hcju/proofwikiqa huggingface dataset:
Checking in turn each of the criteria for equivalence: === Reflexivity === Trivially, any separation of $T$ is a partition of $S$. So any $a \in S$ will be in exactly one of the open sets forming the partition. So $a \sim a$ and $\sim$ is seen to be reflexive. {{qed|lemma}} === Symmetry === If $a \sim b$ then each separation of $T$ includes a single open set $U \ne \O$ which contains both $a$ and $b$. Trivially it follows that $b \sim a$ and so $\sim$ is seen to be symmetric. {{qed|lemma}} === Transitivity === Let $a \sim b$ and $b \sim c$. Then by definition: :Each separation of $T$ includes a single open set $U_1 \ne \O$ which contains both $a$ and $b$. :Each separation of $T$ includes a single open set $U_2 \ne \O$ which contains both $b$ and $c$. So $b \in U_1$ and $b \in U_2$ and so $U_1 \cap U_2 \ne \O$. So by the definition of partition it follows that $U_1 = U_2$ and so $a \sim c$. So $\sim$ has been shown to be transitive. {{qed|lemma}} $\sim$ has been shown to be reflexive, symmetric and transitive. Hence by definition it is an equivalence relation. {{qed}}
Let $C$ be the quasicomponent of $p$. Let $Q$ be the set of clopen sets containing $p$ that are not equal to $S$. First let $x \in C$. Let $U \in Q$. By definition of $Q$: :$p \in U$ Let $V = \relcomp S U$ be the complement of $U$ relative to $S$. By Complement of Clopen Set is Clopen $V$ is also a clopen set of $T$. By Clopen Set and Complement form Separation $U$ and $V$ separate $S$. By the definition of quasicomponent, $U \mid V$ is a separation of $S$ which contains a single open set $U$ containing both $x$ and $p$. Thus: :$x \in U$ As this holds for all such $U$: :$x \in \displaystyle \bigcap Q$ As this holds for all such $x$: :$C \subseteq Q$ {{qed|lemma}} Let $x \in \displaystyle \bigcap Q$. Let $U \mid V$ be a separation of $S$. Let $p \in U$. Then $U \in Q$. Thus $x \in U$. Thus $x$ lies on the same side as $p$ of any separation of $X$. So $x \in C$. {{qed}}
Let $x \in S$. By hypothesis, the quasicomponent of $x$ is $S$ itself. Thus by definition of quasicomponent: :$\forall y \in S: y \sim x$ where $\sim$ is the relation defined on $T$ as: :$x \sim y \iff T$ is connected between the two points $x$ and $y$ Let $K = \displaystyle \bigcap_{x \mathop \in U} U: U$ is clopen in $T$. By Quasicomponent is Intersection of Clopen Sets: :$\displaystyle \bigcap K = S$ Thus there is no non-empty clopen set of $T$ apart from $S$. The result follows by definition of connected space. {{qed}}
Define a relation $\sim$ on the Cartesian product $A \times S$ by: :$\tuple {a, s} \sim \tuple {b, t} \iff \exists u \in S: a t u = b s u$ === Lemma 1 === {{:Localization of Ring Exists/Lemma 1}}{{qed|lemma}} Let $A_S$ be used to denote the  $\paren {A \times S} / \sim$. Let $a / s$ be the equivalence class of $\tuple {a, s}$ in $\paren {A \times S} / \sim$. For $\dfrac a s, \dfrac b t \in A_S$, let the following be defined: :$\dfrac a s + \dfrac b t = \dfrac {a t + b s} {s t}$ :$\dfrac a s \cdot \dfrac b t = \dfrac {a b} {s t}$ === Lemma 2 === {{:Localization of Ring Exists/Lemma 2}}{{qed|lemma}} Now define $\iota: A \to A_S$ by: :$\map \iota a := \dfrac a 1$ It is to be shown that $\struct {A_S, \iota}$ satisfy the universal property for localization. {{MissingLinks|universal property for localization}} Let $B$ be a ring. Let $g: A \to B$ be a mapping such that: :$\map g S \subseteq B^\times$ {{explain|$B^\times$}} Suppose that $h: A_S \to B$ is a ring homomorphism with $h \circ \iota = g$. Then we must have: :$\map h {\dfrac a 1} = \map g a$ and: :$\map h {\dfrac 1 s} \cdot \map h {\dfrac s 1} = 1$ Therefore: :$\map h {\dfrac 1 s} \map g s = 1$ so: :$\map h {\dfrac 1 s} = \map g s^{-1}$ Therefore: :$\map h {\dfrac a s} = \map h {\dfrac a 1} \cdot \map h {\dfrac 1 s} = \map g a \map g s^{-1}$. So if such $h$ exists it must equal $\map g a \map g s^{-1}$, so is unique. {{MissingLinks|Links to results demonstrating the validity of the steps in the above.}} Therefore, to conclude the proof we pull out: === Lemma 3 === {{:Localization of Ring Exists/Lemma 3}}{{qed|lemma}} Hence the result. {{qed}}
First, a lemma: === Lemma === {{:Transitivity of Integrality/Lemma}}{{qed|lemma}} Now let $x \in C$. Thus $x$ is supposed integral over $B$. That is, we can find an expression: :$(1): \quad x^n + b_{n - 1} x^{n - 1} + \dotsb + b_1 x + b_0 = 0, \quad b_i \in B, \ i = 0, \dotsc, n - 1$ Let $D$ be the subring of $C$ generated by $A \cup \set {b_0, \dotsc, b_{n - 1} }$. By the lemma, $D$ is finitely generated over $A$. Moreover, $D \sqbrk x$ is finitely generated over $D$ because of the equation $(1)$. Therefore by Transitivity of Finite Generation: :$D \sqbrk x$ is a finitely generated $A$-module. Finally by Equivalent Definitions of Integral Dependence, $x$ is integral over $A$. {{qed}}
Suppose $x \in B$ is integral over $C$. Certainly $C$ is integral over $A$, so by Transitivity of Integrality, $C[x]$ is integral over $A$. In particular, $x$ is integral over $A$, so $x \in C$. {{Qed}} Category:Algebraic Number Theory Category:Commutative Algebra 1k8rp56mvkw2r0m8ykdktnr7yke9yul
We have by definition that: :$\forall x \in A: f_b \left({x}\right) = b$ So: $f^{-1} \left({b}\right) = A$ For $c \in B: c \ne B$ we have that: $f^{-1} \left({c}\right) = \varnothing$ Let $U \in \tau_B$ such that $b \in U$. Then $f^{-1} \left({U}\right) = A$ Let $V \in \tau_B$ such that $b \notin V$. Then $f^{-1} \left({V}\right) = \varnothing$ From the definition of topology, $A$ is open in $T$. From Empty Set is Element of Topology, $\varnothing$ is also open in $T$. So $f_b$ fulfils the conditions for it to be continuous. {{qed}}
Checking in turn each of the criteria for equivalence: === Reflexivity === From Point is Path-Connected to Itself, we have that $a \sim a$. So $\sim$ is reflexive. {{qed|lemma}} === Symmetry === If $a \sim b$ then $a$ is is path-connected to $b$ by definition. We form the mapping $g: \closedint 0 1 \to \closedint 0 1$: :$\map g x = 1 - x$ which is trivially continuous. By Composite of Continuous Mappings is Continuous $f \circ g$ is continuous. Putting it together we see that $f \circ g$ maps $0$ to $b$ and $1$ to $a$. So $b \sim a$ and $\sim$ has been shown to be symmetric. {{qed|lemma}} === Transitivity === Follows directly from Joining Paths makes Another Path. {{qed|lemma}} $\sim$ has been shown to be reflexive, symmetric and transitive. Hence by definition it is an equivalence relation. {{qed}}
Checking in turn each of the criteria for equivalence: === Reflexivity === The definition of arc-connected specifically makes provision that: :$\forall a \in S: a \sim a$ specifically to ensure that $\sim$ be reflexive. {{qed|lemma}} === Symmetry === If $x \sim y$ then $x$ is is arc-connected to $y$ by definition. In the trivial case of $x = y$, $y \sim x$ by reflexivity. Suppose that $x \ne y$. We form the mapping $g: \R \to \R$: :$\map g x = 1 - x$ which is trivially continuous and injective. By Composite of Continuous Mappings is Continuous, $g \circ f$ is continuous. By Composite of Injections is Injection, $g \circ f$ is also injective. Putting it together we see that $g \circ f$ is a continuous injection which maps $0$ to $y$ and $1$ to $x$. So $y \sim x$ and $\sim$ has been shown to be symmetric. {{qed|lemma}} === Transitivity === Follows directly from Joining Arcs makes Another Arc. {{qed|lemma}} $\sim$ has been shown to be reflexive, symmetric and transitive. Hence by definition it is an equivalence relation. {{qed}}
From Arc in Topological Space is Path, $f$ and $g$ are also paths in $T$. So by Joining Paths makes Another Path it follows that $h$ is a path in $T$. Now if $\operatorname{Im} \left({f}\right) \cap \operatorname{Im} \left({g}\right) = b$ it can be seen that: :$\displaystyle \forall x \in \operatorname{Im} \left({h}\right): x = \begin{cases} f \left({y}\right) & \text{for some } y \in \left[{0 \,.\,.\, \dfrac 1 2}\right], \text{ or} \\ g \left({z}\right) & \text{for some } z \in \left[{\dfrac 1 2 \,.\,.\, 1}\right] \\ \end{cases}$ and it follows that $h$ is an injection, and therefore an arc. {{mistake|that should be an XOR, really. And the intervals for $y$ and $z$ are simply $I$}} On the other hand, suppose: : $\exists y \in \left[{0 \,.\,.\, \dfrac 1 2}\right]: \exists z \in \left[{\dfrac 1 2 \,.\,.\, 1}\right]: f \left({y}\right) = g \left({z}\right)$ such that $f \left({y}\right) \ne b$. {{finish|It remains to show that you can built an arc out of the bits up till where the arcs cross. It's intuitively obvious but requires some analysis work.}} {{qed}} Category:Topology dd45te87wsxmhkn1yz0eacd2javv00r
By definition, an arc from $a$ to $b$ is a path $f: I \to T$ such that $f$ is injective. Hence the result, by definition. {{qed}} Category:Topology k0foahpy900i24trhw6en2y1gfckc4o
Let $f \in A$. By Arc in Topological Space is Path we have that $f \in P$. That is, $A \subseteq P$. {{qed|lemma}} ---- Let $f \in P$. From Path-Connected Space is Connected we have directly that $P \subseteq C$. {{qed|lemma}} ---- Let $f \in C$. From Connected Space is Connected Between Two Points we have directly that $C \subseteq Q$. {{qed|lemma}} Hence the result. {{qed}} {{finish|Plenty to do.}}
=== $(1)$ implies $(2)$ === Suppose $(1)$ holds. Pick $\varnothing \ne S \subseteq P$. Let $x_1 \in S$ be arbitrary. Given $x_k \in S$, pick $x_{k+1} \in S$ strictly bigger than $x_k$. By hypothesis the process must eventually terminate, say $x_n$ is the last element. Then by construction there are no larger elements than $x_n$, i.e. $x_n$ is a maximal element of $S$. {{qed|lemma}} {{improve|Last three sentences can be made more explicit and less handwaving}} === $(2)$ implies $(1)$ === Suppose $(2)$ holds. Let $\left({x_k}\right)_{k \in \N}$ be an increasing sequence of elements of $P$. By hypothesis, the set $\left\{{x_k: k \in \N}\right\}$ has a maximal element, say $x_n$. Then since $x_m \ge x_n$ for all $m \geq n$, we must have $x_n = x_{n+1} = \cdots$. {{qed}} Category:Increasing Mappings Category:Sequences 26tn5iv2h0yqpm30p23o2v3z3lvjqoz
=== Definition 2 iff Definition 3 === This follows by Increasing Sequence in Ordered Set Terminates iff Maximal Element. {{qed|lemma}} === Definition 2 implies Definition 1 === Assume there is an ideal $I$ which is not finitely generated. For any finite set $\left\{ {a_1, \dotsc, a_n}\right \}$ where $n \in \N$, the generated ideal is not equal to $I$. Consider the chain: : $\left\langle{a_1}\right\rangle \subset \left\langle{a_1, a_2}\right\rangle \subset \cdots$ This chain does not satisfy the ascending chain condition (note that $I$ has infinitely many elements by assumption).  {{qed|lemma}} === Definition 1 implies Definition 2 === Let there be a chain of ideals $I_1 \subset I_2 \subset \cdots$. Then $\displaystyle J = \bigcup_{n \mathop \ge 1} I_n$ is an ideal. Let $J$ be finitely generated, by $\left\{ {b_1, \dotsc b_m }\right\}$ for some $m \in \N$. As the chain is ascending, there exists an ideal such that: :$\left\{ {b_1, \dotsc b_m }\right\} \subset I_k$ for some $k \in \N$. It follows that: :$I_k = \left\langle{b_1, \dotsc, b_m}\right\rangle$ Hence: :$\forall l \ge k: I_l = I_k$ {{finish|definition 4}} {{ProofWanted|another proof using Definition:Noetherian Module and definition 4}}
=== $(1)$ iff $(2)$ === Let $T = \struct {S, \tau}$ be ultraconnected by Definition 1: :no two non-empty closed sets of $T$ are disjoint. Let $x, y \in S$. By Topological Closure is Closed, both $\set x^-$ and $\set y^-$ are closed. By hypothesis: :$\set x^- \cap \set y^- \ne \O$ That is, $T = \struct {S, \tau}$ is ultraconnected by Definition 2. {{qed|lemma}} Let $T = \struct {S, \tau}$ be ultraconnected by Definition 2: :$\forall x, y \in S: \set x^- \cap \set y^- \ne \O$ Let $V_1, V_2$ be closed sets of $T$. Let $x \in V_1, y \in V_2$. Then: :$\set x^- \cap \set y^- \ne \O$ But then from Topological Closure of Subset is Subset of Topological Closure we have that: :$\set x^- \subseteq V_1^-$ :$\set y^- \subseteq V_2^-$ But from Closed Set Equals its Closure $V_1^- = V_1, V_2^- = V_2$. So from Intersection is Subset: :$\set x^- \cap \set y^- \subseteq V_1$ :$\set x^- \cap \set y^- \subseteq V_2$ from which: :$V_1 \cap V_2 \ne \O$ As $V_1$ and $V_2$ are arbitrary, it follows that $T$ is ultraconnected by Definition 1. {{qed}} === $(1)$ iff $(3)$ === {{:Equivalence of Definitions of Ultraconnected Space/1 iff 3}}
Let $T = \struct {S, \tau}$ be ultraconnected. Let $a, b \in S$. Let $p \in \set a^- \cap \set b^-$ where $\set a^-$ is the closure of $\set a$. Such a $p$ can be chosen, as $T$ being ultraconnected guarantees that $\set a^- \cap \set b^- \ne \O$. Consider the mapping $f: \closedint 0 1 \to X$ such that: :$\map f x = \begin{cases} a & : x \in \hointr 0 {\dfrac 1 2} \\ p & : x = \dfrac 1 2 \\ b & : x \in \hointl {\dfrac 1 2} 1 \\ \end{cases}$ Then $f$ is continuous. {{explain|The above is a nontrivial claim}} The result follows from the definition of path-connectedness. {{qed}}
By definition of $T = \struct {S, \tau}$ being irreducible: :$\forall U_1, U_2 \in \tau: U_1, U_2 \ne \O\implies U_1 \cap U_2 \ne \O$ For every $x \in \R$, define: :$L_x := f^{-1} \sqbrk {\openint {-\infty} x}$ :$U_x := f^{-1} \sqbrk {\openint x \infty}$ By continuity of $f$, these are open in $T$. They are also disjoint, because for each $s \in S$, $\map f s < x$ and $\map f s > x$ cannot both hold simultaneously. Suppose now that $\map f s = a$ and $\map f {s'} = b$, for some $s, s' \in S$ and $a, b \in \R$ with $a < b$. Define $c = \dfrac {a + b} 2$. Then $a < c < b$, whence: :$s \in L_c$ and $s' \in U_c$ However, because $T$ is irreducible, it follows that either $L_c$ or $U_c$ must be empty. This contradicts the existence of such $a$ and $b$. Hence $f$ must be constant. {{qed}}
We have that Continuous Real-Valued Function on Irreducible Space is Constant. A constant mapping is trivially bounded. Hence the result by definition of pseudocompact. {{qed}}
The relevant justifications are listed as follows: :Ultraconnected Space is Path-Connected :Arc-Connected Space is Path-Connected :Path-Connected Space is Connected :Irreducible Space is Connected {{qed}}
Let $T = \struct {S, \tau}$ be arc-connected. Then $\forall x, y \in S$, there exists a continuous injection $f: \closedint 0 1 \to S$, such that $\map f 0 = x$ and $\map f 1 = y$. As $f$ is a continuous injection, it is also simply a continuous mapping. The result follows from the definition of path-connectedness. {{qed}}
Let $T = \struct {S, \tau}$ be irreducible. Then: :$\forall U_1, U_2 \in \tau: U_1, U_2 \ne \O \implies U_1 \cap U_2 \ne \O$ So trivially there are no two open sets that can form a separation of $T$. The result follows from definition of connected. {{qed}}
$T = \struct {S, \tau}$ be ultraconnected. Thus by definition: :$(1): \quad \forall x, y \in S: \set x^- \cap \set y^- \ne \O$ Let $a, b \in S$ such that $a \ne b$. {{AimForCont}} $T$ is a $T_1$ (Fréchet) space. By definition of $T_1$ Space, $\set a$ and $\set b$ are closed. From Closed Set Equals its Closure we have that $\set a^- = \set a$ and $\set b^- = \set b$. It immediately follows that: :$\set a^- \cap \set b^- = \O$ But that contradicts $(1)$ above.  The result follows by Proof by Contradiction. {{qed}}
Recall the definition of a $T_4$ space: $T = \struct {S, \tau}$ is a $T_4$ space {{iff}}: :for any two disjoint closed sets $A, B \subseteq S$, there exist disjoint open sets $U, V \in \tau$ containing $A$ and $B$ respectively. As no two closed sets of an ultraconnected space are actually disjoint, it follows that $T_4$-ness follows vacuously. {{qed}}
{{begin-eqn}} {{eqn | l = \mathbf u \cdot \mathbf u       | r = \left\Vert{\mathbf u}\right\Vert \left\Vert{\mathbf u}\right\Vert \cos \angle \mathbf u, \mathbf u       | c = Cosine Formula for Dot Product }} {{eqn | r = \left\Vert{\mathbf u}\right\Vert^2 \cos 0       | c = since the angle between a vector and itself is $0$ }} {{eqn | r = \left\Vert{\mathbf u}\right\Vert^2       | c = Cosine of Zero is One }} {{end-eqn}} {{qed}} === Note === Because this theorem is used to prove the general ($n$-dimensional) case of Cosine Formula for Dot Product, this proof is circular the way we have defined the dot product. However, some texts use the cosine formula as the definition of the dot product and derive the sum of products form as a consequence. The two definitions are equivalent, so we have included this proof to show how the statement would be proved from the cosine definition.
Let $T = \struct {S, \tau}$ be irreducible. Then: :$\forall U_1, U_2 \in \tau: U_1, U_2 \ne \O \implies U_1 \cap U_2 \ne \O$ So trivially there are no two open sets that can form a separation of $T$. As a basis consists of open sets, this applies to all sets in a basis for $T$. The result follows from definition of locally connected. {{qed}}
Let $T = \struct {S, \tau}$ be arc-connected. Then $T$ has a basis consisting entirely of arc-connected sets. From Arc-Connected Space is Path-Connected, this basis consisting entirely of path-connected sets. The result follows from definition of locally path-connected. {{qed}}
Let $x \in S$ be any point of $T$. Let $\BB$ be a local basis of path-connected sets for $x$. From Path-Connected Space is Connected, $\BB$ is a local basis of connected sets. Thus, $T$ is locally connected by definition. {{qed}}
Let $m \in M$ be arbitrary. By definition there is an open neighborhood $U$ of $m$, homeomorphic to an open subset of $\R^d$. By the definition of an open set, there is some open ball: :$B = B_\delta \left({\phi \left({m}\right)}\right) = \left\{{x \in \R^d: \left|{x - \phi \left({m}\right)}\right| < \delta}\right\}$ of radius $\delta$ containing $\phi \left({m}\right)$, contained in $U$. By Closure of Open Ball in Metric Space and Topological Closure is Closed, the set: :$C = \left\{{x \in \R^d: \left|{x - \phi \left({m}\right)}\right| \le \dfrac \delta 2}\right\}$ is closed, and $C \subseteq B \subseteq U$. Moreover, $C$ is trivially bounded, hence compact by the Heine-Borel Theorem. Now if $\phi$ is a homeomorphism $U \to \R^d$, then by definition $\phi^{-1}$ is continuous. Therefore by Continuous Image of Compact Space is Compact, $\phi^{-1} \left({C}\right) \subseteq M$ is compact. Furthermore $m \in \phi^{-1} \left({C}\right)$ because $\phi \left({m}\right) \in C$. Thus every point of $M$ has a compact neighborhood. {{Qed}} {{explain|Clarification is needed as to why this result should be categorised in :Category:Manifolds.}} Category:Manifolds atj03zp0r9bpm24yi9z708tx91dy20y
Let $x \divides y$ denote $x$ divides $y$. Let $x, y \in A$, with complete factorizations: :$x = u x_1 \cdots x_r$ :$y = v y_1 \cdots y_s$ where: :$u, v$ are units :the $x_i$, $y_i$ irreducible. We arrange the complete factorizations as follows: :$x = u \paren {x_1 \cdots x_t} x_{t + 1} \cdots x_r$ :$y = v \paren {y_1 \cdots y_t} y_{t + 1} \cdots y_s$ where: :$t \le \min \set {r, s}$ :For $i = 1, \ldots, t$, $x_i$ and $y_i$ are associates :For any $i \in \set {t + 1, \ldots, r}$, $j \in \set {t + 1, \ldots, s}$, $x_i$ and $y_j$ are not associates. Let $d = x_1 \cdots x_t$ (recall that the empty product is $1$, i.e. $d = 1$ when $t = 0$). We claim that $d$ is a greatest common divisor for $x$ and $y$. Certainly $d \divides x$ and $d \divides y$. So, let $f$ be another common divisor of $x$ and $y$. We can find $w, z \in A$ such that $x = f w$, and $y = f z$. If $f$ is a unit, then $f \divides d$ by definition. {{AimForCont}} $f \nmid d$. Then the complete factorization of $f$ must contain an irreducible element that does not divide $d$. Call this irreducible element $g$. We have that: :$g$ must divide some $x_j$ where $j > t$ and :$g$ must divide some $y_k$ where $k > t$. Either: :$g$ is a unit, contradicting its irreducibility or: :$x_j$ and $y_k$ are not irreducible, which is a contradiction also. Hence by Proof by Contradiction: :$f \divides d$ and so $x$ and $y$ have a greatest common divisor. {{qed}} Category:Unique Factorization Domains Category:Factorization k6llamsyc7acqtb6vxxrbunq6tnjqvp
Let $K$ be the field of quotients of $A$. Let $x \in K$ be integral over $A$. Let: $x = a / b$ for $a, b \in A$ with $\gcd \set {a, b} \in A^\times$. This makes sense because a UFD is GCD Domain. There is an equation: :$\paren {\dfrac a b}^n + a_{n - 1} \paren {\dfrac a b}^{n - 1} + \dotsb + a_0$ {{explain|This is not actually an equation, it has no equals sign in it.}} with $a_i \in A$, $i = 0, \dotsc, n - 1$. Multiplying by $b^n$, we obtain: :$a^n + b c = 0$ with $c \in A$. Therefore: :$b \divides a^n$ {{AimForCont}} $b$ is not a unit. Then: :$\gcd \set {a, b} \notin A^\times$ which is a contradiction. So $b$ is a unit, and: :$a b^{-1} \in A$ {{explain|Make it clear why this fulfils the conditions for the statement of the result}} {{qed}} Category:Algebraic Number Theory Category:Factorization Category:Unique Factorization Domains hc338lvrefaby5nmas6fbieb7ik5n10
Let $T = \struct {S, \tau}$ be a totally pathwise disconnected space by definition 1. That is: :all path components of $T$ are singletons. By definition of path component, this means there are no two points $x, y \in S$ such that there exists a path between $x$ and $y$. By definition of path, this means the only paths in $T$ are from a single point to that single point. That is, that the paths in $T$ are constant mappings. By definition of path: :the only continuous mappings from the closed unit interval $\closedint 0 1$ to $T$ are constant mappings. That is the definition of a totally pathwise disconnected space by definition 2. {{qed|lemma}} Let $T = \struct {S, \tau}$ be a totally pathwise disconnected space by definition 2. That is: :the only continuous mappings from the closed unit interval $\closedint 0 1$ to $T$ are constant mappings. That is, that the paths in $T$ are constant mappings. That is, the path components are singletons. That is the definition of a totally pathwise disconnected space by definition 1. {{qed}}
Let $T = \struct {S, \tau}$ be totally disconnected. Then as its components are singletons, it follows that each of its points is closed. From Equivalence of Definitions of $T_1$ Space, it follows that $T$ is a $T_1$ (Fréchet) space. {{qed}}
Let $A_\mathcal B$ and $A_\mathcal C$ be the matrices of $A$ relative to $\mathcal B$ and $\mathcal C$ respectively. Let $\det$ also denote the determinant of a matrix. We are required to show that $\det A_\mathcal B = \det A_\mathcal C$. Let $P$ be the change of basis matrix from $\mathcal B$ to $\mathcal C$. By Change of Coordinate Vectors Under Linear Mapping and since $A_\mathcal B$ and $A_\mathcal C$ represent the same linear operator with respect to different bases, the following diagram commutes: :250px where $u \in V$, and $\left[{u}\right]_\mathcal B$ indicates the coordinate vector of $u$ with repect to $\mathcal B$, and similarly for $\left[{u}\right]_\mathcal C$. That is, $P A_\mathcal B = A_\mathcal C P$. Therefore, because a Change of Basis is Invertible we have: : $A_\mathcal B = P^{-1} A_\mathcal C P$ So: {{begin-eqn}} {{eqn | l = \det \left({A_\mathcal B}\right)       | r = \det \left({P^{-1} A_\mathcal C P}\right) }} {{eqn | r = \det \left({P^{-1} }\right) \det \left({A_\mathcal C}\right) \det \left({P}\right)       | c = Determinant of Matrix Product }} {{eqn | r = \det \left({P}\right)^{-1} \det \left({A_\mathcal C}\right) \det \left({P}\right)       | c = Determinant of Inverse Matrix }} {{eqn | r = \det \left({A_\mathcal C}\right) }} {{end-eqn}} Hence the result. {{qed}} Category:Linear Operators Category:Determinants k7z62jj2gyon3luhlhzgyqlbikf8elv