hcju/proofwikiqa huggingface dataset:
First let $k > 1$. {{begin-eqn}} {{eqn | l = \binom n k       | r = \frac {n!} {k! \, \paren {n - k}!}       | c = Definition of Binomial Coefficient }} {{eqn | o = <       | r = \frac {n!} {\paren {n - k}!}       | c = as $k! > 0$ by definition of factorial }} {{eqn | r = \underbrace {n \paren {n - 1} \cdots (n - k + 1)}_{k \text { factors} }       | c = expanding the factorials }} {{eqn | o = <       | r = n^k       | c = for all $n, k$ here considered }} {{end-eqn}} For $k > 1$, the product has at least two factors. Hence at least one factor is strictly less than $n$. {{qed|lemma}} Let $k < 0$. By definition of binomial coefficient: :$\dbinom n k = 0$ while $n^k > 0$. Hence the result follows for $k < 0$. {{qed|lemma}} Finally let $0 \le k \le 1$. From Binomial Coefficient with Zero: :$\dbinom n 0 = 1 = n^0$ From Binomial Coefficient with One: :$\dbinom n 1 = n = n^1$ So equality holds when $k = 0$ or $k = 1$. {{qed|lemma}} Hence the result has been shown to hold for all $k \in \Z$. {{qed}}
We dismiss the cases where $k < 0$ by observing that in such cases $\dbinom n k = 0$ while $n^k > 0$. Similarly we dismiss $k = 0$: we have $\dbinom n 0 = 1 = n^0$. Let: : $N = \set {1, \ldots, n}$ : $K = \set {1, \ldots, k}$ From Cardinality of Set of Strictly Increasing Mappings, $\dbinom n k$ is the number of strictly increasing mappings from $K$ to $N$. From Cardinality of Set of All Mappings, $n^k$ is the number of all mappings from $K$ to $N$. For $k = 1$ there is exactly 1 mapping from $K$ to $N$ which is trivially strictly increasing. Otherwise we have that $K$ has more than one element, and so therefore does $n$. The mapping $f: K \to N: \forall s \in K: \map f s = 1$ is clearly not a strictly increasing mapping. So not all mappings from $K$ to $N$ are strictly increasing. Hence a strict inequality holds, and so $\dbinom n k < n^k$. {{qed}}
By the definition of absolute convergence, we have that $\displaystyle \sum_{n \mathop = 1}^\infty \size {a_n}$ is convergent. From the Comparison Test we have that $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ converges {{iff}}: : $\forall n \in \N_{>0}: \size {a_n} \le b_n$ where $\displaystyle \sum_{n \mathop = 1}^\infty b_n$ is convergent. So substituting $\size {a_n}$ for $b_n$ in the above, the result follows. {{qed}}
{{begin-eqn}} {{eqn | l = \forall S: S       | o = \subseteq       | r = S       | c = Set is Subset of Itself }} {{eqn | ll= \leadsto       | l = \forall S: S       | o = \in       | r = \powerset S       | c = {{Defof|Power Set}} }} {{end-eqn}} {{qed}}
From Empty Set is Element of Power Set and Set is Element of its Power Set: :$\O \in \powerset \O$ From Empty Set is Subset of All Sets: :$S \subseteq \O \implies S = \O$ That is: :$S \in \powerset \O \implies S = \O$ Hence the only element of $\powerset \O$ is $\O$. {{qed}}
{{begin-eqn}} {{eqn | l = S * T       | r = \paren {S \cup T} \setminus \paren {S \cap T}       | c = {{Defof|Symmetric Difference|index = 2}} }} {{eqn | o = \subseteq       | r = \paren {S \cup T}       | c = Set Difference is Subset }} {{end-eqn}} {{qed}}
=== Necessary Condition === Suppose $P + Q$ is a projection. Then: {{begin-eqn}} {{eqn | l = P + Q       | r = \paren {P + Q}^2       | c = $P + Q$ is an idempotent }} {{eqn | r = P^2 + P Q + Q P + Q^2 }} {{eqn | r = P + P Q + Q P + Q       | c = $P$ and $Q$ are idempotents }} {{eqn | ll= \leadstoandfrom       | l = PQ + QP       | r = 0 }} {{end-eqn}} Now suppose that $h \in \Rng Q$; say $h = Q q$ for $q \in H$. Then it follows that $Q h = Q Q q = Q q = h$ as $Q$ is idempotent. It follows that: :$0 = P Q h + Q P h = P h + Q P h$ From Characterization of Projections, statement $(6)$, $\innerprod {Q P h} {P h}_H \ge 0$. Next, observe: :$0 = \innerprod {P h + Q p h} {P h}_H = \innerprod {P h} {P h}_H + \innerprod {Q P h} {P h}_H$ As the second term is non-negative, the first is non-positive; it follows that $P h = \mathbf 0_H$ from the definition of the inner product. Hence: :$h \in \Ker P = \paren {\Rng P}^\perp$ It follows that $\Rng Q \perp \Rng P$, as asserted. {{qed|lemma}} === Sufficient Condition === Suppose that $\Rng P \perp \Rng Q$. Then as $P, Q$ are projections, have: :$\Rng P \subseteq \Ker Q$ :$\Rng Q \subseteq \Ker P$ That is, for all $h \in H$: :$Q P h = P Q h = \mathbf 0_H$ Hence: {{begin-eqn}} {{eqn | l = \paren {P + Q}^2       | r = P^2 + P Q + Q P = Q^2 }} {{eqn | r = P + Q       | c = $P$ and $Q$ are idempotents }} {{end-eqn}} That is, $P + Q$ is an idempotent. Furthermore, by Adjoining is Linear, have: :$\paren {P + Q}^* = P^* + Q^* = P + Q$ where the latter follows from Characterization of Projections, statement $(4)$. This same statement implies that $P + Q$ is also a projection. {{qed}}
The proof proceeds by first showing that $(1)$ is equivalent to $(2)$. Then, these are combined and shown equivalent to $(3)$. === $(1)$ implies $(2)$ === Suppose $PQ$ is a projection.  Then by Characterization of Projections, statement $(4)$, one has: :$PQ = \left({PQ}\right)^* = Q^* P^* = QP$ where the penultimate equality follows from Adjoint of Composition. {{qed|lemma}} === $(2)$ implies $(1)$ === Suppose that $PQ = QP$. Then $\left({PQ}\right)^2 = PQPQ = P^2 Q^2 = PQ$ as $P, Q$ are projections. Hence $PQ$ is an idempotent. Also, note that $\left({PQ}\right)^* = Q^* P^* = QP = PQ$. Hence, by Characterization of Projections, statement $(4)$, $PQ$ is a projection. {{qed|lemma}} === $(1), (2)$ imply $(3)$ === The above establishes that assuming either of $(1)$ and $(2)$ yields both to hold. So assuming $(1)$, $P, Q$ and $PQ$ are all projections, and $PQ = QP$. Now compute: {{begin-eqn}} {{eqn | l = \left({P + Q - PQ}\right)^2       | r = P^2 + Q^2 + \left({PQ}\right)^2 + PQ + QP - PPQ - PQP - QPQ - PQQ }} {{eqn | r = P + Q - PQ + PQ + PQ - PQ - PQ - PQ - PQ       | c = $P, Q, PQ$ projections, $PQ = QP$ }} {{eqn | r = P + Q - PQ }} {{end-eqn}} It follows that $P + Q - PQ$ is an idempotent. Observe from Adjoining is Linear and Adjoint of Composition: :$\left({P + Q - PQ}\right)^* = P^* + Q^* - Q^* P^* = P + Q - QP = P + Q - PQ$ Now applying Characterization of Projections, statement $(4)$, conclude that $P + Q - PQ$ is a projection. {{qed|lemma}} === $(3)$ implies $(2)$ === Let $P + Q - PQ$ be a projection. Then by Characterization of Projections, statement $(4)$, compute: {{begin-eqn}} {{eqn | l = P + Q - PQ       | r = \left({P + Q - PQ}\right)^* }} {{eqn | r = P^* + Q^* - Q^* P^*       | c = Adjoining is Linear, Adjoint of Composition }} {{eqn | r = P + Q - QP }} {{end-eqn}} Hence necessarily $PQ = QP$. {{qed}}
First it is shown that $(2)$ is equivalent to $(3)$. Then, equivalence to $(1)$ is shown. === $(2)$ iff $(3)$ === Suppose $P Q = Q$. Then $P = P + Q - P Q$, and by Product of Projections, necessarily $P Q = Q P$. That is, $Q P = P Q = Q$. The converse is obtained by swapping the rôles of $P$ and $Q$ in Product of Projections. {{qed|lemma}} === $(2), (3)$ imply $(1)$ === Suppose $P Q = Q P = Q$. Then as $P, Q$ are projections: :$\paren {P - Q}^2 = P^2 + Q^2 - P Q - Q P = P - Q$ That is, $P - Q$ is idempotent. From Adjoining is Linear, $\paren {P - Q}^* = P^* - Q^* = P - Q$. An application of Characterization of Projections, statement $(4)$ shows that $P - Q$ is a projection. {{qed|lemma}} === $(1)$ implies $(2)$ === Let $P - Q$ be a projection. Then by Characterization of Projections, statement $(6)$, one has: :$\innerprod {P h} h_H - \innerprod {Q h}, h_H \ge 0$ Applying this statement on $P, Q$ also, one obtains that: :$P h = 0 \implies Q h = 0$ that is: :$\ker P \subseteq \ker Q$ Orthocomplement Reverses Subset and that $P, Q$ are projections combine to state $\Rng Q \subseteq \Rng P$. So for every $h \in H$, there is a $p \in H$ with $Q h = P p$. It follows that: :$P Q h = P P p = P p = Q h$ Hence $PQ = Q$. {{qed}}
Since $\AA$ is countable, its contents can be arranged in a sequence: :$\AA = \set {A_1, A_2, \ldots}$ Let $B \in \BB$. Consider the sequence of sets: :$\sequence {A_1 \cup B, A_2 \cup B, \ldots}$ We may leave out any possible repetitions, and obtain a countable set: :$\set {A \cup B: A \in \AA}$ for every $B \in \BB$. Thus as $B$ varies over all the elements of $\BB$, we obtain the countable family: :$\sequence {A \cup B: A \in \AA}_{\paren {B \mathop \in \BB} }$ From Countable Union of Countable Sets is Countable, their union is countable. {{qed}} {{AoC|Countable Union of Countable Sets is Countable|proof}}
Both $\mathcal A$ and $\mathcal B$ are countable. Enumerate their elements by the sequences $\sequence {A_n}_{n \mathop \in \N}$ and $\sequence {B_n}_{n \mathop \in \N}$ respectively. When one of $\mathcal A$ and $\mathcal B$ would be finite, achieve such sequences by allowing elements to occur multiple times. Define $\mathcal C := \set {A \cup B: A \in \mathcal A, B \in \mathcal B}$. For each $C \in \mathcal C$, define the set $N_C := \set {\tuple {i, j} \in \N \times \N: A_i \cup B_j = C}$. The definition of $\mathcal C$ ensures that $N_C \ne \o$ for all $C \in \mathcal C$. Consider the lexicographic ordering $\preceq$ on $\N \times \N$. Combining the Well-Ordering Principle with Finite Lexicographic Order on Well-Ordered Sets is Well-Ordering, $\preceq$ is a well-ordering on $\N \times \N$. As $N_C \subseteq \N \times \N$ is nonempty, it thus has a smallest element; call it $n_C$. Define now the map $\phi: \mathcal C \to \N \times \N$ by $\map \phi C = n_C$. Well-Ordering Minimal Elements are Unique ensures that $\phi$ is well-defined. Now suppose that $C_1, C_2 \in \mathcal C$ satisfy: :$\map \phi {C_1} = \map \phi {C_2}$ that is, $n_{C_1} = n_{C_2}$, which in turn equal some $\tuple {i, j} \in \N \times \N$. As $\tuple {i, j} \in N_{C_1}$ and $\tuple {i, j} \in N_{C_2}$, $C_1 = A_i \cup B_j = C_2$. It follows that $\phi$ is an injection. Finally, $\N \times \N$ is countable by Cartesian Product of Countable Sets is Countable. Let $\psi: \N \times \N \to \N$ be an injection. Then $\psi \circ \phi: \mathcal C \to \N$ is also an injection by Composite of Injections is Injection. Hence $\mathcal C$ is countable. {{qed}}
Since $\mathcal A$ and $\mathcal B$ are countable, $\mathcal A \times \mathcal B$ is countable by Cartesian Product of Countable Sets is Countable. Thus by Surjection from Natural Numbers iff Countable there is a surjection $f: \N \to \mathcal A \times \mathcal B$. Let $\mathcal C = \left\{{A \cup B: A \in \mathcal A, B \in \mathcal B}\right\}$. Let $g: \mathcal A \times \mathcal B \to \mathcal C$ be defined by letting $g \left({A, B}\right) = A \cup B$. $g$ is a surjection: Let $C \in \mathcal C$. Then there exist $A \in \mathcal A$ and $B \in \mathcal B$ such that $C = A \cup B$. By the definition of Cartesian product: : $\left({A, B}\right) \in \mathcal A \times \mathcal B$ Then: : $g \left({A, B}\right) = A \cup B = C$ It follows that $g$ is a surjection. By Composite of Surjections is Surjection, $g \circ f: \N \to \mathcal C$ is surjective. Hence $\mathcal C$ is countable. {{qed}}
By definition of a countable set, there is an injection $f: A \to \N$. Let $\le_f$ be the ordering induced by $f$ on $A$. By Injection Induces Well-Ordering, $\le_f$ is a well-ordering. Let $A^{\text{fin}}$ be the set of finite subsets of $A$. For $n \in \N$, denote with $A^{\left({n}\right)}$ the set of subsets of $A$ that have precisely $n$ elements. For $A' \in A^{\left({n}\right)}$, we write $A' = \left\{{a'_1, a'_2, \ldots, a'_n}\right\}$ uniquely so that $i < j$ implies $a'_i <_f a'_j$. Then we define $f^{\left({n}\right)}: A^{\left({n}\right)} \to \N$ by: :$\displaystyle f^{\left({n}\right)} \left({A'}\right) := \prod_{i \mathop = 1}^n p_i^{f \left({a'_i}\right)}$ where $p_i$ denotes the $i$th prime. By the Fundamental Theorem of Arithmetic, if $f^{\left({n}\right)} \left({A'}\right) = f^{\left({n}\right)} \left({A''}\right)$, then: :$\left({f \left({a'_1}\right), \dotsc, f \left({a'_n}\right)}\right) = \left({f \left({a''_1}\right), \dotsc, f \left({a''_n}\right)}\right)$ As $f$ is injective, also: :$\left({a'_1, \dotsc, a'_n}\right) = \left({a''_1, \dotsc, a''_n}\right)$ and in particular $A' = A''$. Thus $f^{\left({n}\right)}: A^{\left({n}\right)} \to \N$ is an injection. So $A^{\left({n}\right)}$ is countable. Now define $f^{\text{fin}}: A^{\text{fin}} \to \N \times \N$ by: :$f^{\text{fin}} \left({A'}\right) := \left({n, f^{\left({n}\right)} \left({A'}\right)}\right)$ where $n = \left\vert{A'}\right\vert$ is the cardinality of $A'$. Suppose that: :$f^{\text{fin}} \left({A'}\right) = f^{\text{fin}} \left({A''}\right)$ Then by Equality of Ordered Pairs: :$n = \left\vert{A'}\right\vert = \left\vert{A''}\right\vert$ and: :$f^{\left({n}\right)} \left({A'}\right) = f^{\left({n}\right)} \left({A''}\right)$ Since $f^{\left({n}\right)}$ was already shown to be an injection, it follows that: :$A' = A''$ Thus $f^{\text{fin}}$ is also an injection. By Cartesian Product of Natural Numbers with Itself is Countable, $\N \times \N$ is countable. The result follows from Domain of Injection to Countable Set is Countable. {{qed}}
By the definition of a countable set, there exists an injection $g: A \to \N$. Let $\mathcal F$ denote the set of all finite subsets of $A$. Let $f: \mathcal F \to \N$ be the mapping defined by: :$\displaystyle \map f F = \prod_{k \mathop \in \map g F} p_{k + 1}$ where $p_n$ denotes the $n$th prime number. We define $\map f \O = 1$, the vacuous product. Let $F, G \in \mathcal F$, and suppose that $\map f F = \map f G$. By Expression for Integer as Product of Primes is Unique: :$\map g F = \map g G$ By Subset equals Preimage of Image iff Mapping is Injection: :$F = \map {g^{-1} } {\map g F} = \map {g^{-1} } {\map g G} = G$ That is, $f$ is an injection. {{qed}}
Let $A^{\paren n}$ be the set of subsets of $A$ with no more than $n$ elements. Thus: :$A^{\paren 0} = \set \O$ :$A^{\paren 1} = A^{\paren 0} \cup \set {\set a: a \in A}$ and $\forall n \ge 0$: :$A^{\paren {n + 1} } = \set {a^{\paren n} \cup a^{\paren 1}: a^{\paren n} \in A^{\paren n} \land a^{\paren 1} \in A^{\paren 1} }$ Let us verify by induction that each $A^{\paren n}$ is countable. Let $A$ be countable. :$A^{\paren 1}$ is countable, as its cardinality is $1 + \card A$. Suppose $A^{\paren n}$ is countable. Then by Union of Countable Sets of Sets, so $A^{\paren {n + 1} }$ also countable. By induction, each $A^{\paren n}$ is countable. Denote with $A^f$ the set of finite subsets of $A$. It is apparent that every finite subset is in some $A^{\paren n}$, and so: :$A^f = \displaystyle \bigcup_{n \mathop \in \N} A^{\paren n}$ The result follows from Countable Union of Countable Sets is Countable. {{qed}}
To establish that $\le_f$ is an ordering on $S$, we need to show that it is reflexive, antisymmetric and transitive. So, checking in turn each of the criteria for an ordering: === Reflexivity === For all $s \in S$, one has $f \left({s}\right) \le f \left({s}\right)$ as $\le$ is an ordering. Hence, by definition of $\le_f$, $s \le_f s$ for all $s \in S$. So $\le_f$ is reflexive. {{qed|lemma}} === Antisymmetry === Suppose that for some $s_1, s_2 \in S$, one has both $s_1 \le_f s_2$ and $s_2 \le_f s_1$. This is equivalent to $f \left({s_1}\right) \le f \left({s_2}\right)$ and $f \left({s_2}\right) \le f \left({s_1}\right)$. As $\le$ is an ordering, this implies $f \left({s_1}\right) = f \left({s_2}\right)$. That $f$ is an injection now implies that $s_1 = s_2$. Hence $\le_f$ is antisymmetric. {{qed|lemma}} === Transitivity === Suppose that for some $s_1, s_2, s_3 \in S$, one has both $s_1 \le_f s_2$ and $s_2 \le_f s_3$. This is equivalent to $f \left({s_1}\right) \le f \left({s_2}\right)$ and $f \left({s_2}\right) \le f \left({s_3}\right)$. As $\le$ is an ordering, this implies $f \left({s_1}\right) \le f \left({s_3}\right)$. The definition of $\le_f$ now implies that $s_1 \le_f s_3$. So $\le_f$ is transitive. {{qed|lemma}} So we have shown that $\le_f$ is an ordering on $S$. {{qed}} Category:Order Theory Category:Injections lldi5ek9vrmgyze2lc9lv2gfmfa18yt
By Ordering Induced by Injection is Ordering, $\le_f$ is an ordering. Let $s_1, s_2 \in S$. Then as $\le$ is a total ordering, at least one of the following holds: :$(1): \qquad f \left({s_1}\right) \le f \left({s_2}\right)$ :$(2): \qquad f \left({s_2}\right) \le f \left({s_1}\right)$ If $(1)$ holds, the definition of $\le_f$ implies that $s_1 \le s_2$. Similarly, if $(2)$ holds, $s_2 \le s_1$. This leads to the conclusion that at least one of $s_1 \le s_2$ and $s_2 \le s_1$ holds. Hence $\le_f$ is a total ordering. {{qed}} Category:Total Orderings Category:Injections 2yzdr7kcr2eanxa3w2vp8budup2vozf
By Ordering Induced by Injection is Ordering, $\le_f$ is an ordering. Let $S' \subseteq S$ be non-empty. Then as $\le$ is a well-ordering, the set: :$f \left({S'}\right) = \left\{{f \left({s}\right): s \in S'}\right\}$ has a minimal element, say $f \left({s_0}\right)$. That is, for all $s \in S', f \left({s_0}\right) \le f \left({s}\right)$. By definition of $\le_f$, this implies: :$\forall s \in S': s_0 \le_f s$ That is, $s_0$ is a minimal element of $S'$. Hence $\le_f$ is a well-ordering, as $S'$ was arbitrary. {{qed}} Category:Well-Orderings Category:Injections 7hfxhks81racj6qp7e5oupmvgi9e4rg
Suppose that $G$ is $k$-connected. Then * $\left|{V \left({G}\right) }\right| > k > l$ * $G$ is connected * If $W$ is a vertex cut of $G$, then $\left|{W}\right| \ge k > l$ so $\left|{W}\right| \ge l$. {{qed}} Category:Graph Theory rfow1tll27dgjiwxtx4rb11zjuf8xut
Define $\alpha = \set {\beta: \text{$\beta$ is an ordinal and there is an injection $\beta \to S$}}$. First of all, it is to be shown that $\alpha$ is a set. To this end, define the set $W$ by: :$W = \set {\paren {S', \preceq}: \text{$S' \subseteq S$ and $\preceq$ well-orders $S'$}}$ By Woset is Isomorphic to Unique Ordinal, each $w \in W$ corresponds to a unique ordinal $\beta_w$. Thus by the Axiom of Replacement, the following is a set: :$\set {\beta: \exists w \in W: \beta = \beta_w}$ It follows from Injection Induces Well-Ordering that this set coincides with $\alpha$. In particular, then, $\alpha$ is a set. Next, to establish $\alpha$ is an ordinal. Suppose $\beta \in \alpha$ and $\gamma < \beta$. Let $i: \beta \to S$ be an injection, and let $\iota: \gamma \to \beta$ be the inclusion of $\gamma$ in $\beta$. Then by Composite of Injections is Injection, $i \circ \iota: \gamma \to S$ is an injection. Hence $\gamma \in \alpha$, and therefore $\alpha$ is an ordinal. Finally, by Ordinal is not Element of Itself, $\alpha \notin \alpha$. That is to say, there is no injection $\alpha \to S$. {{qed}}
Let $W$ be the set of all well-orderings on subsets of $S$. By Woset is Isomorphic to Unique Ordinal, there exists a mapping $F: W \to \On$ defined by letting $\map F s$ be the ordinal which is isomorphic to $s$. {{explain|"isomorphism" between ordinals}} By Mapping from Set to Ordinal Class is Bounded Above, $F \sqbrk W$ has an upper bound $\alpha_0$. Then if $\alpha$ is any ordinal strictly greater than $\alpha_0$, then $\alpha \notin F \sqbrk {W}$. Seeking a contradiction, suppose that there is an injection $g: \alpha \to S$. Then by Injection to Image is Bijection, there is a bijection from $\alpha$ onto $g \sqbrk \alpha$. But this induces a well-ordering on $g \sqbrk \alpha \subseteq S$ which is isomorphic to $\alpha$, contradicting the fact that $\alpha \notin F \sqbrk W$. {{qed}}
Let $\gamma$ be the Hartogs number of $X$. Suppose that $x \in X$. Define $g : \gamma \to X$ by transfinite recursion as follows:  :$\map g 0 = x$ :$\map g {\alpha + 1} = \map f {\map g \alpha}$ :$\map g \alpha = \sup \set {\map g \beta: \beta < \alpha}$ when $\alpha$ is a limit ordinal. That $f$ is inflationary guarantees both that: :$(1) \quad \set {\map g \beta: \beta < \alpha}$ is a chain for each $\alpha < \gamma$ :$(2) \quad g$ is increasing. {{explain|I'm not sure whether further explanation is needed as to why the above should be so, or whether I'm just slow.}} If $g$ is strictly increasing, then $g$ is strictly monotone and, by Strictly Monotone Mapping with Totally Ordered Domain is Injective, an injection. But by definition of Hartogs number, $\gamma$ is the least ordinal such that there is no injection from $\gamma$ to $X$. Therefore there must be some $\alpha < \beta < \gamma$ with $\map g \alpha = \map g \beta$. Then: :$\alpha + 1 \le \beta$ so: :$\map g \alpha \le \map g {\alpha + 1} \le \map g \beta = \map g \alpha$ whence: :$\map f {\map g \alpha} = \map g \alpha$ Because $x = \map g 0$: :$\exists y \in \map g \gamma \subseteq \set {y \in X : x \le y}: \map f y = y$ {{qed}} {{Namedfor|Nicolas Bourbaki|name2 = Ernst Witt|cat = Bourbaki|cat2 = Witt}}
From Zermelo's Theorem, exactly one of the following cases holds: :$(1): \quad S \le T$, that is, $T$ dominates $S$ :$(2): \quad S \equiv T$, that is, $T$ is equivalent to $S$ :$(3): \quad S \ge T$, that is, $S$ dominates $T$. Suppose $(1)$ holds, that is, $T$ dominates $S$. By definition, there exists an injection from $S$ to $T$. Hence in this case the result holds. {{qed|lemma}} Suppose $(2)$ holds, that is, $T$ is equivalent to $S$. By definition, there then exists a bijection from $S$ to $T$. A bijection is, by definition, a mapping which is both an injection and a surjection. Hence in this case the result holds. {{qed|lemma}} Suppose $(3)$ holds, that is, $S$ dominates $T$. By definition, there exists an injection from $S$ to $T$. By Injection has Surjective Left Inverse Mapping, $f$ has a left inverse $g: T \to S$ which is surjective. Hence in this case the result holds. {{qed|lemma}} All cases have been shown to lead to the same conclusion, that is, that there exists either an injection or a surjection or both between $S$ and $T$. Hence the result, by Proof by Cases. {{qed}} {{AoC|Zermelo's Theorem (Set Theory)}} Category:Injections Category:Surjections 7i5scj9fe0as9hfjtfpi0sdn2csd23b
If $S = \varnothing$, the empty mapping suffices, as it is vacuously an injection. Let $f: S \to \mathcal P \left({S}\right)$ be the mapping defined as: :$\forall s \in S: f \left({s}\right) = \left\{{s}\right\}$ Let $s, t \in S$ such that $f \left({s}\right) = f \left({t}\right)$. Then $\left\{{s}\right\} = \left\{{t}\right\}$. By definition of set equality it follows directly that $s = t$. Hence $f$ is the required injection, by definition. {{qed}} Category:Injections Category:Power Set tp08qw91z590k5pnjljyrv6uxsnrtn1
Pick a vertex $v \in G$ with $\deg_G \left({v}\right) = \delta \left({G}\right)$, that is, a vertex with minimum degree. Recall that $\Gamma_G \left({v}\right)$ is the neighborhood of $v$ in $G$. Suppose that $V = \Gamma_G \left({v}\right) \cup \left \{{v}\right\}$. That is, that $v$ is adjacent to all other vertices of $G$. Then: :$\left\vert{V}\right\vert = \left\vert{\Gamma_G \left({v}\right)}\right\vert + 1 = \delta \left({G}\right) + 1$ By definition, $\kappa \left({G}\right) < \left\vert{V}\right\vert$, so $\kappa \left({G}\right) \leq \delta \left({G}\right)$. Otherwise, there is at least one vertex of $G$ not adjacent to $v$. So, deleting $\Gamma_G \left({v}\right)$ from $G$ leaves $v$ isolated. Thus $\Gamma_G\left({v}\right)$ is a vertex cut of size $\delta \left({G}\right)$.  Since $G$ has a vertex cut of size $\delta \left({G}\right)$, it follows that $\kappa \left({G}\right) \leq \delta \left({G}\right)$. {{qed}}
Checking in turn each of the criteria for equivalence: === Reflexivity === Let $T$ be a topological space. From Identity Mapping is Homeomorphism, the identity mapping $I_T: T \to T$ is a homeomorphism. So $T \sim T$, and $\sim$ has been shown to be reflexive. {{qed|lemma}} === Symmetry === Let $T_1$ and $T_2$ be topological spaces such that $T_1 \sim T_2$. By definition, there exists a homeomorphism $f: T_1 \to T_2$. From Inverse of Homeomorphism is Homeomorphism it follows that $f^{-1}: T_2 \to T_1$ is also a homeomorphism. So $T_2 \sim T_1$, and $\sim$ has been shown to be symmetric. {{qed|lemma}} === Transitivity === Let $T_1, T_2, T_3$ be topological spaces such that $T_1 \sim T_2$ and $T_2 \sim T_3$. By definition, there exist homeomorphisms $f: T_1 \to T_2$ and $g: T_2 \to T_3$. From Composite of Homeomorphisms is Homeomorphism it follows that $g \circ f: T_1 \to T_3$ is also a homeomorphism. So $T_1 \sim T_3$, and $\sim$ 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}}