hcju/proofwikiqa huggingface dataset:
=== $\epsilon$-$\delta$ Definition implies $\epsilon$-Ball Definition === Suppose that $f$ is $\tuple {d_1, d_2}$-continuous at $a$ in the sense that: :$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in A_1: \map {d_1} {x, a} < \delta \implies \map {d_2} {\map f x, \map f a} < \epsilon$ where $\R_{>0}$ denotes the set of all strictly positive real numbers. Then: {{begin-eqn}} {{eqn | ll= \text{Let}       | l = \epsilon       | o = \in       | r = \R_{>0} }} {{eqn | ll= \leadsto       | l = \exists \delta \in \R_{>0} : \forall x \in A_1 : \map {d_1} {x, a} < \delta       | o = \implies       | r = \map {d_2} {\map f x, \map f a} < \epsilon       | c = by hypothesis }} {{eqn | ll = \text{Let}       | l = y       | o = \in       | r = f \sqbrk {\map {B_\delta} {a; d_1} }       | c = }} {{eqn | ll= \leadsto       | l = \exists x \in \map {B_\delta} {a; d_1}: y       | r = \map f x       | c = {{Defof|Image of Subset under Mapping|index = 1}} }} {{eqn | ll= \leadsto       | l = \map {d_1} {x, a}       | o = <       | r = \delta       | c = {{Defof|Open Ball}} }} {{eqn | ll= \leadsto       | l = \map {d_2} {\map f x, \map f a}       | o = <       | r = \epsilon       | c = by hypothesis }} {{eqn | ll= \leadsto       | l = \map {d_2} {y, \map f a}       | o = <       | r = \epsilon       | c = as $y = \map f x$ }} {{eqn | ll= \leadsto       | l = y       | o = \in       | r = \map {B_\epsilon} {\map f a; d_2}       | c = {{Defof|Open Ball}} }} {{eqn | ll= \leadsto       | l = f \sqbrk {\map {B_\delta} {a; d_1} }       | o = \subseteq       | r = \map {B_\epsilon} {\map f a; d_2}       | c = by hypothesis }} {{end-eqn}} {{qed|lemma}} === $\epsilon$-Ball Definition implies $\epsilon$-$\delta$ Definition=== Suppose that $f$ is $\tuple {d_1, d_2}$-continuous at $a$ in the sense that: :$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \map f {\map {B_\delta} {a; d_1} } \subseteq \map {B_\epsilon} {\map f a; d_2}$ where $\map {B_\epsilon} {\map f a; d_2}$ denotes the open $\epsilon$-ball of $\map f a$ with respect to the metric $d_2$, and similarly for $\map {B_\delta} {a; d_1}$. Then: {{begin-eqn}} {{eqn | ll= \text{Let}       | l = \epsilon       | o = \in       | r = \R_{>0} }} {{eqn | ll= \leadsto       | l = \exists \delta \in \R_{>0} : \map f {\map {B_\delta} {a; d_1} }       | o = \subseteq       | r = \map {B_\epsilon} {\map f a; d_2}       | c = by hypothesis }} {{eqn | ll= \text{Let}       | l = x \in A_1       | o = \text{ such that }       | r = \map {d_1} {x, a} < \delta       | c = (*) }} {{eqn | ll= \leadsto       | l = x       | o = \in       | r = \map {B_\delta} {a; d_1}       | c = {{Defof|Open Ball|Open $\epsilon$-Ball}} }} {{eqn | ll= \leadsto       | l = \map f x       | o = \in       | r = \map {B_\epsilon} {\map f a; d_2}       | c = by hypothesis }} {{eqn | ll= \leadsto       | l = \map {d_2} {\map f x, \map f a}       | o = <       | r = \epsilon       | c = {{Defof|Open Ball|Open $\epsilon$-Ball}} }} {{end-eqn}} {{qed}} (*) Only consider this case, since otherwise $\map {d_1} {x, a} < \delta \implies \map {d_2} {\map f x, \map f a} < \epsilon$ is vacuously true.
Let $\epsilon \in \R: \epsilon > 0$. Let $\delta = 1$. Let $x \in \R$. Let $y \in \R$ such that $d_2 \left({x, y}\right) < \delta$. That is, $d_2 \left({x, y}\right) < 1$. By the definition of the standard discrete metric on $\R$, that would mean that $d_2 \left({x, y}\right) = 0$ and so $x = y$. Thus $f \left({x}\right) = f \left({y}\right)$. By definition of a metric, that means: : $d_1 \left({f \left({x}\right), f \left({y}\right)}\right) = 0 < \epsilon$ Thus the conditions for $\left({d_2, d_1}\right)$-continuity at a point are fulfilled. This is true for all $x \in \R$. So by definition $f$ is $\left({d_2, d_1}\right)$-continuous on $\R$. {{qed}}
* $(1): \quad p - p = 0$: We have: {{begin-eqn}} {{eqn | l = \left({p - p}\right) + \left({q - p}\right)       | r = \left({p + \left({q - p}\right)}\right) - p }} {{eqn | r = q - p }} {{end-eqn}} From Zero Element is Unique: :$p - p = 0$ * $(2): \quad p + 0 = p$ Using $(1)$ we see that: {{begin-eqn}} {{eqn | l = p + 0       | r = p + \left({p - p}\right) }} {{eqn | r = p }} {{end-eqn}} * $(3): \quad p + u = p + v \iff u = v$ Let $u = v$. By definition a mapping has a unique image point on a given element. It follows that: :$p + u = p + v$ Let $p + u = p + v$. We have: {{begin-eqn}} {{eqn | l = p + u       | r = p + v }} {{eqn | ll= \implies       | l = \left({p + u}\right) - p       | r = \left({p + v}\right) - p }} {{eqn | ll= \implies       | l = \left({p - p}\right) + u       | r = \left({p - p}\right) + v }} {{eqn | ll= \implies       | l = u       | r = v       | c = by $(1)$ }} {{end-eqn}} * $(4): \quad q - p = r - p \iff q = r$: Let $q = r$. By definition a mapping has a unique image point on a given element. It follows that: :$q - p = r - p$ Let $q - p = r - p \in V$. Then {{begin-eqn}} {{eqn | l = q - p       | r = r - p }} {{eqn | ll= \implies       | l = p + \left({q - p}\right)       | r = p + \left({r - p}\right) }} {{eqn | ll= \implies       | l = q       | r = r       | c = By the hypothesis $q - p = r - p$ }} {{end-eqn}} {{Qed}} Category:Affine Geometry kobid36sl87s3p0qgkw4z9o710zoxw4
=== Necessary Condition === Suppose that $g: S \to \N$ is a right inverse of $f$. That is, let $f \circ g = I_S$, the identity mapping on $S$. We have that $I_S$ is a surjection. By Surjection if Composite is Surjection, it follows that $f$ is a surjection. {{qed|lemma}} === Sufficient Condition === Now, suppose that $f: \N \to S$ is a surjection. By the definition of a surjection, it follows that: :$\forall x \in S: f^{-1} \left({x}\right)$ is non-empty where $f^{-1} \left({x}\right)$ denotes the preimage of $x$ under $f$. From the Well-Ordering Principle, $\N$ is well-ordered. Hence, we can define the mapping $g: S \to \N$ as: :$\forall x \in S: g \left({x}\right) = \min \left( f^{-1} \left({x}\right) \right)$ By the definition of a smallest element, it follows that: :$\forall x \in S: g \left({x}\right) \in f^{-1} \left({x}\right)$ That is: :$\forall x \in S: f \left({g \left({x}\right)}\right) = \left({f \circ g}\right) \left({x}\right) = x$ In other words, $g$ is a right inverse of $f$. {{qed}}
For brevity, let us write $\displaystyle \sum$ for $\displaystyle \sum_{i \mathop = 1}^n$. Then: {{begin-eqn}} {{eqn | l = \sum \left({x_i - \overline x}\right)^2       | r = \sum \left({x_i - \overline x}\right)\left({x_i - \overline x}\right) }} {{eqn | r = \sum x_i\left({x_i - \overline x}\right) - \overline x \sum \left({x_i - \overline x}\right)       | c = Summation is Linear }} {{eqn | r = \sum x_i\left({x_i - \overline x}\right) - 0       | c = Sum of Deviations from Mean }} {{eqn | r = \sum x_i\left({x_i - \overline x}\right) + 0 }} {{eqn | r = \sum x_i\left({x_i - \overline x}\right) + \overline x \sum \left({x_i - \overline x}\right)       | c = Sum of Deviations from Mean }} {{eqn | r = \sum \left({x_i + \overline x}\right)\left({x_i - \overline x}\right)       | c = Summation is Linear }} {{eqn | r = \sum \left({x_i^2 - \overline x^2}\right) }} {{end-eqn}} {{qed}}
In this context, $x_1, x_2, \ldots, x_n$ are instances of a discrete random variable. Hence the result Variance as Expectation of Square minus Square of Expectation can be applied: :$\var X = E \paren {X^2} - \paren {E \paren X}^2$ which means the same as this but in the language of probability theory. {{qed}}
=== Necessary Condition === Suppose that $f: \N \to S$ is a surjection. By Surjection from Natural Numbers iff Right Inverse, $f$ admits a right inverse $g: S \to \N$. We have that $g$ is an injection by Right Inverse Mapping is Injection. Hence the result, by the definition of a countable set. {{qed|lemma}} === Sufficient Condition === Suppose that $S$ is countable and non-empty. Then there is an injection $g: S \to \N$. By Injection has Surjective Left Inverse Mapping, there is a surjection $f: \N \to S$. {{qed}} {{LEM|Injection has Surjective Left Inverse Mapping}}
=== Necessary Condition === Follows directly from the definitions of the generated topology and an analytic sub-basis. {{qed|lemma}} === Sufficient Condition === Suppose that $\mathcal S$ is an analytic sub-basis for $\tau$. Let $\tau'$ be the topology on $X$ generated by the synthetic sub-basis $\mathcal S$. By definition $1$ of the generated topology and the definition of an analytic sub-basis, we have $\tau \subseteq \tau'$. By definition $2$ of the generated topology, it follows that $\tau' \subseteq \tau$. By definition of set equality: : $\tau = \tau'$ {{qed}} Category:Topological Bases hv5mlxe6etmjncam5p72a5gh3a78g00
Define: :$\displaystyle \BB = \left\{{\bigcap \AA: \AA \subseteq \SS, \AA \text{ is finite}}\right\} \subseteq \powerset {X_2}$ Let $B \in \BB$. Then there exists a finite subset $\AA \subseteq \SS$ such that: :$\displaystyle B = \bigcap \AA$ Hence: {{begin-eqn}} {{eqn | l = f^{-1} \sqbrk B       | r = f^{-1} \sqbrk {\bigcap \AA} }} {{eqn | r = \bigcap_{S \mathop \in \AA} f^{-1} \sqbrk S       | c = Preimage of Intersection under Mapping: General Result }} {{eqn | o = \in       | r = \tau_1       | c = General Intersection Property of Topological Space }} {{end-eqn}} Define: :$\displaystyle \tau = \set {\bigcup \AA: \AA \subseteq \BB} \subseteq \powerset {X_2}$ By the definition of an analytic sub-basis, we have $\tau_2 \subseteq \tau$. Let $U \in \tau_2$. Then $U \in \tau$; therefore: :$\displaystyle \exists \AA \subseteq \BB: U = \bigcup \AA$ Hence: {{begin-eqn}} {{eqn | l = f^{-1} \sqbrk U       | r = f^{-1} \sqbrk {\bigcup \AA} }} {{eqn | r = \bigcup_{S \mathop \in \AA} f^{-1} \sqbrk S       | c = Preimage of Union under Mapping: General Result }} {{eqn | o = \in       | r = \tau_1       | c = because $\forall B \in \BB: f^{-1} \sqbrk B \in \tau_1$, and by open set axiom $({1})$ applied to $\tau_1$ }} {{end-eqn}} That is, $f$ is continuous. {{qed}}
Let $\tau$ be the final topology on $X_2$ with respect to $f$. By hypothesis, $\SS \subseteq \tau$. By Synthetic Sub-Basis and Analytic Sub-Basis are Compatible, we have that $\tau_2$ is the topology generated by the synthetic sub-basis $\SS$. By the definition of the generated topology, we have $\tau_2 \subseteq \tau$. By the definition of the final topology, it follows that $f$ is continuous. {{qed}}
Recall the definition: :$\displaystyle \map {L_j} X = \prod_{\substack {0 \mathop \le i \mathop \le n \\ i \mathop \ne j}} \frac {X - x_i} {x_j - x_i} \in \R \sqbrk X$ {{mistake|Not sure if it's a mistake or a different way of defining it, but {{BookReference|Dictionary of Mathematics|1989|Ephraim J. Borowski|author2 = Jonathan M. Borwein|entry = Lagrange interpolation formula}} has the above wrapped up in another product symbol}} From this we see that: :$\map {L_j} {x_i} = \delta_{i j}$ Therefore: :$\displaystyle \map P{x_i} = \sum_{j \mathop = 0}^n a_i \delta_{i j} = a_i$ Moreover, by Degree of Product of Polynomials over Integral Domain and Degree of Sum of Polynomials, the degree of $P$ as defined above is at most $n$. It remains to show that the choice of $P$ is unique. {{AimForCont}} $\tilde P$ is another polynomial with the required properties. Let $\Delta = P - \tilde P$. By Degree of Sum of Polynomials, the degree of $\Delta$ is at most $n$. Now we see that for $i = 0, \ldots, n$: :$\map \Delta {x_i} = \map P {x_i} - \map {\tilde P} {x_i} = a_i - a_i = 0$ Since by hypothesis the $x_i$ are distinct, $\Delta$ has $n + 1$ distinct zeros in $\R$. But by the corollary to the Polynomial Factor Theorem this shows that: :$\displaystyle \map \Delta X = \alpha \prod_{i \mathop = 0}^n \paren {X - x_i}$ If $\alpha \ne 0$, then this shows that the degree of $\Delta$ is $n+1$, a contradiction. Therefore: :$\Delta = 0$ and so: :$P = \tilde P$ This establishes uniqueness. {{Qed}} {{Namedfor|Joseph Louis Lagrange|cat = Lagrange}}
We have: {{begin-eqn}} {{eqn | l = 2 \sin \theta       | r = \frac {e^{i \theta} - e^{-i \theta} } i       | c = Sine Exponential Formulation }} {{eqn | r = \frac {e^{i \theta} e^{-i \theta} - e^{-i \theta} e^{-i \theta} } {i e^{-i \theta} }       | c =  }} {{eqn | r = \left({1 - e^{-2 i \theta} }\right) \left({-i e^{i \theta} }\right)       | c =  }} {{eqn | r = \left({1 - e^{-2 i \theta} }\right) \left({e^{i \theta} e^{-i \pi / 2} }\right)       | c = Euler's Formula/Examples/e^-i pi by 2 }} {{eqn | r = \left({1 - e^{-2 i \theta} }\right) \left({e^{i \theta - i \pi / 2} }\right)       | c =  }} {{eqn | ll= \leadsto       | l = \prod_{k \mathop = 0}^{n - 1} \left({2 \sin \pi \left({x + \frac k n}\right)}\right)       | r = \prod_{k \mathop = 0}^{n - 1} \left({1 - e^{-2 i \pi \left({x + k / n}\right)} }\right) \left({e^{i \pi \left({x - \left({1 / 2}\right) + \left({k / n}\right)}\right)} }\right)       | c =  }} {{eqn | r = \left({1 - e^{-2 i \pi n x} }\right) \left({e^{i \pi \left({n x - 1 / 2}\right)} }\right)       | c =  }} {{eqn | r = 2 \sin \pi n x       | c =  }} {{end-eqn}} {{finish|Work to be done to establish the identities used here}} Category:Sine Function q4axqqt74dpubye91mw95qozyaxc7st
Define: :$\forall i \in I: \vartheta_i = \set {U \subseteq X: \map {f_i^{-1} } U \in \tau_i} \subseteq \powerset X$ Then, by the definition of intersection: :$\ds \tau = \bigcap_{i \mathop \in I} \vartheta_i$ From the Intersection of Topologies is Topology, it suffices to show, for all $i \in I$, that $\vartheta_i$ is a topology on $X$. We now verify the axioms for $\vartheta_i$ to be a topology on $X$. === $(\text O 1)$: Union of Open Sets === Let $\AA \subseteq \vartheta_i$. Then, by Preimage of Union under Mapping: General Result and by the definition of a topology, it follows that: :$\ds f_i^{-1} \sqbrk {\bigcup \AA} = \bigcup_{U \mathop \in \AA} f_i^{-1} \sqbrk U \in \tau_i$ That is, $\ds \bigcup \AA \in \vartheta_i$. {{qed|lemma}} === $(\text O 2)$: Pairwise Intersection of Open Sets === Let $U, V \in \vartheta_i$. Then, by Preimage of Intersection under Mapping and by the definition of a topology, it follows that: :$f_i^{-1} \sqbrk {U \cap V} = f_i^{-1} \sqbrk U \cap f_i^{-1} \sqbrk V \in \tau_i$ That is, $U \cap V \in \vartheta_i$. {{qed|lemma}} === $(\text O 3)$: Set Itself === By the definition of a topology, it follows that: :$f_i^{-1} \sqbrk X = Y_i \in \tau_i$ That is, $X \in \vartheta_i$. {{qed}} Category:Final Topology m3sq09107jj4b9qt4ld7f4ldgjoh652
By Initial Topology with respect to Mapping equals Set of Preimages, it follows that: :$\tau_Y = \set {\iota^{-1} \sqbrk U: U \in \tau}$ From Preimage of Subset under Inclusion Mapping, we have: :$\forall S \subseteq X: \iota^{-1} \sqbrk S = S \cap Y$ Hence the result. {{qed}} Category:Topological Subspaces Category:Inclusion Mappings Category:Initial Topology nsroz5lsoiw4pheya2emqo1xhxt6rip
Follows directly from Subspace Topology is Initial Topology with respect to Inclusion Mapping. {{qed}}
Define: :$\tau = \set {f^{-1} \sqbrk U: U \in \tau_Y}$ By definition, $\tau_X$ is the topology generated by $\tau$. Therefore: :$\tau \subseteq \tau_X$ If $\tau$ is a topology on $X$, then it follows from the definition of the generated topology that: :$\tau_X \subseteq \tau$ By definition of set equality: :$\tau_X = \tau$ Hence, it suffices to prove that $\tau$ is a topology on $X$. We now verify the open set axioms for $\tau$ to be a topology on $X$. === $({\text O 1})$: Union of Open Sets === Let $\AA \subseteq \tau$. It is to be shown that: :$\displaystyle \bigcup \AA \in \tau$ Define: :$\displaystyle \AA' = \set {V \in \tau_Y: f^{-1} \sqbrk V \subseteq \bigcup \AA} \subseteq \tau_Y$ Let: :$\displaystyle U = \bigcup \AA'$ By the definition of a topology, we have $U \in \tau_Y$. By Preimage of Union under Mapping: General Result and Union is Smallest Superset: Family of Sets: :$\displaystyle f^{-1} \sqbrk U = \bigcup_{V \mathop \in \AA'} f^{-1} \sqbrk V \subseteq \bigcup \AA$ By the definition of $\tau$ and by Set is Subset of Union: General Result, we have: :$\displaystyle \forall S \in \AA: \exists V \in \tau_Y: S = f^{-1} \sqbrk V \subseteq \bigcup \AA$ That is: :$\forall S \in \AA: \exists V \in \AA': S = f^{-1} \sqbrk V$ By Set is Subset of Union: General Result, we have: :$\forall V \in \AA': V \subseteq U$ By Image of Subset under Relation is Subset of Image: Corollary 3, it follows that: :$\forall S \in \AA: S \subseteq f^{-1} \sqbrk U$ By Union is Smallest Superset: General Result, we conclude that: :$\displaystyle \bigcup \AA \subseteq f^{-1} \sqbrk U$ Hence, by definition of set equality: :$\displaystyle \bigcup \AA = f^{-1} \sqbrk U \in \tau$ {{qed|lemma}} === $({\text O 2})$: Pairwise Intersection of Open Sets === Let $A, B \in \tau$. Let $U, V \in \tau_Y$ be such that $A = f^{-1} \sqbrk U$ and $B = f^{-1} \sqbrk V$. By the definition of a topology, we have $U \cap V \in \tau_Y$. Then, by Preimage of Intersection under Mapping, $A \cap B = f^{-1} \sqbrk {U \cap V} \in \tau$. {{qed|lemma}} === $({\text O 3})$: Set Itself === By the definition of a topology, we have $Y \in \tau_Y$. Hence, by Preimage of Mapping equals Domain, it follows that $X = f^{-1} \sqbrk Y \in \tau$. {{qed}} Category:Topology ol4uqec4eg7sdbsk7loqwssuqejdv5r
We know from the Residue Theorem: {{begin-eqn}} {{eqn | l = \oint_C f \left({z}\right) \, \mathrm d z       | r = 2 \pi i \, \sum \limits_{z_k} R_k(z_k)       | c =  }} {{eqn | r = 2 \pi i \,\sum_{z_k} \lim_{z \to z_k} \left({\left({z - z_k}\right) \frac {f \left({z}\right)} {\sin \pi z} }\right)       | c =  }} {{end-eqn}} This is for poles $z_k$ at order $N = 1$ because we say that simple poles exist for $\dfrac {f \left({z}\right)} {\sin \pi z}$. {{explain|Rewrite the above comprehensibly}} Using l'Hôpital's rule: {{begin-eqn}} {{eqn | l = \oint_C f \left({z}\right) \, \mathrm d z       | r = 2 \pi i \sum_{z_k} \lim_{z \to z_k} \left({\frac {\partial_z \left({z - z_k}\right) f \left({z}\right)} {\partial_z \sin \pi z} }\right)       | c =  }} {{eqn | r = 2 \pi i \sum_{z_k} \lim_{z \to z_k} \left({\frac {f \left({z}\right) + \left({z - z_k}\right) f' \left({z}\right)}{\pi \cos \pi z} }\right)       | c =  }} {{end-eqn}} {{explain|$\partial_z$}} But $\sin \pi z$ has poles at $z_k = n$ for some $n \in \Z$ which implies: {{begin-eqn}} {{eqn | l = \oint_C f \left({z}\right) \, \mathrm d z       | r = 2 \pi i \sum_{n \mathop = -\infty}^\infty \lim_{z \mathop \to n} \left({\frac {f \left({z}\right) + \left({z - n}\right) f' \left({z}\right)} {\pi \cos \pi z} }\right)       | c =  }} {{eqn | r = 2 i \sum_{n \mathop = -\infty}^\infty \left({\frac {f \left({n}\right)} {\cos \pi n} }\right)       | c =  }} {{end-eqn}} Finally: :$\dfrac 1 {\cos \pi n} = \cos \pi n = \left({-1}\right)^n$ Therefore: :$\displaystyle \frac 1 {2 i} \oint_C f \left({z}\right) \, \mathrm d z = \sum_{n \mathop = -\infty}^\infty \left({-1}\right)^n f \left({n}\right)$ {{qed}} {{namedfor|Arnold Johannes Wilhelm Sommerfeld|name2 = George Neville Watson|cat = Sommerfeld|cat2 = Watson}} Category:Complex Analysis e7l37dl35t151zrsescfcarvn9fp62v
Let $S \subseteq \N$ be the set of all natural numbers $n \in \N$ such that: :For any finite generator $F$ of $V$ over $R$, if $\card {F \cap L} \ge n$, then $\card L \le \card F$. It is to be demonstrated that $S = \N$. That is, that $\card {F \cap L} \ge n \implies \card L \le \card F$ for all $n \in \N$. By Intersection is Subset and Cardinality of Subset of Finite Set, we have $\card {F \cap L} \le \card F$. Hence, it is vacuously true that $\card F + 1 \in S$. Therefore, $S$ is non-empty. From the well-ordering principle, $S$ has a smallest element $N$. If $N = 0$, the theorem immediately follows. {{AimForCont}} $N \ge 1$. Let $\card {F \cap L} \ge N - 1$. If $L \subseteq F$, the theorem follows from Cardinality of Subset of Finite Set. Otherwise, there exists a $v \in L$ such that $v \notin F$. Let $F' = F \cup \set v$. By Intersection is Subset, we have $F' \cap L \subseteq L$, so it follows by Subset of Linearly Independent Set that $F' \cap L$ is linearly independent over $R$. Also, by Intersection is Subset: :$F' \cap L \subseteq F'$ We have that $F'$ is a generator of $V$ over $R$. By Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set, there exists a basis $B$ for $V$ such that: :$F' \cap L \subseteq B \subseteq F'$ Since $v \notin F$ is a linear combination of $F$, it follows that $F'$ is linearly dependent over $R$. By the definition of a basis: :$B \subsetneq F'$ By Cardinality of Subset of Finite Set and Cardinality is Additive Function: :$\card B < \card {F'} = \card F + 1$ Hence: :$\card B \le \card F$ We have that: {{begin-eqn}} {{eqn | l = \card {B \cap L}       | o = \ge       | r = \card {\paren {F' \cap L} \cap L}       | c = Set Intersection Preserves Subsets and Cardinality of Subset of Finite Set }} {{eqn | r = \card {F' \cap L}       | c = Intersection is Associative and Intersection is Idempotent }} {{eqn | r = \card {\paren {F \cap L} \cup \paren {\set v \cap L} }       | c = Intersection Distributes over Union }} {{eqn | r = \card {\paren {F \cap L} \cup \set v}       | c = Intersection with Subset is Subset }} {{eqn | o = \ge       | r = N       | c = Cardinality is Additive Function }} {{end-eqn}} Since $N \in S$, it follows by the definition of a basis that: :$\card L \le \card B \le \card F$ Hence: :$N - 1 \in S$ But this contradicts the assumption that $N$ is the smallest element of $S$. Therefore: :$N = 0$ Hence the result. {{qed}}
The axioms $(C1)$ to $(C3)$ are checked for a metacategory. Pick any two morphisms $f : \left({A, a}\right) \to \left({B, b}\right)$ and $g : \left({B, b}\right) \to \left({C, c}\right)$ from $\mathbf{Set}_*$. By the definition of composition in the category of pointed sets: :$\left({g \circ f}\right) \left({a}\right) = g \left({f \left({a}\right)}\right) = g \left({b}\right) = c$ whence $g \circ f$ is a pointed mapping from $\left({A, a}\right)$ to $\left({C, c}\right)$. That composition of pointed mappings is associative follows from Composition of Mappings is Associative. For any object $\left({A, a}\right)$, the identity mapping $\operatorname{id}_A$ induces a pointed map $\operatorname{id}_{\left({A, a}\right)}: \left({A, a}\right) \to \left({A, a}\right)$, as $\operatorname{id}_A \left({a}\right) = a$. By Identity Mapping is Left Identity and Identity Mapping is Right Identity, this is the identity morphism for $\left({A, a}\right)$. {{qed}} Category:Category Theory 508169qexbek7gz2m6l5hfg83j45x57
Let $V \subseteq T'$ be closed in $T'$. Then, from Closed Set in Topological Subspace, $V = T' \cap V$ is closed in $T$. If $V$ is closed in $T$ then $V = T' \cap W$ where $W$ is closed in $T$. Since $T'$ is closed in $T$, it follows by Topology Defined by Closed Sets that $V$ is closed in $T$. {{qed}}
Follows directly from Subset of Real Numbers is Path-Connected iff Interval. {{qed}}
=== Necessary Condition === Let $S$ be a path-connected metric subspace of $\R$. From Path-Connected Space is Connected, it follows that $S$ is connected. From Subset of Real Numbers is Interval iff Connected, it follows that $S$ is a real interval. {{qed|lemma}} === Sufficient Condition === Let $S$ be a real interval. {{proof wanted|We have Connected Open Subset of Euclidean Space is Path-Connected but need to show that this still holds when $S$ is closed.}}