hcju/proofwikiqa huggingface dataset:
Let $\alpha_1, \alpha_2, \ldots, \alpha_n$ be a generator of $V$. Let $\xi_1, \xi_2, \ldots, \xi_r$ be a linearly independent set of elements of $V$. Hence the sequence $\sequence {\xi_1, \alpha_1, \alpha_2, \ldots, \alpha_n}$ is a linearly dependent sequence of elements of $V$. One of these elements, which cannot be $\xi_1$, is a linear combination of the preceding elements. Let this elements be $\alpha_i$. So we can omit $\alpha_i$ from that sequence, and the remaining set is still a generator of $V$. Therefore $\xi_2$ is a linear combination of these. Thus $\sequence {\xi_2, \xi_1, \alpha_1, \alpha_2, \ldots, \alpha_{i - 1}, \alpha {i + 1}, \ldots, \alpha_n}$ is a linearly dependent sequence of elements of $V$. Again, one of them is a linear combination of the preceding elements. This cannot be $\xi_2$, as $\xi_2$ has no preceding elements. Neither can it be $\xi_1$, as $\xi_1$ and $\xi_2$ are linearly independent. Thus we can omit whichever $\alpha_j$ it is, and we have a new set which is a generator of $V$. This consists of $\xi_1$, $\xi_2$ and whichever $n - 2$ of the remaining elements of $\set {\alpha_1, \alpha_2, \ldots, \alpha_n}$. After $p$ such steps, we have a set which is a generator of $V$ which consists of: :$\xi_1, \xi_2, \ldots, \xi_p$ and $n - p$ of the elements of $\set {\alpha_1, \alpha_2, \ldots, \alpha_n}$. {{AimForCont}} suppose $n < r$. Then when $p = n$, the remaining set which is a generator of $V$ consists of: :$\xi_1, \xi_2, \ldots, \xi_n$ and there is at least one more element $\xi_{n + 1}$. This is a linear combination of $\set {\xi_1, \xi_2, \ldots, \xi_n}$. But this contradicts the supposition that $\set {\xi_1, \xi_2, \ldots, \xi_n, \xi_{n + 1} }$ is a linearly independent set. Hence, by Proof by Contradiction, $n \ge r$. The result follows. {{qed}}
Follows directly from the Intermediate Value Theorem and from the definition of root. {{qed}}
We have that $x - 1$ and $x + 1$ are both real numbers, so: :$x - 1 \in \R \setminus \set x$ :$x + 1 \in \R \setminus \set x$ Let $\mathbb I := \closedint 0 1$ be the closed unit interval. Suppose there exists a path $f: \mathbb I \to \R \setminus \set x$ from $x - 1$ to $x + 1$. Then by Image of Interval by Continuous Function is Interval, it follows that $x = \map f x$ for some $s \in \mathbb I$. But $x \notin \R \setminus \set x$ by definition of set difference. Hence such an $f$ can not exist. The result follows by definition of path-connected. {{qed}}
Let $\mathbf x, \mathbf y \in \R^n$ be arbitrary points of $\R^n$. Define $l: \closedint 0 1 \to \R^n$ by: :$\map l t = \paren {1 - t} \mathbf x + t \mathbf y$ Then $\map l 0 = 1 \mathbf x + 0 \mathbf y = \mathbf x$, whereas $\map l 1 = 0 \mathbf x + 1 \mathbf y = \mathbf y$. Finally, it remains to show that $l$ is continuous.  Fix $\epsilon > 0$ and suppose that $t, t' \in \closedint 0 1$ are such that $\size {t - t'} < \dfrac {\epsilon} {1 + \norm {\mathbf x} + \norm {\mathbf y} }$. Then: {{begin-eqn}} {{eqn | l = \map l t - \map l {t'}       | r = \paren {1 - t} \mathbf x + t \mathbf y - \paren {\paren {1 - t'} \mathbf x + t' \mathbf y} }} {{eqn | r = \paren {\paren {1 - t} - \paren {1 - t'} } \mathbf x + \paren {t - t'} \mathbf y }} {{eqn | r = \paren {t' - t} \mathbf x + \paren {t - t'} \mathbf y }} {{end-eqn}} We can now estimate the norm of this last expression: {{begin-eqn}} {{eqn | l = \norm {\paren {t' - t} \mathbf x + \paren {t - t'} \mathbf y}       | o = \le       | r = \norm {\paren {t' - t} \mathbf x} + \norm {\paren {t - t'} \mathbf y}       | c = Triangle Inequality for Vectors in Euclidean Space }} {{eqn | r = \size {t' - t} \norm {\mathbf x} + \size {t - t'} \norm {\mathbf y}       | c = Axiom $(N2)$ for norms }} {{eqn | r = \size {t' - t} \paren {\norm {\mathbf x} + \norm {\mathbf y} } }} {{eqn | o = <       | r = \frac \epsilon {1 + \norm {\mathbf x} + \norm {\mathbf y} } \paren {\norm {\mathbf x} + \norm {\mathbf y} } }} {{eqn | o = <       | r = \epsilon }} {{end-eqn}} Since $\epsilon$ was arbitrary, we conclude that $l$ is continuous. Therefore, it forms a path from $\mathbf x$ to $\mathbf y$. Since $\mathbf x$ and $\mathbf y$ were arbitrary, it follows that $\R^n$ is path-connected. {{qed}}
Follows directly from Equalizer is Monomorphism and the Duality Principle. {{qed}}
Let $\le$ denote the ordering on integers. Let $\preceq$ denote the ordeing on multiindices. Recall that if $k = \family {k_j}_{j \mathop \in J}$ and $\ell = \family {\ell_j}_{j \mathop \in J}$ are multiindices, then $k \le \ell$ if $k_j \preceq \ell_j$ for all $j \in J$. Let $k$ and $\ell$ be as above, and let $m = \family {m_j}_{j \mathop \in J} \in Z$. From Integers under Addition form Totally Ordered Group, the integers are totally ordered by $\le$. Hence the ordering on the integers is reflexive, antisymmetric and transitive. === Reflexivity === By reflexivity of $\le$, we have for all $j \in J$: :$k_j \le k_j$ This shows that $\preceq$ is reflexive. {{Qed|lemma}} === Antisymmetry === Suppose that $k \preceq \ell$ and $\ell \preceq k$. Then for all $j \in J$, we have: :$k_j \le \ell_j \le k_j$ By antisymmetry of $\le$, we have $k_j = \ell_j$ for all $j$, that is, $k = \ell$. This shows $\preceq$ to be antisymmetric. {{Qed|lemma}} === Transitivity === Suppose that $k \preceq \ell$ and $ell \preceq m$. Then for all $j \in J$, we have: :$k_j \le \ell_j$ and $\ell_j \le m_j$ By transitivity of $\le$, this shows that for all $j \in J$, $k_j \le m_j$. Therefore $k \preceq m$, and $\preceq$ is shown to be transitive. {{Qed}} Category:Polynomial Theory gxk9vz00a40xc0aumtkgkmpi67ocfky
We check each of the axioms in turn. Let $k = \left \langle {k_j}\right \rangle_{j \in J}$, $\ell = \left \langle {\ell_j}\right \rangle_{j \in J}$ and $m = \left \langle {m_j}\right \rangle_{j \in J}$ be multiindices. === Closure === Trivially we have that: :$j \mapsto \left({ k_j + \ell_j }\right)$ is a sequence of integers indexed by $J$. We know that finitely many of the $k_j$ are non-zero, and finitely many of the $\ell_j$ are non-zero. Therefore finitely many of the $k_j + \ell_j$ are non-zero. This shows that $k + \ell$ is a multiindex and so $+$ is closed. {{qed|lemma}} === Associativity === For all $j \in J$, we have: {{begin-eqn}} {{eqn|l = \left(\left(\ell + k\right) + m \right)_j      |r = \left(\ell + k\right)_j + m_j      |c =  }} {{eqn|r = \left(\ell_j + k_j\right) + m_j      |c =  }} {{eqn|r = \ell_j + \left(k_j + m_j\right)      |c = by Integer Addition is Associative }} {{eqn|r = \ell_j + \left(k + m\right)_j      |c =  }} {{eqn|r = \left(\ell + \left(k + m\right) \right)_j      |c =  }} {{end-eqn}} Thus $+$ is shown to be associative. {{qed|lemma}} === Commutativity === For all $j \in J$, we have: {{begin-eqn}} {{eqn|l = \left(\ell + k\right)_j      |r = \ell_j + k_j      |c =  }} {{eqn|r = k_j + \ell_j      |c = By Integer Addition is Commutative }} {{eqn|r = \left(k + \ell\right)_j      |c =  }} {{end-eqn}} Thus $+$ is shown to be commutative. {{qed|lemma}} === Identity Element === Let $0_Z$ be the multiindex defined by: :$\left(0_Z\right)_j = 0$ for all $j \in J$. Then we have, for all $j \in J$: {{begin-eqn}} {{eqn|l = \left(0_Z + \ell\right)_j      |r = 0 + \ell_j      |c =  }} {{eqn|r = \ell_j      |c =  }} {{end-eqn}} Since we have seen that $\left({Z, +}\right)$ is commutative, this shows that $0_Z$ is an identity element for $Z$. {{Qed}} Category:Polynomial Theory qe0w3f8egjspdnmdpiopykvs42e2a9q
These follow directly from: * Real Number Line is Metric Space * Complex Plane is Metric Space * Metric Induces Topology {{qed}}
First of all, by definition of multiindexed powers: :$\displaystyle \left(x + y\right)^\alpha = \prod_{k \mathop = 1}^n\left(x_k + y_k\right)^{\alpha_k}$ Then: {{begin-eqn}} {{eqn | l = \left(x + y\right)^\alpha       | r = \prod_{k \mathop = 1}^n \sum_{\beta_k \mathop = 0}^{\alpha_k} {\alpha_k \choose \beta_k} x_k^{\alpha_k - \beta_k} y_k^{\beta_k}       | c = Binomial Theorem for Integral Indices }} {{eqn | r = \sum_{\beta_1 \mathop = 0}^{\alpha_1} \cdots \sum_{\beta_n \mathop = 0}^{\alpha_n} {\alpha_1 \choose \beta_1} \cdots {\alpha_n \choose \beta_n} x_1^{\alpha_1 - \beta_1} \cdots x_n^{\alpha_n - \beta_n} y_1^{\beta_1} \cdots y_n^{\beta_n}       | c = expanding out the product }} {{end-eqn}} On the other hand: {{begin-eqn}} {{eqn | l = \sum_{0 \mathop \le \beta \mathop \le \alpha} {\alpha \choose \beta} x^\beta y^{\alpha - \beta}       | r = \sum_{0 \mathop \le \beta \mathop \le \alpha} {\alpha_1 \choose \beta_1} \cdots {\alpha_n \choose \beta_n} x_1^{\alpha_1 - \beta_1} \cdots x_n^{\alpha_n - \beta_n} y_1^{\beta_1} \cdots y_n^{\beta_n}       | c = by definition of Power and Binomial Coefficient }} {{eqn | r = \sum_{\beta_1 \mathop = 0}^{\alpha_1} \cdots \sum_{\beta_n \mathop = 0}^{\alpha_n} {\alpha_1 \choose \ell_1}  \cdots {\alpha_n \choose \ell_n} x_1^{\alpha_1 - \ell_1} \cdots x_n^{\alpha_n - \ell_n} y_1^{\ell_1} \cdots y_n^{\ell_n}       | c = by definition of Ordering on Multiindices }} {{end-eqn}} {{Mistake|In the second of the equations above, $\ell$ has not been introduced.}} This shows that: :$\displaystyle \left({x + y}\right)^\alpha = \sum_{0 \mathop \le \beta \mathop \le \alpha} {\alpha \choose \beta} x^\beta y^{\alpha - \beta}$ as required. {{Qed}} {{Proofread}} Category:Binomial Theorem gxvapm0g7zqevbmfqp4ml39ojxrnoxg
Let $\map f s, \map f {s'} \in f \sqbrk S$, for some $s, s' \in S$. Let $\mathbb I$ denote the closed unit interval: :$\mathbb I = \closedint 0 1$ Let $p: \mathbb I \to S$ be a continuous mapping such that: :$\map p 0 = s, \map p 1 = s'$ Such a $p$ exists since $S$ is a path-connected subspace of $M_1$. Now define $q: \mathbb I \to f \sqbrk S$ by: :$\map q t := f \circ \map p t$ Then $q$ is a continuous mapping, being the composition of such, and: :$\map q 0 = \map f {\map p 0} = \map f s$ :$\map q 1 = \map f {\map p 1} = \map f {s'}$ Thus $\map f s$ and $\map f {s'}$ are connected by a path in $f \sqbrk S$. Since these points were arbitrary in $f \sqbrk S$, the result follows. {{qed}} {{expand|This result immediately generalises to topological spaces; the definition of path-conn'ness for metric spaces is instantiated from that for top. spaces<br>... I agree - a proof should also be included. This one can then link to that one as an instantiation, as a second proof.}}
First, inserting the definitions, the statement of the theorem reads: :$\displaystyle \map {\partial_1^{\alpha_1} \partial_2^{\alpha_2} \cdots \partial_n^{\alpha_n} } {f g} = \sum_{\beta_1 = 0}^{\alpha_1} \cdots \sum_{\beta_n \mathop = 0}^{\alpha_n} \binom {\alpha_1} {\beta_1} \cdots \binom {\alpha_n} {\beta_n} \paren {\partial_1^{\beta_1} \cdots \partial_n^{\beta_n} f} \paren {\partial_1^{\alpha_1 - \beta_1} \cdots \partial_n^{\alpha_n - \beta_n} g}$ We prove this by induction over $n \ge 1$. === Basis for the Induction === If $n = 1$, the result is a simple restatement of Leibniz's Rule in one variable: :$\displaystyle \paren {\map f x \, \map g x}^{\paren n} = \sum_{k \mathop = 0}^n \binom n k \map {f^{\paren k} } x \, \map {g^{\paren {n - k} } } x$ This is the basis for the induction. === Induction Step === Suppose now that the result is true for functions of $n - 1$ variables. In particular, we suppose that: :$\displaystyle \map {\partial_2^{\alpha_2} \cdots \partial_n^{\alpha_n} } {f g} = \sum_{\beta_2 \mathop = 0}^{\alpha_2} \cdots \sum_{\beta_n \mathop = 0}^{\alpha_n} \binom {\alpha_2} {\beta_2} \cdots \binom {\alpha_n} {\beta_n} \paren {\partial_2^{\beta_2} \cdots \partial_n^{\beta_n} f} \paren {\partial_2^{\alpha_2 - \beta_2} \cdots \partial_n^{\alpha_n - \beta_n} g}$ Now let us apply $\partial_1^{\alpha_1}$. Using the linearity of derivatives: :$\displaystyle \map {\partial_1^{\alpha_1} \partial_2^{\alpha_2} \cdots \partial_n^{\alpha_n} } {f g} = \sum_{\beta_2 \mathop = 0}^{\alpha_2} \cdots \sum_{\beta_n \mathop = 0}^{\alpha_n} \binom {\alpha_2} {\beta_2} \cdots \binom {\alpha_n} {\beta_n} \partial_1^{\alpha_1} \paren {\paren {\partial_2^{\beta_2} \cdots \partial_n^{\beta_n} f} \paren {\partial_2^{\alpha_2 - \beta_2} \cdots \partial_n^{\alpha_n - \beta_n} g} }$ Applying Leibniz's Rule in one variable we have: :$\displaystyle \map {\partial_1^{\alpha_1} \partial_2^{\alpha_2} \cdots \partial_n^{\alpha_n} } {f g} = \sum_{\beta_2 \mathop = 0}^{\alpha_2} \cdots \sum_{\beta_n \mathop = 0}^{\alpha_n} \binom {\alpha_2} {\beta_2} \cdots \binom {\alpha_n} {\beta_n} \sum_{\beta_1 \mathop = 0}^{\alpha_1} \binom {\alpha_1} {\beta_1} \paren {\partial_1^{\beta_1} \partial_2^{\beta_2} \cdots \partial_n^{\beta_n} f} \paren {\partial_1^{\alpha_1 - \beta_1} \partial_2^{\alpha_2 - \beta_2} \cdots \partial_n^{\alpha_n - \beta_n} g}$ Now, all the sums are finite, thus trivially absolutely convergent. By Manipulation of Absolutely Convergent Series and Summation is Linear we can move the inner sum to the far left, giving: :$\displaystyle \map {\partial_1^{\alpha_1} \partial_2^{\alpha_2} \cdots \partial_n^{\alpha_n} } {f g} = \sum_{\beta_1 \mathop = 0}^{\alpha_1} \cdots \sum_{\beta_n \mathop = 0}^{\alpha_n} \binom {\alpha_1} {\beta_1} \cdots \binom {\alpha_n} {\beta_n} \paren {\partial_1^{\beta_1} \cdots \partial_n^{\beta_n} f} \paren {\partial_1^{\alpha_1 - \beta_1} \cdots \partial_n^{\alpha_n - \beta_n} g}$ The result now follows by the Principle of Mathematical Induction. {{Qed}}
By definition, $\tau$ is the product topology on $X \times D$. That is, $\tau$ has as a basis sets of the form: :$U \times V$ with $U \in \tau$ and $V$ open in $D$. Since $D$ is endowed with the indiscrete topology, either $V = \O$ or $V = D$. In the former case, by Cartesian Product is Empty iff Factor is Empty, $U \times V = \O$ for all $U \in \tau$. In particular, then, it follows that $\tau$ has the basis: :$\BB = \set {U \times D: U \in \tau}$ since $U = \O \in \tau$, and so the basis elements with $V = \O$ are also accounted for. It is therefore clear that all $U \times D$ with $U \in \tau$ are open in $\tau$. That open sets of this form constitute all of $\tau$ amounts to showing that: :$\tau = \BB$ that is, that $\BB$ is already a topology on $S \times D$. {{finish|show this, maybe transfer/replace this result to/with general "Topology of Product with Indiscrete Space" or s.t. like that}} {{qed}} Category:Double Pointed Topologies pj1sa27w606egyd5snjm2u8k9soo5cb
By definition, $X$ is closed {{iff}} its complement $\map \complement X$ is open. By Open Sets of Double Pointed Topology, it follows by that for some $U \in \tau$: :$\map \complement X = U \times D$ Then by Cartesian Product Distributes over Set Difference and Complement of Complement, we have that: {{improve|rather, Cartesian Product Distributes over Complement (and for Relative Complement), but those don't exist}} :$X = \paren {S \times D} \setminus \paren {U \times D} = \paren {S \setminus U} \times D$ Since $S \setminus U$ is a closed set of $\tau$, the result follows. {{qed}} Category:Double Pointed Topologies 54k40gvbl6yivcnruerescni5s08tly
So $\Q \sqbrk {\sqrt 2} \subseteq \R$. From Numbers of Type Rational a plus b root 2 form Field, $\struct {\Q \sqbrk {\sqrt 2}, +, \times}$ is a field. As stated in the proof of the Numbers of Type Rational a plus b root 2 form Field, numbers of the form $a + b \sqrt 2$ are real. Hence the result by definition of subfield. {{qed}}
Follows directly from Open Set Less One Point is Open and Finite Intersection of Open Sets of Metric Space is Open. Let: : $U_1 = U \setminus \left\{{\alpha_1}\right\}, U_2 = U \setminus \left\{{\alpha_2}\right\}, \ldots, U_n = U \setminus \left\{{\alpha_n}\right\}$ From the above, $U_1, U_2, \ldots, U_n$ are all open in $M$. From Finite Intersection of Open Sets of Metric Space is Open $\displaystyle \bigcap_{i \mathop = 1}^n U_i$ is open. {{qed}} Category:Open Sets 8nbeenl0n28j14eo4h2bsnrbaocfu2c
Let $a < c < b$. Let $\epsilon < \min \left({b - c, c - a}\right)$. From the definition of positive it follows that $\epsilon \in \R_{>0}$. Let $B_\epsilon \left({c}\right) = \left({c - \epsilon \,.\,.\, c + \epsilon}\right)$ be the open $\epsilon$-ball of $c$. We have that $c + \epsilon < b$ and $a < c - \epsilon$. Thus: : $B_\epsilon \left({c}\right) \subseteq \left[{a \,.\,.\, b}\right]$ {{explain|There needs to be a result linked to that proves that $B_\epsilon \left({c}\right)$ is in fact a subset of $\left[{a \,.\,.\, b}\right]$.}} It follows that, by definition, $\left[{a \,.\,.\, b}\right]$ is a neighborhood of $c$. Now consider $a$. Let $\epsilon \in \R_{>0}$. We have that $a - \epsilon < a$ and so $B_\epsilon \left({a}\right) = \left({a - \epsilon \,.\,.\, a + \epsilon}\right)$ does not lie entirely in $\left[{a \,.\,.\, b}\right]$. Similarly, consider $b$. Let $\epsilon \in \R_{>0}$. We have that $b < b + \epsilon$ and so $B_\epsilon \left({b}\right) = \left({b - \epsilon \,.\,.\, b + \epsilon}\right)$ does not lie entirely in $\left[{a \,.\,.\, b}\right]$. It follows that, by definition, $\left[{a \,.\,.\, b}\right]$ is a neighborhood of neither $a$ nor $b$. {{qed}}
Let $c \in \R$ such that $a < c < b$. Let $\epsilon = \min \set {b - c, c - a}$. From the definition of positive it follows that $\epsilon \in \R_{>0}$. Let $\map {B_\epsilon} c = \openint {c - \epsilon} {c + \epsilon}$ be the open $\epsilon$-ball of $c$. Let $y \in \map {B_\epsilon} c$. Then: :$\size {c - y} < \epsilon $ implies :$-\epsilon < c - y < \epsilon$ We have that: {{begin-eqn}} {{eqn | l = a       | r = c - \paren {c - a} }} {{eqn | o = \le       | r = c - \epsilon }} {{eqn | o = <       | r = y }} {{eqn | o = <       | r = c + \epsilon }} {{eqn | o = \le       | r = c + \paren {b - c} }} {{eqn | r = b }} {{end-eqn}} In other words: :$y \in \map {B_\epsilon} c \implies y \in \openint a b$ Thus: : $\map {B_\epsilon} c \subseteq \openint a b$ It follows that, by definition, $\openint a b$ is a neighborhood of $c$. The result follows by definition of open set. {{qed}}
From Closed Real Interval is Neighborhood Except at Endpoints, $a$ and $b$ have no open $\epsilon$-ball lying entirely in $\left[{a \,.\,.\, b}\right]$. The result follows by definition of open set. {{qed}}
Consider the Euclidean space $\R^2$. Let: :$U \subseteq \R^2: U = \set {\tuple {x_1, x_2}: a < x_1 < b, c < x_2 < d}$ Let $x = \tuple {x_1, x_2} \in U$. Then $\map {B_\epsilon} x \subseteq U$ when $\epsilon = \min \set {x_1 - a, b - x_1, x_2 - c, d - x_2}$: :310px So by definition, $U$ is open in $M$. However, $U$ is not an open ball. {{qed}}
Let $f: S \to T$ be a mapping as in the following commutative diagram: ::$\begin{xy}\xymatrix{  \mathcal R    \ar[r]<2pt>^*{r_1}   \ar[r]<-2pt>_*{r_2} &   S   \ar[r]^*{q}   \ar[rd]_*{f} &  S / \mathcal R   \ar@{.>}[d]^*{\bar f} \\ & &  T }\end{xy}$ This translates to, for $s_1, s_2 \in S$ with $s_1 \mathcal R s_2$: :$f \circ r_1 \left({s_1, s_2}\right) = f \circ r_2 \left({s_1, s_2}\right)$ i.e. $f \left({s_1}\right) = f \left({s_2}\right)$. The commutativity of the diagram implies that we must define $\bar f: S / \mathcal R \to T$ by: :$\bar f \left({\left[\!\left[{s_1}\right]\!\right]}\right) = f \left({s_1}\right)$ since $q \left({s_1}\right) = \left[\!\left[{s_1}\right]\!\right]$. The above condition precisely states that $\bar f$ is well-defined. In conclusion, for any $f$ with $f \circ r_1 = f \circ r_2$, there is a unique $\bar f$ making the diagram commute. That is, $q$ is a coequalizer of $r_1$ and $r_2$. {{qed}}
{{begin-eqn}} {{eqn | l = \closedint a b       | r = \set {x \in \R: x \ge a \land x \le b}       | c = {{Defof|Closed Real Interval}} }} {{eqn | ll= \leadsto       | l = \R \setminus \closedint a b       | r = \R \setminus \set {x \in \R: x \ge a \land x \le b}       | c =  }} {{eqn | r = \set {x \in \R: x < a \lor x > b}       | c = De Morgan's Laws: Disjunction of Negations }} {{eqn | r = \openint {-\infty} a \cup \openint b \infty       | c = {{Defof|Open Real Interval}} }} {{end-eqn}} From the corollary to Open Real Interval is Open Set, both $\openint {-\infty} a$ and $\openint b \infty$ are open sets in $M$. From Union of Open Sets of Metric Space is Open it follows that $\openint {-\infty} a \cup \openint b \infty$ is open in $\R$. But $\openint {-\infty} a \cup \openint b \infty$ is the relative complement of $\closedint a b$ in $\R$. The result follows by definition of closed set. {{qed}}
{{MissingLinks}} {{proofread|Use of injectivity of $g$ here is implicit; but I ''think'' it's still rigorous without details: second opinion wecome}} By the definition of the matrix $A$: : $A \circ f = g \circ \phi$. Therefore if $x \in \ker \phi$ we have: :$A f \left({x}\right) = g \left({\phi \left({x}\right)}\right) = g \left({0}\right) = 0$ This shows that: :$f \left[{\ker \phi}\right] \subseteq N \left({A}\right)$ Now let $x = \left({x_1, \ldots, x_n}\right) \in N \left({A}\right)$. Let $y = x_1 e_1 + \cdots + x_n e_n \in V$. We have: :$g \circ \phi \left({y}\right) = A \circ f \left({y}\right) = A \left({x_1, \ldots, x_n}\right)^T = 0$ so $\phi \left({y}\right) = 0 f_1 + \cdots + 0 f_m = 0$. This shows that $y \in \ker \phi$. Since $x = f \left({y}\right)$, we have shown that $N \left({A}\right) \subseteq f \left({\ker \phi}\right)$. Therefore $f \left({\ker \phi}\right) = N \left({A}\right)$. We deduce immediately from the definitions that $\ker \phi \subseteq f^{-1} \left[{N \left({A}\right)}\right]$. Now suppose that $x \in f^{-1} \left[{N \left({A}\right)}\right]$. Then: :$f \left({x}\right) \in N \left({A}\right)$ Therefore: :$g \circ \phi \left({x}\right) = A \circ f \left({x}\right) = 0$ so $\phi \left({x}\right) = 0 f_1 + \cdots + 0 f_m = 0$. This shows that: :$f^{-1} \left[{N \left({A}\right)}\right] \subseteq \ker \phi$ as required. {{Qed}} Category:Null Spaces e2zc7tcpsda1ptoyvlgvmp6erkhx24e
From Half-Open Real Interval is not Open Set we have that $\left[{a \,.\,.\, b}\right)$ is not an open set of $\R$. From Half-Open Real Interval is not Closed in Real Number Line we have that $\left[{a \,.\,.\, b}\right)$ is not a closed set of $\R$. {{qed}}
=== Necessary Condition === Suppose that $X \subseteq S \times D$ is a compact set in $\vartheta$. It is to be shown that $C = \map {\pr_1} X$ is compact in $\tau$. This follows from Compactness Properties Preserved under Projection Mapping. {{qed|lemma}} === Sufficient Condition === Suppose $\map {\pr_1} X = C$ for some compact $C \subseteq S$. Let $\UU$ be an open cover for $X$, that is: :$\displaystyle \bigcup \UU \supseteq X$ By Open Sets of Double Pointed Topology, each $U \in \UU$ is of the form: :$U' \times D$ with $U'$ open in $\vartheta$; note that $U' = \map {\pr_1} U$. By Cartesian Product Distributes over Union, it follows that: :$\displaystyle \bigcup \UU = \paren {\bigcup_{U \mathop \in \UU} \map {\pr_1} U} \times D$ Since $X \subseteq \displaystyle \bigcup \UU$, it follows by Image Preserves Subsets and Image of Union under Relation that: :$C = \map {\pr_1} X \subseteq \bigcup_{U \mathop \in \UU} \map {\pr_1} U$ showing that: :$\VV := \set {\map {\pr_1} U: U \in \UU}$ is an open cover for $C$. Since $C$ is compact, we find a finite subcover $\VV'$ of $\VV$ such that: :$C \subseteq \displaystyle \bigcup \VV'$ Finally, we will show that: :$\UU' := \set {V \times D: V \in \VV'} \subseteq \UU$ forms a finite subcover of $\UU$. That is, that: :$X \subseteq \displaystyle \bigcup \UU'$ So let $\tuple {s, d} \in X$. By assumption on $X$, we have that $s \in C$, and since $\VV'$ is a open cover for $C$: :$\exists V \in \VV': s \in V$ It is immediate that $\tuple {s, d} \in V \times D$ by definition of Cartesian product. Since $V \times D \in \UU'$, it follows by definition of set union that: :$\tuple {s, d} \in \displaystyle \bigcup \UU'$ That is, $\UU'$ is a finite subcover for $X$, and hence $X$ is compact. {{qed}} Category:Double Pointed Topologies drzbnbwfo99no39rnfvwn950cdd0uug