hcju/proofwikiqa huggingface dataset:
A measure is an additive function, and, by definition, nowhere negative. So Additive Nowhere Negative Function is Subadditive applies. Hence the result directly: :$\mu \left({E \cup F}\right) \le \mu \left({E}\right) + \mu \left({F}\right)$ {{qed}}
By the definition of degree of a linear representation, it is known that $\dim \left({V}\right) = 1$. Let $W$ be a proper vector subspace of $V$. It follows from Dimension of Proper Subspace is Less Than its Superspace that: :$\dim \left({W}\right) < 1$ and hence $\dim \left({W}\right) = 0$. Now from Trivial Vector Space iff Zero Dimension, it follows that:  :$W = \left\{{\mathbf 0}\right\}$ But this is not a non-trivial proper subspace of $V$. Thus $V$ has no non-trivial proper vector subspaces. Hence, by definition, $\rho$ is an irreducible linear representation. {{qed}} Category:Representation Theory danaed4hxwyj92gbmkbtb2z59clixql
=== Sufficient Condition === Suppose that $\dim \left({V}\right) = 1$. That $\rho$ is irreducible is shown on Representation of Degree One is Irreducible. {{qed|lemma}} === Necessary Condition === Suppose that $\rho$ is an irreducible linear representation. Let $g \in G$ be arbitrary. Now, for all $h \in G$, have: {{begin-eqn}} {{eqn|l = \rho \left({g}\right) \rho \left({h}\right)      |r = \rho \left({g h}\right)      |c = $\rho$ is a group homomorphism }} {{eqn|r = \rho \left({h g}\right)      |c = $G$ is an abelian group }} {{eqn|r = \rho \left({h}\right) \rho \left({g}\right)      |c = $\rho$ is a group homomorphism }} {{end-eqn}} Now, combining Commutative Linear Transformation is G-Module Homomorphism and Corollary to Schur's Lemma (Representation Theory) yields that: :$\exists \lambda_g \in k: \rho \left({g}\right) = \lambda_g \operatorname{Id}_V$ That is, there is a $\lambda_g \in k$ such that $\rho \left({g}\right)$ is the linear mapping of multiplying by $\lambda_g$. Hence, for all $v \in V$, $\rho \left({g}\right) \left({v}\right) = \lambda_g v$. It follows that any vector subspace of $V$ of dimension $1$ is invariant.  So, had $V$ any proper vector subspace of dimension $1$, $\rho$ would not be irreducible. Since $V$ is non-null, it follows from Trivial Vector Space iff Zero Dimension that $\dim \left({V}\right) > 0$. Hence necessarily $\dim \left({V}\right) = 1$. {{qed}} Category:Representation Theory fogs6da21i3iult165g0jn3efol0b3p
From Kernel is G-Module, $\ker \left({f}\right)$ is a $G$-submodule of $V$. From Image is G-Module, $\operatorname{Im} \left({f}\right)$ is a $G$-submodule of $V'$. By the definition of irreducible: :$\ker \left({f}\right) = \left\{{0}\right\}$ or: :$\ker \left({f}\right) = V$ {{explain|Link to a result which shows this. While it does indeed follow from the definition, it would be useful to have a page directly demonstrating this.}} If $\ker \left({f}\right) = V$ then by definition: :$f \left({v}\right) = 0$ for all $v \in V$ Let $\ker \left({f}\right) = \left\{{0}\right\}$. Then from Linear Transformation is Injective iff Kernel Contains Only Zero: :$f$ is injective. {{explain|Establish whether the above result (which discusses linear transformations on $R$-modules, not $G$ modules) can be directly applied. If so, amend its wording so as to make this clear.}} It also follows that: :$\operatorname{Im} \left({f}\right) = V'$ {{Explain|Prove this}} Thus $f$ is surjective and injective. Thus by definition $f$ is a bijection and thence an isomorphism. {{qed}} {{namedfor|Issai Schur|cat = Schur}} Category:Representation Theory bythnhbim7dkufkqe8z7ggoqxlq6p70
=== Necessary Condition === Assume that $V$ is an irreducible $G$-module, but it has a non-trivial proper $G$-submodule. By the definition irreducible, its associated representation is irreducible. Let this representation be denoted $\tilde \phi = \rho: G \to \operatorname{GL} \left({V}\right)$). In Correspondence between Linear Group Actions and Linear Representations it is defined as: :$\rho \left({g}\right) \left({v}\right) = \phi \left({g, v}\right)$ where $g \in G$ and $v \in V$. Since $V$ has a proper  $G$-submodule, there exists $W$ a non-trivial proper vector subspace which $\phi \left({G, W}\right) \subseteq W$ and so $\rho \left({G}\right) W \subseteq W$. Hence $W$ is invariant by every linear operators in $\left\{ {\rho \left({g}\right): g \in G}\right\}$. By definition, $\rho$ cannot be irreducible. Thus we have reached a contradiction, and $V$ has then no non-trivial  proper $G$-submodules. {{qed|lemma}} === Sufficient Condition === Assume now that $V$ has no proper $G$-submodules, but it is a reducible $G$-module. By the definition of reducible $G$-module, it follow that its associated representation is reducible. Let this representation be denoted $\tilde \phi = \rho: G \to \operatorname{GL} \left({V}\right)$). From the definition of reducible representation, it follows that there exists a vector space $W$ of $V$. This is invariant under all the linear operators in $\left\{ {\rho \left({g}\right): g \in G}\right\}$. {{explain|Why is it invariant?}} Then: :$\phi \left({G, W}\right) = \rho \left({G}\right) W \subseteq W$ which is the definition of a $G$-submodule of $V$. By our assumption, $V$ has no non-trivial proper $G$-submodules. Thus we have reached a contradiction and $V$ must be then an irreducible $G$-module. {{qed}} Category:Representation Theory 73n1827vgl6vqae7ngeyjo0kv1fpmmu
From G-Submodule Test it suffices to prove that $\phi \sqbrk {\struct {G, \map \ker f} } \subseteq \map \ker f$.  That is, it is to be shown that, if $g \in G$ and $v \in \map \ker f$, then $\map \phi {g, v} \in \map \ker f$. Assume that $g \in G$ and $v \in \map \ker f$. {{begin-eqn}} {{eqn | l = \map f {\map \phi {g, v} }       | r = \map \mu {g, \map f v}       | c = $f$ is a $G$-module homomorphism }} {{eqn | r = \map \mu {g, 0}       | c = $v \in \map \ker f$ }} {{eqn | r = 0       | c = $\mu$ is a linear action }} {{end-eqn}} Thus $\map \phi {g, v} \in \map \ker f$. Hence $\map \ker f$ is a $G$-submodule of $V$. {{qed}} Category:Representation Theory 2woti1bqyrmd6d01v2dsexhnha80exm
From G-Submodule Test it suffices to prove that $\mu \left({G, \operatorname{Im} \left({f}\right) }\right) \subseteq \operatorname{Im} \left({f}\right)$. In other words: for any $g \in G$ and $w \in \operatorname{Im} \left({f}\right)$, it is to be shown that $\mu \left({g, w}\right) \in \operatorname{Im} \left({f}\right)$. Assume that $g \in G$ and $w\in\operatorname{Im} \left({f}\right)$. Then there exists a $v \in V$ such that $f \left({v}\right) = w$. By definition of homomorphism, have: :$\mu \left({g, w}\right) = \mu \left({g, f \left({v}\right) }\right) = f \left({\phi \left({g, v}\right) }\right)$ Hence, for all $g \in G$ and $w\in\operatorname{Im} \left({f}\right)$, $\mu \left({g, w}\right) \in \operatorname{Im} \left({f}\right)$. By G-Submodule Test, it follows that $\operatorname{Im} \left({f}\right)$ is a $G$-submodule of $V'$. {{qed}} Category:Representation Theory lrqe10cldex5s9jftgmv10fkv100lhw
{{begin-eqn}} {{eqn | l = x       | o = \in       | r = \bigcap_{\alpha \mathop \in I} A_\alpha       | c =  }} {{eqn | lll=\leadsto       | ll= \forall \alpha \in I:       | l = x       | o = \in       | r = A_\alpha       | c = {{Defof|Intersection of Family}} }} {{eqn | lll=\leadsto       | ll= \forall \alpha \in I:       | l = x       | o = \in       | r = B_\alpha       | c = {{Defof|Subset}} }} {{eqn | lll=\leadsto       | l = x       | o = \in       | r = \bigcap_{\alpha \mathop \in I} B_\alpha       | c = {{Defof|Intersection of Family}} }} {{end-eqn}} By definition of subset: :$\displaystyle \bigcap_{\alpha \mathop \in I} A_\alpha \subseteq \bigcap_{\alpha \mathop \in I} B_\alpha$ {{qed}}
Let $A \subseteq B$ and $S \subseteq T$. Then: {{begin-eqn}}      {{eqn| l = x \in A      | o = \implies      | r = x \in B      | c = {{Defof|Subset}} }} {{eqn| l = x \in S      | o = \implies      | r = x \in T      | c = {{Defof|Subset}} }} {{end-eqn}} Now we invoke the Praeclarum Theorema of propositional logic: :$\paren {p \implies q} \land \paren {r \implies s} \vdash \paren {p \land r} \implies \paren {q \land s}$ applying it as: :$\paren {x \in A \implies x \in B, \ x \in S \implies x \in T} \leadsto \paren {x \in A \land x \in S \implies x \in B \land x \in T}$ The result follows directly from the definition of set intersection: :$\paren {x \in A \implies x \in B, \ x \in S \implies x \in T} \leadsto \paren {x \in A \cap S \implies x \in B \cap T}$ and from the definition of subset: :$A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$ {{qed}} Category:Set Intersection Category:Subsets Category:Set Intersection Preserves Subsets ji875znv1qa80dwb7ktw1erjpbhjpbv
{{begin-eqn}} {{eqn | l = \mathbf A \mathbf 0       | r = \begin {bmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\ a_{2 1} & a_{2 2} & \cdots & a_{2 n} \\  \vdots &  \vdots & \ddots &  \vdots \\ a_{m 1} & a_{m 2} & \cdots & a_{m n} \\ \end {bmatrix} \begin {bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end {bmatrix} }} {{eqn | r = \begin {bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end {bmatrix} }} {{eqn | r = \mathbf 0 }} {{end-eqn}} The order is correct by hypothesis. The result follows by the definition of null space. {{qed}}
From Matrix Product as Linear Transformation, $\mathbf {Ax} = \mathbf 0$ defines a linear transformation from $\R^m$ to $\R^n$. The result then follows from Linear Transformation Maps Zero Vector to Zero Vector.
By the definition of null space, $\mathbf 0$ is a solution {{iff}} the null space contains the zero vector. The result follows from Null Space Contains Zero Vector. {{qed}} Category:Linear Algebra Category:Null Spaces mp2ftq9clu23hsbzit9h6vwj15zfend
Let: :$\forall g \in G: \rho \left({g}\right) \circ f = f \circ \rho \left({g}\right)$ Let $v$ be a vector $v \in V$. Then: : $\rho \left({g}\right) \left({f \left({v}\right)}\right) = f \left({\rho \left({g}\right) \left({v}\right)}\right)$ Using the properties from Correspondence between Linear Group Actions and Linear Representations: :there exists a $G$-module $\left({V, \phi}\right)$  associated with $\rho$ such that: ::$\phi \left({g, v}\right) = \rho \left({g}\right) \left({v}\right)$ Applying the last formula: :$\rho \left({g}\right) \left({f \left({v}\right)}\right) = \phi \left({g, f \left({v}\right)}\right)$ and: :$f \left({\phi \left({g, v}\right)}\right) = f \left({\rho \left({g}\right) \left({v}\right)}\right)$ Thus our assumption is equivalent to: :$f \left({\phi \left({g, v}\right)}\right) = \phi \left({g, f \left({v}\right)}\right)$ Hence, by definition of $G$-module homomorphism, $f: V \to V$ is a $G$-module homomorphism. {{qed}} Category:Representation Theory seiecfhsn2xwcc0n9y12p95801ucz5a
Let $\left({E_n}\right)_{n \in \N}$ be a sequence of sets in $\Sigma$. It is required to show that: :$\displaystyle \mu \left({\bigcup_{n \mathop \in \N} E_n}\right) \le \sum_{n \mathop \in \N} \mu \left({E_n}\right)$ Now define the sequence $\left({F_n}\right)_{n\in\N}$ in $\Sigma$ by: :$F_n := \displaystyle \bigcup_{k \mathop = 1}^n E_n$ By Subset of Union, it follows that, for all $n \in \N$, $F_n \subseteq F_{n+1}$. Hence, $\left({F_n}\right)_{n\in\N}$ is increasing. It is immediate that $F_n \uparrow \displaystyle \bigcup_{n \mathop \in \N} E_n$, where $\uparrow$ signifies the limit of an increasing sequence of sets. Now reason as follows: {{begin-eqn}} {{eqn|l = \mu \left({\bigcup_{n \mathop \in \N} E_n}\right)      |r = \lim_{n \to \infty} \mu \left({F_n}\right)      |c = Characterization of Measures, $(3)$ }} {{eqn|r = \lim_{n \to \infty} \mu \left({E_1 \cup \cdots \cup E_n}\right)      |c = Definition of $F_n$ }} {{eqn|o = \le      |r = \lim_{n \to \infty} \sum_{k \mathop = 1}^n \mu \left({E_k}\right)      |c = Measure is Subadditive: Corollary }} {{eqn|r = \sum_{k \mathop \in \N} \mu \left({E_k}\right) }} {{end-eqn}} Hence the result. {{qed}}
Let us verify in turn that $\delta_x$ satisfies the axioms for a measure. === Axiom $(1)$ === By definition of the Dirac measure, $\delta_x \left({E}\right) \ge 0$ for all $E \in \Sigma$. {{qed|lemma}} === Axiom $(2)$ === Let $\left({E_n}\right)_{n \in \N}$ be a sequence of pairwise disjoint sets. It follows that if for some $m \in \N$, $x \in E_m$, it must be that $n \ne m$ implies $x \notin E_n$. Now suppose $x \in E_m$ for some $m \in \N$. Then by definition of set union, $x \in \displaystyle \bigcup_{n \mathop \in \N} E_n$. Thus: {{begin-eqn}} {{eqn|l = \delta_x \left({\bigcup_{n \mathop \in \N} E_n}\right)      |r = 1 }} {{eqn|r = \sum_{n \mathop \in \N} \delta_x \left({E_n}\right) }} {{end-eqn}} because $\delta_x \left({E_n}\right) = 0$ iff $n \ne m$, and $1$ otherwise. Finally, if $x \notin E_n$ for all $n \in \N$, then by definition of set union: :$x \notin \displaystyle \bigcup_{n \mathop \in \N} E_n$ so that: {{begin-eqn}} {{eqn|l = \delta_x \left({\bigcup_{n \mathop \in \N} E_n}\right)      |r = 0 }} {{eqn|r = \sum_{n \mathop \in \N} 0 }} {{eqn|r = \sum_{n \mathop \in \N} \delta_x \left({E_n}\right) }} {{end-eqn}} Hence, from Proof by Cases: :$\displaystyle \sum_{n \mathop \in \N} \delta_x \left({E_n}\right) = \delta_x \left({\bigcup_{n \mathop \in \N} E_n}\right)$ {{qed|lemma}} === Axiom $(3)$ === By definition of the Dirac measure, $\delta_x \left({X}\right) = 1$. Hence there is an $E \in \Sigma$ such that $\delta_x \left({E}\right)$ is finite. {{qed|lemma}} Thus, $\delta_x$, satisfying all the axioms, is a measure. {{qed}} Category:Measure Theory 6y5upvpzfpb3c45fniy5wpmiryuduh6
By Dirac Measure is Measure, $\delta_x$ is a measure. Also, $\delta_x \left({X}\right) = 1$ because $x \in X$. Hence $\delta_x$ is a probability measure. {{qed}} Category:Measure Theory 3rtjm4gw46znkd67a9hzhx1ka4odmr0
Let us verify the measure axioms $(1)$, $(2)$ and $(3')$ for $\mu$. === Proof of $(1)$ === For all $S \in \Sigma$, $\mu \left({S}\right)$ is $0$ or $1$. In either case, $\mu \left({S}\right) \ge 0$. {{qed|lemma}} === Proof of $(2)$ === It is to be shown that (for a sequence $\left({S_n}\right)_{n \in \N}$ of pairwise disjoint sets): :$\displaystyle \sum_{n \mathop = 1}^\infty \mu \left({S_n}\right) = \mu \left({\bigcup_{n \mathop = 1}^\infty S_n}\right)$ Suppose that at least one $S_n$ is co-countable, say $S_N$. Since the $S_n$ are pairwise disjoint, it follows that, for all $n \in \N$, $n \ne N$: :$S_n \subseteq X \setminus S_N$ by Empty Intersection iff Subset of Complement. From Subset of Countably Infinite Set is Countable, the $S_n$ with $n \ne N$ are all countable, as $S_N$ is co-countable. Therefore: :$\mu \left({S_n}\right) = \begin{cases}1 & \text{if $n = N$}\\ 0 & \text{if $n \ne N$}\end{cases}$ and subsequently: :$\displaystyle \sum_{n \mathop = 1}^\infty \mu \left({S_n}\right) = 1$ From Superset of Co-Countable Set, $\displaystyle \bigcup_{n \mathop = 1}^\infty S_n$ is co-countable. Hence: $\displaystyle \mu \left({\bigcup_{n \mathop = 1}^\infty S_n}\right) = 1$ verifying $(2)$ for $\mu$ in this case. If on the other hand, all $S_n$ are countable, then for all $n \in \N$: :$\mu \left({S_n}\right) = 0$ and so: :$\displaystyle \sum_{n \mathop = 1}^\infty \mu \left({S_n}\right) = 0$ From Countable Union of Countable Sets is Countable: :$\displaystyle \bigcup_{n \mathop = 1}^\infty S_n$ is also countable, hence: :$\displaystyle \mu \left({\bigcup_{n \mathop = 1}^\infty S_n}\right) = 0$ verifying $(2)$ for $\mu$ in this case as well. Hence $(2)$ holds for $\mu$, from Proof by Cases. {{qed|lemma}} === Proof of $(3')$ === Note that $\varnothing \in \Sigma$ as $\Sigma$ is a $\sigma$-algebra. By Empty Set is Countable, $\varnothing$ is countable, whence: :$\mu \left({\varnothing}\right) = 0$ {{qed|lemma}} Having verified the axioms, it follows that $\mu$ is a measure. {{qed}} Category:Measure Theory 0llfqdlmt7mmdp5v3m9qucpw4xh0uvk
By Co-Countable Measure is Measure, $\mu$ is a measure. By Relative Complement with Self is Empty Set, have $\complement_X \left({X}\right) = \varnothing$. As $\varnothing$ is countable, it follows that $X$ is co-countable. Hence $\mu \left({X}\right) = 1$, and so $\mu$ is a probability measure. {{qed}} Category:Measure Theory g4nimxfe31rmdrmhywpyp7ufg4kpxhc
Let us verify the measure axioms $(1)$, $(2)$ and $(3')$ for $\left\vert{\cdot}\right\vert$. === Proof of $(1)$ === The values that $\left\vert{\cdot}\right\vert$ can take are the natural numbers $\N$ and $+\infty$. All of these are positive, whence: :$\forall S \in \Sigma: \left\vert{S}\right\vert \ge 0$ {{qed|lemma}} === Proof of $(2)$ === It is to be shown that (for a sequence $\left({S_n}\right)_{n \in \N}$ of pairwise disjoint sets): :$\displaystyle \sum_{n \mathop = 1}^\infty \left\vert{S_n}\right\vert = \left\vert{\bigcup_{n \mathop = 1}^\infty S_n}\right\vert$ Suppose that the cardinality of at least one $S_i$ is infinite. Then the cardinality of: :$\displaystyle \bigcup_{n \mathop = 1}^\infty S_n$ is infinite by Subset of Finite Set is Finite. Hence: :$\displaystyle \mu \left({\bigcup_{n \mathop = 1}^\infty S_n}\right) = +\infty$ Now as: :$\mu \left( {S_i} \right) = +\infty$ it follows by the definition of extended real addition that: :$\displaystyle \sum_{n \mathop = 1}^\infty \mu \left({S_n}\right) = +\infty$ Suppose now that all $S_i$ are finite, and $\displaystyle \sum_{n \mathop = 1}^\infty \left\vert{S_n}\right\vert$ converges. Then by Convergent Series of Natural Numbers, it follows that for some $N \in \N$: :$\forall m \ge N: \left\vert{S_m}\right\vert = 0$ which by Cardinality of Empty Set means $S_m = \varnothing$. Therefore we conclude that: :$\displaystyle \sum_{n \mathop = 1}^\infty \left\vert{S_n}\right\vert = \sum_{n \mathop = 1}^N \left\vert{S_n}\right\vert$ and by Union with Empty Set: :$\displaystyle \bigcup_{n \mathop = 1}^\infty S_n = \bigcup_{n \mathop = 1}^N S_n$ By Cardinality of Set Union: Corollary: :$\displaystyle \left\vert{\bigcup_{n \mathop = 1}^N S_n}\right\vert = \sum_{n \mathop = 1}^N \left\vert{S_n}\right\vert$ and combining this with the above yields the desired identity. Suppose finally that $\displaystyle \sum_{n \mathop = 1}^\infty \left\vert{S_n}\right\vert$ diverges. To establish the desired identity, it is to be shown that $\displaystyle \bigcup_{n \mathop = 1}^\infty S_n$ is infinite. Suppose to the contrary that it has finite cardinality, say $k$. By Cardinality of Set Union: Corollary, for each $N \in \N$: :$\displaystyle \sum_{n \mathop = 1}^N \left\vert{S_n}\right\vert = \left\vert{\bigcup_{n \mathop = 1}^N S_n}\right\vert$ Now since: :$\displaystyle \bigcup_{n \mathop = 1}^N S_n \subseteq \bigcup_{n \mathop = 1}^\infty S_n$ it follows by Cardinality of Subset of Finite Set that, for each $N \in \N$: :$\displaystyle \left\vert{\bigcup_{n \mathop = 1}^N S_n}\right\vert \le k$ which contradicts the assumption that $\displaystyle \sum_{n \mathop = 1}^\infty \left\vert{S_n}\right\vert$ diverges. Therefore $\displaystyle \bigcup_{n \mathop = 1}^\infty S_n$ is infinite and the identity follows. {{qed|lemma}} === Proof of $(3')$ === By Cardinality of Empty Set: :$\left\vert{\varnothing}\right\vert = 0$ {{qed|lemma}} Having verified the axioms, it follows that $\left\vert{\cdot}\right\vert$ is a measure. {{qed}} Category:Measure Theory 34c73wo8vgoojdpu83oxz25nxmh5b6w
=== Necessary Condition === Suppose $V = \left\{{\mathbf 0}\right\}$. We have that $V$ has no $\mathbf v \in V, \mathbf v \ne \mathbf 0$. Thus there exists no $\left\{{\mathbf v}\right\} \subseteq V$ such that $\mathbf v \ne \mathbf 0$. Such a $\left\{{\mathbf v}\right\}$ would be a linearly independent set. Hence $\varnothing$ is the only possible basis for $V$. Hence $\dim \left({V}\right) = \left|{\varnothing}\right| = 0$. {{qed|lemma}} === Sufficient Condition === Suppose $\dim \left({V}\right) = 0$. Then by definition of dimension, $0 = \dim \left({V}\right) = \left\vert|{B}\right\vert|$, where $B$ is a basis for $V$. Thus $B = \varnothing$. Suppose there are sets $\left\{{\mathbf v}\right\}$ such that $\mathbf v \in V, \mathbf v \ne \mathbf 0$. By Singleton is Linearly Independent, any such $\left\{{\mathbf v}\right\}$ would be a linearly independent set. So $V$ has no such $\mathbf v \ne \mathbf 0$. Thus $V = \left\{{\mathbf 0}\right\}$. {{qed}} {{SUBPAGENAME}} {{SUBPAGENAME}} jvrj05y0ei1p0owjuzmpa1n0j08l0h8
=== Necessary Condition === Assume that $W$ is a $G$-submodule of $V$. Hence by definition $\phi_W: G \times W \to W$ is a linear action on $W$. Also by definition, $\phi_W \left({G, W}\right) = \phi \left({G, W}\right) \subseteq W$. {{qed|lemma}} === Sufficient Condition === Assume now that $\phi \left({G, W}\right) = \phi_W \left({G, W}\right) \subseteq W$. Then it is correct to define $\phi_W: G \times W \to W$; it is a well-defined mapping. We need to check if $\phi_W$ is a linear action on $W$: Assume $a,b \in W$ and $g \in G$; in particular, then, $a,b \in V$ and: {{begin-eqn}} {{eqn|l = \phi_W \left({g, a + b}\right)      |r = \phi \left({g, a + b}\right)|      |c = Definition of $\phi_W$ }} {{eqn|r = \phi \left({g, a}\right) + \phi \left({g, b}\right)      |c = $\phi$ is a linear action on $V$ }} {{eqn|r = \phi_W \left({g, a}\right) + \phi_W \left({g, b}\right)      |c = Definition of $\phi_W$ }} {{end-eqn}} Further, assume $\lambda \in k$ and $g\in G$, then: {{begin-eqn}} {{eqn|l = \phi_W \left({g, \lambda b}\right)      |r = \phi \left({g, \lambda b}\right)      |c = Definition of $\phi_W$ }} {{eqn|r = \lambda \phi \left({g, b}\right)      |c = $\phi$ is a linear action on $V$ }} {{eqn|r = \lambda \phi_W \left({g, b}\right)      |c = Definition of $\phi_W$ }} {{end-eqn}} Thus $W$ is a $G$-submodule of $V$. {{qed}} {{SUBPAGENAME}} kjp6yspncfxbw87ek5l5gkmlxk27le1
Let $N$ be any neighborhood of $\map {\paren {g \circ f} } x$. By the definition of continuity at a point: :there exists a neighborhood $L$ of $\map f x$ such that $g \sqbrk L \subseteq N$ and :there exists a neighborhood $M$ of $x$ such that $f \sqbrk M \subseteq L$. Thus $\paren {g \circ f} \sqbrk M \subseteq g \sqbrk L \subseteq N$, as desired. {{qed}} Category:Composite Mappings Category:Continuous Mappings 8lye5zl5n78g8xgbl2jcbv6wnj0x0dz
=== Sufficient Condition === Suppose that: :$U \in \tau_2 \implies f^{-1} \sqbrk U \in \tau_1$ Let $x \in S_1$. Let $N \subseteq S_2$ be a neighborhood of $\map f x$. By the definition of a neighborhood, there exists a $U \in \tau_2$ such that $U \subseteq N$. Now, $f^{-1} \sqbrk U$ is a neighborhood of $x$. Then: :$f \sqbrk {f^{-1} \sqbrk U} = U \subseteq N$ as desired. {{qed|lemma}} === Necessary Condition === Now, suppose that $f$ is continuous at every point in $S_1$. We wish to show that: :$U \in \tau_2 \implies f^{-1} \sqbrk U \in \tau_1$ So, let $U \in \tau_2$. Assume that $f^{-1} \sqbrk U$ is non-empty, otherwise $f^{-1} \sqbrk U = \O \in \tau_1$ by Empty Set is Element of Topology. Let $x \in f^{-1} \sqbrk U$. By the definition of continuity at a point, there exists a neighborhood $N$ of $x$ such that $f \sqbrk N \subseteq U$. By the definition of a neighborhood, there exists a $X \in \tau_1$ such that $x \in X \subseteq N$. By Image of Subset under Mapping is Subset of Image: :$f \sqbrk X \subseteq f \sqbrk N$ This gives: :$f \sqbrk X \subseteq f \sqbrk N \subseteq U$ Let $\CC = \set {X \in \tau_1: f \sqbrk X \subseteq U}$. Let $\ds H = \bigcup \CC$. From the above argument: :$f^{-1} \sqbrk U \subseteq H$ It follows directly from the definition of $H$ (and the definition of $\CC$) that: :$H \subseteq f^{-1} \sqbrk U$ Hence: :$H = f^{-1} \sqbrk U$. By definition, $H$ is the union of open sets (of $S_1$). Hence $H$ is open by the definition of a topology. {{qed}}
Let $\mathbf v, \mathbf w \in \map {\mathrm N} {\mathbf A}$. By the definition of null space: {{begin-eqn}} {{eqn | l = \mathbf A \mathbf v       | r = \mathbf 0 }} {{eqn | l = \mathbf A \mathbf w       | r = \mathbf 0 }} {{end-eqn}} Next, observe that: {{begin-eqn}} {{eqn | l = \mathbf A \paren {\mathbf v + \mathbf w}       | r = \mathbf A \mathbf v + \mathbf A \mathbf w       | c = Matrix Multiplication Distributes over Matrix Addition }} {{eqn | r = \mathbf 0 + \mathbf 0 }} {{eqn | r = \mathbf 0 }} {{end-eqn}} The order is correct, by hypothesis. Hence the result, by the definition of null space. {{qed}}