hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^n \paren {a_k + b_k}^2       | r = \sum_{k \mathop = 1}^n \paren {a_k^2 + 2 a_k b_k + b_k^2} }} {{eqn | r = \sum_{k \mathop = 1}^n a_k^2 + 2 \sum_{k \mathop = 1}^n a_k b_k + \sum_{k \mathop = 1}^n b_k^2 }} {{eqn | o = \le       | r = \sum_{k \mathop = 1}^n a_k^2 + 2 \paren {\sum_{k \mathop = 1}^n a_k^2}^{1/2} \paren {\sum_{k \mathop = 1}^n b_k^2}^{1/2} + \sum_{k \mathop = 1}^n b_k^2       | c = Cauchy's Inequality }} {{eqn | r = \paren {\paren {\sum_{k \mathop = 1}^n a_k^2}^{1/2} + \paren {\sum_{k \mathop = 1}^n b_k^2}^{1/2} }^2 }} {{end-eqn}} The result follows from Order is Preserved on Positive Reals by Squaring.
Without loss of generality, assume that: :$\displaystyle \sum_{k \mathop = 1}^n \left({a_k + b_k}\right)^p \ne 0$ Define: :$q = \dfrac p {p-1}$ Then: :$\dfrac 1 p + \dfrac 1 q = \dfrac 1 p + \dfrac {p-1} p = 1$ It follows that: {{begin-eqn}} {{eqn|l = \sum_{k \mathop = 1}^n \left({a_k + b_k}\right)^p      |r = \sum_{k \mathop = 1}^n a_k \left({a_k + b_k}\right)^{p-1} + \sum_{k \mathop = 1}^n b_k \left({a_k + b_k}\right)^{p-1} }} {{eqn|o = \le      |r = \left({\sum_{k \mathop = 1}^n a_k^p}\right)^{1/p} \left({\sum_{k \mathop = 1}^n \left({a_k + b_k}\right)^p}\right)^{1/q}       |c = by Hölder's Inequality for Sums, and because $\left({p - 1}\right) q = p$, by hypothesis }} {{eqn|o =       |ro= +      |r = \left({\sum_{k \mathop = 1}^n b_k^p}\right)^{1/p} \left({\sum_{k \mathop = 1}^n \left({a_k + b_k}\right)^p}\right)^{1/q} }} {{eqn|r = \left({\left({\sum_{k \mathop = 1}^n a_k^p}\right)^{1/p} + \left({\sum_{k \mathop = 1}^n b_k^p}\right)^{1/p} }\right) \left({\sum_{k \mathop = 1}^n \left({a_k + b_k}\right)^p}\right)^{1/q} }} {{end-eqn}} {{explain|Too many steps are being taken at the crucial point where it needs to be explained carefully, at "by Hölder's Inequality for Sums, and because $\left({p - 1}\right) q {{=}} p$, by hypothesis. Remedy needed: (a) separate out the manipulation of the indices according to the relation between $p$ and $q$ as one step, (b) use an instantiation of Hölder's Inequality for Sums which explains it in terms of summations (at the moment it is expressed in the compact and gnomic notation of $\left\Vert{x}\right\Vert$, which requires one to go to a separate page again for the explanation of this notation) then (c) use as many steps of Hölder's Inequality for Sums (I'm not sure, but I believe there are two separate applications of it) as is necessary to provide the solution.}} The result follows by dividing both sides of the above inequality by $\displaystyle \left({\sum_{k \mathop = 1}^n \left({a_k + b_k}\right)^p}\right)^{1/q}$, and using the equation $\displaystyle 1 - \frac 1 q = \frac 1 p$.
In this case, $p$ and $q$ have opposite sign. The proof then follows the same lines as the proof for $p > 1$, except that the Reverse Hölder's Inequality for Sums is applied instead.
We consider the case where $f$ is increasing; the case where $f$ is decreasing is handled similarly. Let $\epsilon \in \R_{>0}$ be a strictly positive real number. By the Archimedean Principle, there exists a natural number $n$ such that: :$n > \dfrac {\paren {b - a} \paren {\map f b - \map f a} } {\epsilon}$ For $k \in \set {0, 1, 2, \ldots, n}$, define: :$x_k = a + k \dfrac {b - a} n$ Then: :$a = x_0 < x_1 < \cdots < x_n = b$ and hence $P = \set {x_0, x_1, \ldots, x_n}$ is a subdivision of $\closedint a b$. The lower sum and upper sum of $f$ belonging to the subdivision $P$ are, respectively: :$\displaystyle \map L P = \sum_{k \mathop = 1}^n \map f {x_{k - 1} } \paren {x_k - x_{k - 1} } = \frac {b - a} n \sum_{k \mathop = 1}^n \map f {x_{k - 1} }$ :$\displaystyle \map U P = \sum_{k \mathop = 1}^n \map f {x_k} \paren {x_k - x_{k - 1} } = \frac {b - a} n \sum_{k \mathop = 1}^n \map f {x_k}$ Therefore: {{begin-eqn}} {{eqn | l = 0       | o = \le       | r = \overline {\int_a^b} \map f x \rd x - \underline {\int_a^b} \map f x \rd x       | c =  }} {{eqn | o = \le       | r = \map U P - \map L P       | c =  }} {{eqn | r = \frac {\paren {b - a} \paren {\map f b - \map f a} } n       | c =  }} {{eqn | o = <       | r = \epsilon       | c =  }} {{end-eqn}} Since $\epsilon \in \R_{>0}$ is arbitrary, the result follows. {{qed}} {{link wanted|proof of $\displaystyle \underline {\int_a^b} \map f x \rd x \le \overline {\int_a^b} \map f x \rd x$}}
Let $a, b \in S$. Let: : $\tau_a$ be the particular point topology with respect to $a$ on $S$ : $\tau_b$ be the particular point topology with respect to $b$ on $S$ From Homeomorphic Non-Comparable Particular Point Topologies: :neither $\tau_a \le \tau_b$ nor $\tau_b \le \tau_a$. Hence the result. {{qed}}
We will show that the axioms for an abstract geometry hold. === Axiom 1 === Let $P = \left({x_1, y_1}\right)$ and $Q = \left({x_2, y_2}\right)$ be two distinct points in $\left({\R^2, L_E}\right)$. If $x_1 = x_2 = a$ then $P, Q \in L_a$. If $x_1 \ne x_2$ then let: :$m = \dfrac {y_2 - y_1} {x_2 - x_1}$ :$b = y_2 - m x_2$ Then $P, Q \in L_{m,b}$, since: :$b + m x_1 = y_2 - m \left({x_2 - x_1}\right) = y_2 - \left({y_2 - y_1}\right) = y_1$ :$b + m x_2 = y_2 - m \left({x_2 - x_2}\right) = y_2$ So any two points in $\R^2$ lie on a line in $L_E$. {{qed|lemma}} === Axiom 2 === For any $a \in \R$: :$\left({a, 0}\right), \left({a, 1}\right) \in L_a$ Also for any $m, b \in \R$: :$\left({0, b}\right),\left({1, m + b}\right) \in L_{m,b}$ So every line in $L_E$ has at least two distinct points. {{qed|lemma}} Hence $\left({\R^2, L_E}\right)$ is an abstract geometry. {{qed}}
From the definition of pullback, given a commutative diagram: ::$\begin{xy}\xymatrix{  Q   \ar[r]^*+{q_1}   \ar[d]_*+{q_2} &  V   \ar[d]^*+{i} \\  A   \ar[r]_*+{f} &  B }\end{xy}$ we need to find a mapping $u: Q \to f^{-1} \left({V}\right)$ such that $\bar f \circ u = q_1$ and $j \circ u = q_2$. Let $x \in Q$; then $q_1 \left({x}\right) \in V \subseteq B$. Note that the commutativity of the diagram implies that $q_1 \left({x}\right) = f \left({q_2 \left({x}\right)}\right)$ since $i$ is an inclusion. Hence also $f \left({q_2 \left({x}\right)}\right) \in V$. It follows by definition of preimage that $q_2 \left({x}\right) \in f^{-1} \left({V}\right)$. Thus define $u: Q \to f^{-1} \left({V}\right)$ by restriction of $q_2$: :$u \left({x}\right) := q_2 \restriction_{Q \times f^{-1} \left({V}\right)} \left({x}\right)$ Then we have: :$j \circ u \left({x}\right) = u \left({x}\right) = q_2 \left({x}\right)$ since $j$ is an inclusion mapping. Hence $q_2 = j \circ u$ by Equality of Mappings. Furthermore: :$\bar f \circ u \left({x}\right) = \bar f \circ q_2 \left({x}\right) = f \left({q_2 \left({x}\right)}\right) = q_1 \left({x}\right)$ since $\bar f$ and $u$ are restrictions of $f$ and $q_2$, respectively. Hence $\bar f \circ u = q_1$ by Equality of Mappings. Observe that any $u': Q \to f^{-1} \left({V}\right)$ must satisfy: :$j \circ u' = q_2 = j \circ u$ By Inclusion Mapping is Injection, $j$ is an injection. From Injection iff Monomorphism in Category of Sets, we conclude $j$ is a monomorphism, and so $u = u'$. This establishes that $u: Q \to f^{-1} \left({V}\right)$ is unique. Therefore, we have established that: ::$\begin{xy}\xymatrix{  f^{-1} \left({V}\right)   \ar[r]^*+{\bar f}   \ar[d]_*+{j} &  V   \ar[d]^*+{i} \\  A   \ar[r]_*+{f} &  B }\end{xy}$ is a pullback diagram in $\mathbf{Set}$. {{qed}}
We will show that the axioms for an abstract geometry hold. === Axiom 1 === Let $P = \tuple {x_1, y_1}$ and $Q = \tuple {x_2, y_2}$ be two points in $\struct {\H, L_H}$. If $x_1 = x_2 = a$ then $P, Q \in {}_a L$. If $x_1 \ne x_2$ then let: :$c = \dfrac {y_2^2 - y_1^2 + x_2^2 - x_1^2} {2 \paren {x_2 - x_1} }$ :$r = \sqrt{\paren{x_1 - c}^2 + y_1^2}$ From definition of ${}_c L_r$: :${}_c L_r := \set {\tuple {x, y} \in \H : \paren {x - c}^2 + y^2 = r^2}$ We have: {{begin-eqn}} {{eqn | l = \paren{x_1 - c}^2 + y_1^2        | r = r^2 }} {{eqn | l = \paren{x_2 - c}^2 + y_2^2        | r = y_2^2 + x_2^2 - 2 x_2 c + c^2 }} {{eqn | r = y_2^2 + x_2^2 - 2 x_2 \dfrac {y_2^2 - y_1^2 + x_2^2 - x_1^2} {2 \paren {x_2 - x_1} } + c^2 }} {{eqn | r = c^2 + \dfrac 1 {x_2 - x_1} \paren {y_2^2 x_2 - y_2^2 x_1 + x_2^3 - x_2^2 x_1 - x_2 y_2^2 + x_2 y_1^2 - x_2^3 + x_2 x_1^2} }} {{eqn | r = c^2 + \dfrac 1 {x_2 - x_1} \paren {- y_2^2 x_1 - x_2^2 x_1 + x_2 y_1^2 + x_2 x_1^2} }} {{eqn | r = c^2 + \dfrac 1 {x_2 - x_1} \paren {- y_2^2 x_1 - x_2^2 x_1 + x_2 y_1^2 + x_2 x_1^2 + y_1^2 x_1 - y_1^2 x_1 + x_1^3 - x_1^3} }} {{eqn | r = c^2 + \dfrac 1 {x_2 - x_1} \paren {x_2 y_1^2 - y_1^2 x_1 + x_2 x_1^2 - x_1^3 - y_2^2 x_1 + y_1^2 x_1 - x_2^2 x_1 + x_1^3} }} {{eqn | r = y_1^2 + x_1^2 - x_1 \dfrac {y_2^2 - y_1^2 + x_2^2 - x_1^2} {x_2 - x_1} + c^2 }} {{eqn | r = y_1^2 + \paren {x_1 - c}^2 }} {{eqn | r = r^2 }} {{end-eqn}} Thus $P, Q \in {}_c L_r$. So any two points in $\H$ lie on a line in $L_H$. {{qed|lemmma}} === Axiom 2 === For any $a \in \R$: :$\tuple {a, 1}, \tuple {a, 2} \in {}_a L$ Also for any $c \in \R$ and $r \in \R_{>0}$: :$\tuple {c + \dfrac 1 2 r, \dfrac {\sqrt 3} 2 r}, \tuple {c - \dfrac 1 2 r, \dfrac {\sqrt 3} 2 r} \in {}_c L_r$ So every line in $L_H$ has at least two points. {{qed|lemma}} Hence $\struct {\H, L_H}$ is an abstract geometry. {{qed}}
=== Necessary Condition === Suppose that the left square is a pullback. To verify the outer rectangle is a pullback, suppose we are given the following commutative diagram: ::$\begin{xy}\xymatrix@+0.5em{  P   \ar@/^1em/[rrrd]^*+{p_1}   \ar@/_1em/[ddr]_*+{p_2} \\ & & &  D   \ar[d]^*+{h} \\ &  A   \ar[r]_*+{f} &  B   \ar[r]_*+{g} &  C }\end{xy}$ The right square being a pullback implies there is a unique $v: P \to E$ making the following commute: ::$\begin{xy}\xymatrix@+0.5em{  P   \ar@/^1em/[rrrd]^*+{p_1}   \ar@/_1em/[ddr]_*+{p_2}   \ar@{-->}@/^/[rrd]_*+{v} \\ & &  E   \ar[d]^*+{h'}   \ar[r]^*+{g'} &  D   \ar[d]^*+{h} \\ &  A   \ar[r]_*+{f} &  B   \ar[r]_*+{g} &  C }\end{xy}$ Since the left square is also a pullback, we subsequently find a unique $u: P \to F$ making the following commute: ::$\begin{xy}\xymatrix@+0.5em{  P   \ar@/^1em/[rrrd]^*+{p_1}   \ar@/_1em/[ddr]_*+{p_2}   \ar@/^/[rrd]_*+{v}   \ar@{-->}[rd]_*+{u} \\ &  F   \ar[d]_*+{h''}   \ar[r]^*+{f'} &  E   \ar[d]^*+{h'}   \ar[r]^*+{g'} &  D   \ar[d]^*+{h} \\ &  A   \ar[r]_*+{f} &  B   \ar[r]_*+{g} &  C }\end{xy}$ In conclusion, the outer rectangle is a pullback. {{qed|lemma}} === Sufficient Condition === Suppose that the outer rectangle is a pullback. To verify the left square is a pullback, suppose we are given the following commutative diagram: ::$\begin{xy}\xymatrix@+0.5em{  P   \ar@/^/[rrd]^*+{p_1}   \ar@/_1em/[ddr]_*+{p_2} \\ & &  E   \ar[d]^*+{h'} \\ &  A   \ar[r]_*+{f} &  B }\end{xy}$ Evidently, we can amend this diagram as follows: ::$\begin{xy}\xymatrix@+0.5em{  P   \ar@/^/[rrd]_*{p_1}   \ar@/^1em/[rrrd]^*+{g' \circ p_1}   \ar@/_1em/[ddr]_*+{p_2} \\ & &  E   \ar[d]^*+{h'}   \ar[r]^*+{g'} &  D   \ar[d]^*+{h} \\ &  A   \ar[r]_*+{f} &  B   \ar[r]_*+{g} &  C }\end{xy}$ Because the outer rectangle is a pullback, we thus obtain a unique $u: P \to F$ making the following commute: ::$\begin{xy}\xymatrix@+0.5em{  P   \ar@/^1em/[rrrd]^*+{g' \circ p_1}   \ar@/_1em/[ddr]_*+{p_2}   \ar@{-->}[rd]_*+{u} \\ &  F   \ar[d]_*+{h''}   \ar[r]^*+{f'} &  E   \ar[d]^*+{h'}   \ar[r]^*+{g'} &  D   \ar[d]^*+{h} \\ &  A   \ar[r]_*+{f} &  B   \ar[r]_*+{g} &  C }\end{xy}$ If we now can prove that $f' \circ u = p_1$, then we're done. To this end, remark that they both complete the following commutative diagram: ::$\begin{xy}\xymatrix@+0.5em{  P   \ar@{-->}[rrd]   \ar@/^1em/[rrrd]^*+{g' \circ p_1}   \ar@/_1em/[ddr]_*+{p_2} \\ & &  E   \ar[d]^*+{h'}   \ar[r]^*+{g'} &  D   \ar[d]^*+{h} \\ &  A   \ar[r]_*+{f} &  B   \ar[r]_*+{g} &  C }\end{xy}$ Since the right square is a pullback, there is only one such morphism. Therefore, we conclude $f' \circ u = p_1$. Hence the left square is also a pullback. {{qed}}
Since $\mathbb x \in \map {\operatorname{co} } E$, $\mathbf x$ is a convex combination of points in $E$. From the definition of convex combination: :$\displaystyle \mathbf x = \sum_{i \mathop = 1}^k \gamma_i \mathbf y_i$ with: :$\gamma_i \ge 0$ :$\displaystyle \sum_{i \mathop = 1}^k \gamma_i = 1$ and: :$\mathbf y_i \in E$ Let $K \subset \N$ be the set of all possible $k$ such that $\mathbf x$ is a convex combination of $k$ elements of $E$. By the Well-Ordering Principle, $K$ is well-ordered.  $K$ has a smallest element $k_s$ by the definition of well-ordered sets. Suppose that the set $\set {\mathbf y_i: i \le k_s}$ is affinely dependent. From the condition for affinely dependent set, there exists a set $\set {\alpha_i: i \le k_s}$ such that for some $\alpha_i > 0$: :$\displaystyle \sum_{i \mathop = 1}^{k_s} \alpha_i \mathbf y_i = 0$ and: :$\displaystyle \sum_{i \mathop = 1}^{k_s} \alpha_i = 0$ Pick the smallest $\dfrac {\gamma_j} {\alpha_j} > 0$. Then: :$\displaystyle \mathbf x = \paren {\sum_{i \mathop = 1}^{k_s} \gamma_i \mathbf y_i} + 0 = \paren {\sum_{i \mathop = 1}^{k_s} \gamma_i \mathbf y_i} - \paren {\frac {\gamma_j} {\alpha_j} \sum_{i \mathop = 1}^{k_s} \alpha_i \mathbf y_i}$  Rearrange to get: :$\displaystyle \mathbf x = \paren {\sum_{i \mathop = 1}^{k_s} \paren {\gamma_i - \frac {\gamma_j} {\alpha_j} \alpha_i} \mathbf y_i}$ Since $\gamma_j > 0$ is made the smallest and $\alpha_j > 0$ is made the greatest: :$\dfrac {\alpha_i} {\alpha_j} \gamma_j \le \gamma_j \le \gamma_i$ for all $i \le k_s$. Thus: :$\paren {\gamma_i - \dfrac {\gamma_j} {\alpha_j} \alpha_i} \ge 0$ For $i = j$: :$\paren {\gamma_i - \dfrac {\gamma_j} {\alpha_j} \alpha_i} = 0$ For $i \ne j$, let: :$\gamma'_{i'} \equiv \gamma_i - \dfrac {\gamma_j} {\alpha_j} \alpha_i$ We can express: :$\displaystyle x = \sum_{i' \mathop = 1}^{k_s-1} \gamma'_{i'} \mathbf y_i$ which contradicts the assumption that $k_s = \min K$. {{qed}} {{Namedfor|Constantin Carathéodory|cat = Carathéodory}}
We have the inequality: :$1 \le \dfrac \theta {\sin \theta} \le \dfrac 1 {\cos \theta}$ for all $\theta \in \left({-\dfrac \pi 2 \,.\,.\, 0}\right) \cup \left({0 \,.\,.\, \dfrac \pi 2}\right)$. {{explain|Find a link for this}} Taking the limit of the leftmost term and the rightmost term: :$\displaystyle \lim_{\theta \mathop \to 0} \ 1 = 1$ :$\displaystyle \lim_{\theta \mathop \to 0} \frac 1 {\cos\theta} = 1$ So by the Squeeze Theorem: :$\displaystyle \lim_{\theta \mathop \to 0} \frac \theta {\sin\theta} = 1$ {{qed}} Category:Differential Calculus Category:Sine Function Category:Limit of Sine of X over X 917kwt3kc1unm5swydrnipque9ae2rk
As the quaternion group $Q_4$ is an example of the dicyclic group with $4$ elements, is also a non-abelian group. {{qed}}
Let $f_c: \R \to \R$ be the constant function. By definition: : $\forall x \in \R: f_c \left({x}\right) = c$ Thus: : $\sup \left({f_c}\right) = \inf \left({f_c}\right) = c$ So from Upper and Lower Bounds of Integral, we have: : $\displaystyle c \left({b-a}\right) \le \int_a^b c \ \ \mathrm dx \le c \left({b-a}\right)$ Hence the result. {{qed}}
Let: :$\ds \map F x = \int c \rd x$ From the definition of primitive: :$\map {F'} x = c$ From Derivative of Function of Constant Multiple: :$\map {\dfrac \d {\d x} } {c x} = c$ From Primitives which Differ by Constant: :$\map {\dfrac \d {\d x} } {c x + C} = c$ Hence the result. {{qed}}
From Derivative of Constant Multiple: :$\map {\dfrac \d {\d x} } {c \map f x} = c \map {\dfrac \d {\d x} } {\map f x}$ The result follows from the definition of primitive.
From: * Real Number Line is Metric Space * Compact Metric Space is Totally Bounded * Totally Bounded Metric Space is Bounded the result follows by the rule of transposition. {{qed}}
Let $\left\langle{x_n}\right\rangle_{n \ge 1}$ be a sequence in $X$ converging to $x$. Let $V$ be a neighborhood of $f \left({x}\right)$ in $Y$. We are required to show that there exists $N \in \N$ such that $f \left({x_n}\right) \in V$ for all $n \ge N$. By continuity of $f$, choose a neighborhood $U$ of $x$ in $X$ such that $f \left({U}\right) \subseteq V$. Since $\left\langle{x_n}\right\rangle_{n \ge 1}$ converges, there exists $N \in \N$ such that $x_n \in U$ for all $n \ge N$. Therefore we must have that $f \left({x_n}\right) \in f \left({U}\right) \subseteq V$ for all $n \ge N$, as required. {{qed}} Category:Continuous Mappings niicbib9rld1rf8s68zdpaatvlg3sns
We have that a Continuous Mapping is Sequentially Continuous. To prove the converse, by the Rule of Transposition we may prove the contrapositive: :If $f$ is not continuous at $x$, then $f$ is not sequentially continuous at $x$. We suppose therefore that there exists $\epsilon_0 > 0$  such that for all $\delta > 0$ there exists $y \in X$ such that $d \left({x, y}\right) < \delta$ and $e \left({f \left({x}\right), f \left({y}\right)}\right) \ge \epsilon_0$. For $n \ge 1$, define $\delta_n = \dfrac 1 n$. For $n \ge 1$, we may choose $y_n \in X$ such that $d \left({x, y_n}\right) < \delta_n$ and $e \left({f \left({x}\right), f \left({y_n}\right)}\right) \ge \epsilon_0$. Therefore, by definition the sequence $\left\langle{y_n}\right\rangle_{n \ge 1}$ converges to $x$. However, by definition the sequence $\left\langle{f \left({y_n}\right)}\right\rangle_{n \ge 1}$ does not converge to $f \left({x}\right)$. That is, $f$ is not sequentially continuous at $x$. {{Qed}} Category:Metric Spaces Category:Continuous Mappings on Metric Spaces p90ycd9zttj9bebjtd673xiqu3exi30
The first diagram above can be distorted into: ::$\begin{xy}\xymatrix@+.5em@L+2px{  A'   \ar@/^/[rrd]^*{\gamma \circ h_\alpha}   \ar@/_/[ddr]_*{\alpha'} \\ &  B'   \ar[r]^*{h_\beta}   \ar[d]^*{\beta'} &  B   \ar[d]^*{\beta} \\ &  C'   \ar[r]_*{h} &  C }\end{xy}$ and since the square is a pullback, there is a unique $\gamma': A' \to B'$ as desired. That: ::$\begin{xy}\xymatrix@+.5em@L+2px{  A'   \ar[r]^*{h_\alpha}   \ar[d]_*{\gamma'} &  A   \ar[d]^*{\gamma} \\  B'   \ar[r]_*{h_\beta} &  B }\end{xy}$ is a pullback follows immediately from the Pullback Lemma. {{qed}}
{{begin-eqn}} {{eqn | l = \left({a \circ b}\right) \circ \left({a \circ c}\right)       | r = a \circ \left({b \circ a}\right) \circ c       | c = $\circ$ is associative }} {{eqn | r = a \circ \left({a \circ b}\right) \circ c       | c = $\circ$ is commutative }} {{eqn | r = \left({a \circ a}\right) \circ b \circ c       | c = $\circ$ is associative }} {{eqn | r = a \circ b \circ c       | c = $\circ$ is idempotent }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn | l = \left({a \circ c}\right) \circ \left({b \circ c}\right)       | r = a \circ \left({c \circ b}\right) \circ c       | c = $\circ$ is associative }} {{eqn | r = a \circ \left({b \circ c}\right) \circ c       | c = $\circ$ is commutative }} {{eqn | r = a \circ b \circ \left({c \circ c}\right)       | c = $\circ$ is associative }} {{eqn | r = a \circ b \circ c       | c = $\circ$ is idempotent }} {{end-eqn}} {{qed}} Category:Abstract Algebra Category:Commutativity Category:Associativity Category:Idempotence Category:Distributive Operations 8kmodbsjdxvs9f00hcmjpy84c56ct7e
We have: :Intersection is Associative :Intersection is Commutative :Intersection is Idempotent The result follows from Associative Commutative Idempotent Operation is Distributive over Itself. {{qed}} Category:Set Intersection Category:Distributive Operations Category:Intersection Distributes over Intersection sm1iv7tccv98f1b00fj1mhvpqppzw5x
By definition of inclusion mapping: :$i_S: S \to T: \forall x \in S: \map {i_S} x = x$ By definition of identity mapping: :$I_T: T \to T: \forall x \in T: \map {I_T} x = x$ The result follows by definition of restriction of mapping. {{qed}}
Let $x \in i_H^{-1} \left({T}\right)$. By the definition of inclusion mapping: :$i_H \left({x}\right) = x$ By definition of preimage of $T$: :$i_H \left({x}\right) \in T$ Thus $x \in T$. By definition of inclusion mapping: :$x \in \operatorname{Dom} \left({i_H}\right) = H$ So $x \in T$ and $x \in H$ and so by definition of set intersection: :$x \in T \cap H$ Thus by definition of subset: :$i_H^{-1} \left({T}\right) \subseteq T \cap H$ Now let $x \in T \cap H$. As $H = \operatorname{Dom} \left({i_H}\right)$ it follows that: : $x \in \operatorname{Dom} \left({i_H}\right)$ As $x \in T \cap H$ it follows by definition of set intersection that $x \in T$. Thus: : $i_H \left({x}\right) \in T$ and so by definition of preimage of $T$: : $x \in i_H^{-1} \left({T}\right)$ Thus by definition of subset: :$T \cap H \subseteq i_H^{-1} \left({T}\right)$ The result follows by definition of set equality. {{qed}} Category:Inclusion Mappings Category:Subsets 5qor5qq5mjgdiuhzmypk9lb04k7ew86
Let $U \in \tau$. Then from Preimage of Subset under Inclusion Mapping: :$\map {i_H^{-1} } U = U \cap H$ From the definition of the subspace topology: :$U \cap H \in \tau_H$ Hence the result by definition of a continuous mapping. {{qed}} Category:Continuous Mappings Category:Inclusion Mappings qr62gddavqaem41gk9q2zvk9fjbyzex
From Inclusion Mapping is Continuous, $i$ is $\tuple {\tau_H, \tau}$-continuous. It follows from Continuity of Composite Mapping that $f \circ i$ is $\tuple {\tau_H, \tau'}$-continuous. {{qed}}
=== Necessary Condition === Suppose $g$ is $\left({\tau', \tau_H}\right)$-continuous. From Inclusion Mapping is Continuous, $i$ is $\left({\tau_H, \tau}\right)$-continuous. It follows from Continuity of Composite Mapping that $i \circ g$ is $\left({\tau', \tau}\right)$-continuous. === Sufficient Condition === Suppose $i \circ g$ is $\left({\tau', \tau}\right)$-continuous. Let $V \in \tau_H$. Then from the definition of topological subspace: : $V = U \cap H$ for some $U \in \tau$ and by Preimage of Subset under Inclusion Mapping: : $U \cap H = i^{-1} \left({U}\right)$ So by Preimage of Subset under Composite Mapping: : $g^{-1} \left({V}\right) = g^{-1} \left({i^{-1} \left({U}\right)}\right) = \left({i \circ g}\right)^{-1} \left({U}\right)$ Thus: : $g^{-1} \left({V}\right) \in \tau'$ by continuity of $i \circ g$. Hence $g$ is $\left({\tau', \tau_H}\right)$-continuous. {{qed}}
Suppose $\tau''$ is a topology on $H$ such that: :$(1) \quad$ For any topological space $T' = \struct {A', \tau'}$, and :$(2) \quad$ For any mapping $g: A' \to H$: $g$ is $\tuple {\tau', \tau''}$-continuous {{iff}} $i \circ g$ is $\tuple {\tau', \tau}$-continuous. It needs to be shown that $\tau''$ must be the same as $\tau_H$. Let $A' = H$ and $\tau' = \tau''$. Let $g$ be the identity mapping on $H$. From Identity Mapping is Continuous, $g$ is $\tuple {\tau'', \tau''}$-continuous Thus from Continuity of Composite Mapping, $i \circ g$ is $\tuple {\tau'', \tau}$-continuous. Hence for any $U \in \tau$: :$\map {\paren {i \circ g}^{-1} } U \in \tau''$ But: :$\map {\paren {i \circ g}^{-1} } U = \map {i^{-1} } U = U \cap H$ Hence $\tau_H \subseteq \tau''$. Next, take take $A' = H$ and $\tau' = \tau_H$. Let $g$ be the identity mapping on $H$. We have that $i \circ g = i$ is $\tuple {\tau_H, \tau}$-continuous, From Continuity of Composite with Inclusion: Inclusion on Mapping, it follows that $g$ is $\tuple {\tau_H, \tau''}$-continuous. But by definition of continuity, this is the same as saying $\tau'' \subseteq \tau_H$. So $\tau'' = \tau_H$, as required. {{qed}}
Let $W \in \tau$ such that $\tuple{x, y} \in W$. From Natural Basis of Tychonoff Topology of Finite Product, $\tau$ is the topology with basis: : $\BB = \set{U_1 \times U_2: U_1 \in \tau_1, U_2 \in \tau_2}$ Thus $W$ is the union of sets of the form $U_1 \times U_2$ where $U_1 \in \tau_1$ and $U_2 \in \tau_2$. That is: :$\exists U_1 \in \tau_1, U_2 \in \tau_2: \tuple{x, y} \in U_1 \times U_2 \subseteq W$ which is what was to be proved. {{qed}}
By definition of open real interval, for $I_1$ and $I_2$ to be non-empty it must be the case that $a < b$ and $c < d$. In particular it is noted that $a \ne b$ and $c \ne d$. Thus $a - b \ne 0$ and $c - d \ne 0$. Consider the real function $f: I_1 \to I_2$ defined as: :$\forall x \in I_1: f \left({x}\right) = c + \dfrac {\left({d - c}\right) \left({x - a}\right)} {b - a}$ Then after some algebra: :$\forall x \in I_2: f^{-1} \left({x}\right) = a + \dfrac {\left({b - a}\right) \left({x - c}\right)} {d - c}$ Both of these are defined as $a - b \ne 0$ and $c - d \ne 0$. By the Combination Theorem for Continuous Functions, both $f$ and $f^{-1}$ are continuous on the open real intervals on which they are defined. Hence the result by definition of homeomorphic. {{qed}}