === Let $x$ be such that for every open set $U$ of $T$: :$x \in U \implies A \cap U \ne \varnothing$ {{AimForCont}} $x \notin A^-$. Then: :$x \in \complement_S \left({A^-}\right)$ Then by assumption: :$A \cap \complement_S \left({A^-}\right) \ne \varnothing$ as $\complement_S \left({A^-}\right)$ is open. By definition of complement: :$A \cap \complement_S \left({A}\right) = \varnothing$ So by Empty Intersection iff Subset of Complement: :$A \not \subseteq A^-$ From Set is Subset of its Topological Closure: :$A \subseteq A^-$ But from Set Complement inverts Subsets: :$\complement_S \left({A^-}\right) \subseteq \complement_S \left({A}\right)$ from which by Empty Intersection iff Subset of Complement: :$A \cap \complement_S \left({A^-}\right) = \varnothing$ Hence by Proof by Contradiction the assumption that $x \notin A^-$ was false. So $x \in A^-$. {{qed}}