hcju/proofwikiqa huggingface dataset:
We have that $\card S = n$. Then by definition of cardinality: :there is a surjection from $S$ to $T$ {{iff}}: :there is a surjection from $\N_n$ to $T$. So we need consider the case only when $S = \N_n$. By definition of surjection: :$\forall x \in T: f^{-1} \sqbrk {\set x} \ne \O$ where $f^{-1} \sqbrk {\set x}$ denotes the preimage of $\set x$ under $f$. From the Well-Ordering Principle, $\N$ is well-ordered. Therefore by Subset of Well-Ordered Set is Well-Ordered, $S = \N_n$ is well-ordered. We have that $f^{-1} \sqbrk {\set x} \subseteq S$. Therefore $f^{-1} \sqbrk {\set x}$ is also well-ordered. Therefore, by definition of well-ordered set, $f^{-1} \sqbrk {\set x}$ has a smallest element. Let $g: T \to S$ be the mapping defined as: :$\forall x \in T: g \paren x =$ the smallest element of $f^{-1} \sqbrk {\set x}$ Let $x \in T$. Then: :$g \paren x \in f^{-1} \sqbrk {\set x}$ so: :$f \paren {g \paren x} = x$ Thus: :$f \circ g = I_T$ and by Identity Mapping is Bijection: :$I_T: T \to g \sqbrk T$ is a bijection. By Cardinality of Subset of Finite Set: :$\card {g \sqbrk T} \le n$ Let $\card {g \sqbrk T} = m$. By Set Equivalence is Equivalence Relation: :$\card T = m$ elements. Suppose $m = n$. Then by Cardinality of Subset of Finite Set: :$g \sqbrk T = S$ so $g: T \to S$ is a bijection. Therefore: {{begin-eqn}} {{eqn | l = f       | r = f \circ I_S       | c =  }} {{eqn | r = f \circ g \circ g^{-1}       | c =  }} {{eqn | r = I_S \circ g^{-1}       | c =  }} {{eqn | r = g^{-1}       | c =  }} {{end-eqn}} So $f: S \to T$ is a bijection. {{Qed}}
$(2)$ implies $(3)$: Let $f$ be an injection. Then by Cardinality of Image of Injection: :$\card S = \card {f \sqbrk S}$ where $f \sqbrk S$ denotes the image of $S$ under $f$. Therefore the subset $f \sqbrk S$ of $T$ has the same number of elements as $T$. Therefore: : $f \sqbrk S = T$ and so $f$ is a surjection. {{qed|lemma}} $(3)$ implies $(1)$: Let $f$ be surjective. We have that $\card S = \card T$. By Cardinality of Surjection, $f$ is bijective. {{qed|lemma}} $(1)$ implies $(2)$: Let $f$ be a bijection. By definition of bijection, $f$ is also an injection. {{qed}}
Let the mapping $s: \N \to \N$ be defined as: :$\forall n \in \N: \map s n = n + 1$ $s$ is clearly an injection. {{AimForCont}} $\N$ were finite. By Equivalence of Mappings between Sets of Same Cardinality it follows that $s$ is a surjection. But: :$\forall n \in \N: \map s n \ge 0 + 1 > 0$ So: :$0 \notin s \sqbrk \N$ and $s$ is not a surjection. From this contradiction it is seen that $\N$ cannot be finite. So, by definition, $\N$ is infinite. {{Qed}}
Let $A \subseteq \N_{>0}$ such that every $B \subseteq S$ such that $\left|{B}\right| = n$ has a greatest and a smallest element. Any $B \subseteq S$ such that $\left|{B}\right| = 1$ has $1$ element, $b \in B$ say. Then $b$ is both the greatest and smallest element of $B$. So $1 \in A$. Let $n \in A$. Let $B \subseteq S$ such that $\left|{B}\right| = n + 1$. Then $\exists b \in B$, and $\left|B \setminus \left\{{b}\right\}\right| = n$ elements by Cardinality Less One. So, by the induction hypothesis, $B \setminus \left\{{b}\right\}$ has a greatest element $c$ and a smallest element $a$, as $n \in A$. Note that $b \ne c$ as $c \in B \setminus \left\{{b}\right\}$ but $b \notin B \setminus \left\{{b}\right\}$. So as $\left({S, \preceq}\right)$ is totally ordered, either $b \prec c$ or $c \prec b$ as $b \ne c$. If $b \prec c$ then $c$ is the greatest element of $B$, otherwise it's $b$. Similarly, either $b \prec a$ or $a \prec b$, and thus either $a$ or $b$ is the smallest element of $B$. Either way, $B$ has both a greatest and a smallest element, and therefore $n + 1 \in A$. Therefore, by the Principle of Mathematical Induction, $A = \N_{>0}$ and the proof is complete. {{qed}}
The result follows from: : Finite Non-Empty Subset of Ordered Set has Maximal and Minimal Elements : Minimal Element of Chain is Smallest Element : Maximal Element of Chain is Greatest Element {{qed}}
It is sufficient to consider the case where $\struct {T, \preccurlyeq}$ is $\struct {\N_n, \le}$ for some $n \in \N$. Let $A$ be the set of all $n \in \N$ such that if: :$(1): \quad S$ is any set such that $\card S = n$, and :$(2): \quad \preceq$ is any total ordering on $S$ then there is exactly one isomorphism from $\struct {S, \preceq}$ to $\struct {\N_n, \le}$. $\O \in A$ from Empty Mapping is Mapping and Equivalence of Mappings between Sets of Same Cardinality. Now let $n \in A$. Let $\struct {S, \preceq}$ be a totally ordered set with $n + 1$ elements. By Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements, $S$ has a greatest element $b$. Then $\card {S \setminus \set b} = n$ by Cardinality Less One. The total ordering on $S \setminus \set b$ is the one induced from that on $S$. So as $n \in A$, there exists a unique order isomorphism $f: S \setminus \set b \to \struct {\N_n, \le}$. Let us define the mapping $g: S \to \N_{n + 1}$ as follows: :$\forall x \in S: \map g x = \begin{cases} \map f x: & x \in S \setminus \set b \\ n: & x = b \end{cases}$ This is the desired order isomorphism from $\struct {S, \preceq}$ to $\struct {\N_{n + 1}, \le}$. Now let $h: \struct {S, \preceq}$ to $\struct {\N_{n + 1}, \le}$ be an order isomorphism. Then $\map h b = n$ (it has to be). So the restriction of $h$ to $S \setminus \set b$ is an isomorphism from $S \setminus \set b$ to $\struct {\N_n, \le}$, and hence $h = f$ (as $n \in A$, any such order isomorphism is unique). Thus: :$\forall x \in S \setminus \set b: \map h x = \map f x = \map g x$ and: :$\map h b = n = \map g b$ thus $h = g$. Therefore $n + 1 \in A$. The result follows from the Principle of Mathematical Induction. {{qed}}
From Unique Isomorphism between Finite Totally Ordered Sets, it suffices to show that $h$ is an order isomorphism. To this end, remark that, for all $x, y \in \N_{n - m}$: {{begin-eqn}} {{eqn | l = h \left({x}\right)       | r = h \left({y}\right) }} {{eqn | ll= \iff       | l = x + m + 1       | r = y + m + 1 }} {{eqn | ll = \iff       | l = x       | r = y       | c = Natural Number Addition is Cancellable }} {{end-eqn}} proving $h$ is an injection, and so a bijection, from Equivalence of Mappings between Sets of Same Cardinality. By Ordering on Natural Numbers is Compatible with Addition and Natural Number Addition is Cancellable for Ordering, it follows that: :$x \le y \iff h \left({x}\right) \le h \left({y}\right)$ so $h$ is an order isomorphism. {{qed}}
By Cancellable iff Regular Representations Injective, $\lambda_a$ and $\rho_a$ are injections. We have that $S$ is finite. From Injection from Finite Set to Itself is Surjection, both $\lambda_a$ and $\rho_a$ are surjections. Thus $\lambda_a$ and $\rho_a$ are injective and surjective, and therefore bijections. {{qed}} Category:Semigroups Category:Regular Representations Category:Cancellability k62vgwldwv4np50anvdp0y1qep88fep
There is no bijection from a set to its power set. From Injection from Set to Power Set, we have that there exists an injection $f: \N \to \powerset \N$. From the Cantor-Bernstein-Schröder Theorem, there can be no injection $g: \powerset \N \to \N$. So, by definition, $\powerset \N$ is not countable. {{qed}}
Let $S = \set {a_0, a_1, a_2, \ldots}$ be countably infinite. Let $T \subseteq S = \set {a_{n_0}, a_{n_1}, a_{n_2}, \ldots}$, where $a_{n_0}, a_{n_1}, a_{n_2}, \ldots$ are the elements of $S$ also in $T$. If the set of numbers $\set {n_0, n_1, n_2, \ldots}$ has a largest number, then $T$ is finite. Otherwise, consider the bijection $i \leftrightarrow n_i$. This leads to the bijection $i \leftrightarrow a_{n_i}$ This latter bijection is the required one-to-one correspondence between the elements of $T$ and those of $\N$, showing that $T$ is indeed countable. {{finish|formal justification}}
Let $\left\langle{S_n}\right\rangle_{n \in \N}$ be a sequence of countable sets. Define: :$\displaystyle S = \bigcup_{n \mathop \in \N} S_n$. For all $n \in \N$, let $\mathcal F_n$ be the set of all surjections from $\N$ to $S_n$. Since $S_n$ is countable, it follows by Surjection from Natural Numbers iff Countable that $\mathcal F_n$ is non-empty. Using the axiom of countable choice, there exists a sequence $\left\langle{f_n}\right\rangle_{n \in \N}$ such that $f_n \in \mathcal F_n$ for all $n \in \N$. Let $\phi: \N \times \N \to S$, where $\times$ denotes the cartesian product, be the surjection defined by: :$\phi \left({m, n}\right) = f_m \left({n}\right)$ Since $\N \times \N$ is countable, it follows by Surjection from Natural Numbers iff Countable that there exists a surjection $\alpha: \N \to \N \times \N$. Since the composition of surjections is a surjection, the mapping $\phi \circ \alpha: \N \to S$ is a surjection. By Surjection from Natural Numbers iff Countable, it follows that $S$ is countable. {{qed}}
Let $S$ be an infinite set. We use Between Two Sets Exists Injection or Surjection. Suppose that there exists an injection $\psi: \N \to S$. Let $T$ be the image of $\psi$. From Injection to Image is Bijection, it follows that $\psi^{-1}: T \to \N$ is a bijection. Hence, $T$ is a countably infinite subset of $S$. Now, suppose that that there exists a surjection $\phi: \N \to S$. From Surjection from Natural Numbers iff Countable, it follows that $S$ is countably infinite. So, from Set is Subset of Itself, we have that $S$ is a countably infinite subset of $S$. {{qed}}
Let $S$ be an infinite set. First an injection $f: \N \to S$ is constructed. Let $g$ be a choice function on $\powerset S \setminus \set \O$. Then define $f: \N \to S$ as follows: :$\forall n \in \N: f \left({n}\right) = \begin{cases} \map g S & : n = 0 \\ \map g {S \setminus \set {\map f 0, \ldots, \map f {n - 1} } } & : n > 0 \end{cases}$ Since $S$ is infinite, $S \setminus \set {\map f 0, \ldots, \map f {n - 1} }$ is non-empty for each $n \in \N$. Therefore $f \sqbrk \N$ is infinite. To show that $f$ is injective, let $m, n \in \N$, say $m < n$. Then: :$\map f m \in \set {\map f 0, \ldots, \map f {n - 1} }$ but: :$\map f n \in S \setminus \set {\map f 0, \ldots, \map f {n - 1} }$ Hence $\map f m \ne \map f n$. Since $f$ is injective, it follows from Injection to Image is Bijection that $f$ is a bijection from $\N$ to $f \sqbrk \N$. Thus $f \sqbrk \N$ is a countable subset of $S$. {{Qed}}
Let $S$ be an infinite set. First an injection $f: \N \to S$ is constructed. Let $f$ be a choice function on $\powerset S \setminus \set \O$. That is: :$\forall A \in \powerset S \setminus \set \O: \map f A \in A$ This is justified only if the Axiom of Choice is accepted. Let $\CC$ be the set of all finite subsets of $S$. Let $A \in \CC$. Since $S$ is infinite it follows that $S \setminus A \ne \O$. So $S \setminus A \in \Dom f$. Let $g: \CC \to \CC$ be the mapping defined as: :$\map g A = A \cup \set {\map f {S \setminus A} }$ That is, $\map g A$ is constructed by joining $A$ with the element that $f$ chooses from $S \setminus A$. Consider the Recursion Theorem applied to $g$, starting with the set $\O$. We obtain a mapping $U: \N \to \CC$ such that: :$\map U x = \begin{cases} \O & : x = 0 \\ \map U n \cup \set {\map f {S \setminus \map U n} } & : x = n^+ \end{cases}$ where here $\N$ is considered as the set of elements of the von Neumann construction of natural numbers. Consider the mapping $v: \N \to S$, defined as: :$\forall n \in \N: \map v n = \map f {S \setminus \map U n}$ We have that, by definition of $v$: :$(1): \quad \forall n \in \N: \map v n \notin \map U n$ :$(2): \quad \forall n \in \N: \map v n \in \map U {n^+}$ :$(3): \quad \forall m, n \in \N: n \le m \implies \map U n \subseteq \map U m$ Then because $\map v n \in \map U m$ but $\map v m \notin \map U m$: :$(4): \quad \forall m, n \in \N: n < m \implies \map v n \ne \map v m$ So $(4)$ implies that $v$ maps distinct elements of $\N$ onto distinct elements of $S$. Thus $v: \N \to S$ is an injection. It follows from Injection to Image is Bijection that $v$ is a bijection from $\N$ to $\map v \N$. Thus $\map v \N$ is the countable subset of $S$ that was required. {{Qed}}
Let $S$ be an infinite set. For all $n \in \N$, let: :$\mathcal F_n = \left\{{T \subseteq S : \left\vert{T}\right\vert = n}\right\}$ where $\left\vert{T}\right\vert$ denotes the cardinality of $T$. From Set is Infinite iff exist Subsets of all Finite Cardinalities: :$\mathcal F_n$ is non-empty. Using the axiom of countable choice, we can obtain a sequence $\left\langle{S_n}\right\rangle_{n \in \N}$ such that $S_n \in \mathcal F_n$ for all $n \in \N$. Define: :$\displaystyle T = \bigcup_{n \mathop \in \N} S_n \subseteq S$ For all $n \in \N$, $S_n$ is a subset of $T$ whose cardinality is $n$. From Set is Infinite iff exist Subsets of all Finite Cardinalities: :$T$ is infinite. Because the countable union of finite sets is countable, $T$ is countable. {{qed}}
Let $S$ be a finite set, and let $T$ be a proper subset of $S$. Let $f: T \to S$ be an injection. By Cardinality of Image of Injection and Cardinality of Subset of Finite Set: :$\size {\Img f} = \size T < \size S$ Here, $\Img f$ denotes the image of $f$. Thus $\Img f \ne S$, and so $f$ is not a bijection. {{qed}}
Let $T$ be an infinite set. By Infinite Set has Countably Infinite Subset, it is possible to construct a countably infinite subset of $T$. Let $S = \set {a_1, a_2, a_3, \ldots}$ be such a countably infinite subset of $T$. Create a Partition of $S$ into: :$S_1 = \set {a_1, a_3, a_5, \ldots}, S_2 = \set {a_2, a_4, a_6, \ldots}$ Let a bijection be established between $S$ and $S_1$, by letting $a_n \leftrightarrow a_{2 n - 1}$. This is extended to a bijection between $S \cup \paren {T \setminus S} = T$ and $S_1 \cup \paren {T \setminus S} = T \setminus S_2$ by assigning each element in $T \setminus S$ to itself. So a bijection has been demonstrated between $T$ and one of its proper subsets $T \setminus S_2$. That is, if $T$ is infinite, it is equivalent to one of its proper subsets. Now, let $T_0 \subsetneq T$ be a proper subset of $T$, and $f: T \to T_0$ be a bijection. It follows from No Bijection between Finite Set and Proper Subset that $T$ must be infinite. {{qed}}
Let $S$ be a set. Suppose $S$ is finite. From No Bijection between Finite Set and Proper Subset we have that $S$ can not be equivalent to one of its proper subsets. Suppose $S$ is infinite. From Infinite Set has Countably Infinite Subset, we can construct $v: \N \to S$ such that $v$ is an injection. We now construct the mapping $h: S \to S$ as follows. :$\map h x = \begin {cases} \map v {n + 1} & : \exists n \in \N: x = \map v n \\ x & : x \notin \Img v \end {cases}$ It is clear that $h$ is an injection. But we have that $\map v 0 \notin \Img h$ and so $\Img h \subsetneq S$. The result follows from Injection to Image is Bijection. {{qed}}
Let $X$ be a set which has a proper subset $Y$ such that: :$\card X = \card Y$ where $\card X$ denotes the cardinality of $X$. Then: :$\exists \alpha \in \complement_X \paren Y$ and :$Y \subsetneqq Y \cup \set \alpha \subseteq X$ The inclusion mappings: :$i_Y: Y \to X: \forall y \in Y: i \paren y = y$ :$i_{Y \cup \set \alpha}: Y \cup \set \alpha \to X: \forall y \in Y: i \paren y = y$ give: :$\card X = \card Y \le \card Y + \mathbf 1 \le \card X $ from which: :$\card X = \card Y + \mathbf 1 = \card X + \mathbf 1$ So by definition $X$ is infinite. {{qed|lemma}} Now suppose $X$ is infinite. That is: :$\card X = \card X + \mathbf 1$ Let $\alpha$ be any object such that $\alpha \notin X$. Then there is a bijection $f: X \cup \set \alpha \to X$. Let $f_{\restriction X}$ be the restriction of $f$ to $X$. Then from Injection to Image is Bijection: :$\image {f_{\restriction X} } = X \setminus \set {f \paren \alpha}$ which is a proper subset of $X$ which is equivalent to $X$. {{qed}}
Let $S = \set {s_0, s_1, s_2, \dotsc}$ and $T = \set {t_0, t_1, t_2, \dotsc}$ be countable sets. If both $S$ and $T$ are finite, the result follows immediately. Suppose either of $S$ or $T$ (or both) is countably infinite. We can write the elements of $S \times T$ in the form of an infinite table: :$\begin{array} {*{4}c}  \tuple {s_0, t_0} & \tuple {s_0, t_1} & \tuple {s_0, t_2} & \cdots \\  \tuple {s_1, t_0} & \tuple {s_1, t_1} & \tuple {s_1, t_2} & \cdots \\  \tuple {s_2, t_0} & \tuple {s_2, t_1} & \tuple {s_2, t_2} & \cdots \\  \vdots  & \vdots  & \vdots & \ddots \\ \end{array}$ This table clearly contains all the elements of $S \times T$. Now we can count the elements of $S \times T$ by processing the table diagonally. First we pick $\tuple {s_0, t_0}$. Then we pick $\tuple {s_0, t_1}, \tuple {s_1, t_0}$. Then we pick $\tuple {s_0, t_2}, \tuple {s_1, t_1}, \tuple {s_2, t_0}$. We can see that all the elements of $S \times T$ will (eventually) be listed, and there is a specific number (element of $\N$) to index each of its elements with. Thus we have the required one-to-one correspondence between $S \times T$ and $\N$, and our assertion is proved. {{qed}}
Let $S, T$ be countable sets. From the definition of countable, there exists a injection from $S$ to $\N$, and similarly one from $T$ to $\N$. Hence there exists an injection $g$ from $S \times T$ to $\N^2$. Now let us investigate the cardinality of $\N^2$. From the Fundamental Theorem of Arithmetic, every natural number greater than $1$ has a unique prime decomposition. Thus, if a number can be written as $2^n 3^m$, it can be done thus in only one way. So, consider the function $f: \N^2 \to \N$ defined by: :$\map f {n, m} = 2^n 3^m$. Now suppose $\exists m, n, r, s \in \N$ such that $\map f {n, m} = \map f {r, s}$. Then $2^n 3^m = 2^r 3^s$ so that $n = r$ and $m = s$. Thus $f$ is an injection, whence $\N^2$ is countably infinite. Now we see that as $g$ and $f$ are injective, it follows from Composite of Injections is Injection that $f \circ g: S \times T \to \N$ is also injective. Hence the result. {{qed}}
Let $S, T$ be countable sets. Let $S = \set {s_1, s_2, s_3, \dotsc}$ and $T = \set {t_1, t_2, t_3, \dotsc}$ be enumerations of $S$ and $T$ respectively. Let $f: S \times T: \N$ be the mapping defined as: :$\forall \tuple {s_k, t_l} \in S \times T: \map f {s_k, t_l} = \dfrac {\paren {k + l - 1} \paren {k + l - 2} } 2 + \dfrac {l + \paren {-1}^{k + 1} } 2 k + \dfrac {1 + \paren {-1}^{k + l - 1} } 2 l$ Then $f$ gives an enumeration of $S \times T$. This enumeration can be depicted schematically as: :$\begin {array} {} \tuple {s_1, t_1} & & \tuple {s_1, t_2} & \to  & \tuple {s_1, t_3} & & \tuple {s_1, t_4} & \to & \dotsc \\ \downarrow & \nearrow & & \swarrow & & \nearrow  & \dotsc \\ \tuple {s_2, t_1} & & \tuple {s_2, t_2} & & \tuple {s_2, t_3} & \dotsc \\ & \swarrow & & \nearrow & \dotsc \\ \tuple {s_3, t_1} & & \tuple {s_2, t_3} & \dotsc \\ \downarrow & \nearrow & \dotsc \\ \tuple {s_3, t_1} & \dotsc \\ \vdots \\ \end{array}$ Hence the result. {{qed}}
By definition, a sequence is a mapping whose domain is a subset of $\N$. Let the range of $\left \langle {a_j} \right \rangle_{j \in B}$ be $S$. Thus $\left \langle {a_j} \right \rangle_{j \in B}$ can be expressed using the mapping $f: B \to S$ as: :$\forall j \in B: f \left({j}\right) = a_j$ Let $k \in A$. Then $\sigma \left({k}\right) \in B$. By definition of composition of mappings: {{begin-eqn}} {{eqn | l=f \circ \sigma \left({k}\right)       | r=f \left({\sigma \left({k}\right)}\right)       | c= }} {{eqn | ll=\implies       | l=\left \langle {a_j} \right \rangle_{j \in B}       | r=\left \langle {a_{\sigma \left({k}\right)} } \right \rangle_{\sigma \left({k}\right) \in B}       | c= }} {{eqn | r=\left \langle {a_{\sigma \left({k}\right)} } \right \rangle_{k \in A}       | c=as $\sigma \left({k}\right) \in B \implies k \in A$ }} {{end-eqn}} {{qed}} Category:Sequences 2wdusrdz15r73kktgzfr0g4a18i31vr
First we show that the elements of $\left\{{A_k: k \in \left[{1 \,.\,.\, n}\right]}\right\}$ are disjoint. Let $j \in \left[{1 \,.\,.\, n}\right]$. We have that: : $\left \langle {r_k} \right \rangle_{0 \mathop \le k \mathop \le n}$ is strictly increasing and: : $0 \le j - 1 < j \le n$ Thus by Sum with One is Immediate Successor in Naturally Ordered Semigroup: : $r_0 \le r_{j-1} < r_j \le r_n \implies \left({r_0 + 1}\right) \le \left({r_{j-1} + 1}\right) \le r_j \le r_n$ So: : $\varnothing \subset A_j \subseteq \left[{r_0 + 1 \,.\,.\, r_n}\right]$ Because $\left \langle {r_k} \right \rangle_{0 \mathop \le k \mathop \le n}$ is strictly increasing: :$1 \le j < k \le n \implies A_j \cap A_k = \varnothing$ So the elements of $\left\{{A_k: k \in \left[{1 \,.\,.\, n}\right]}\right\}$ are disjoint, as we were to show. It remains to be established that: :if $m \in \left[{r_0 + 1 \,.\,.\, r_n}\right]$ then: : $m \in A_k$ for some $k \in \left[{1\,.\,.\, n}\right]$. Let $m \in \left[{r_0 + 1 \,.\,.\, r_n}\right]$. Consider the set: :$J = \left\{{j \in \left[{0 \,.\,.\, n}\right]: m \le r_j}\right\}$ Clearly $j \ne \varnothing$ as $n \in J$. Let $k$ be the smallest element of $J$. Then because $r_0 < m$: : $k \ne 0$ Thus: : $k \in \left[{1 \,.\,.\, n}\right]$ By its definition: : $r_{k-1} < m \le r_k$ Thus, by Sum with One is Immediate Successor in Naturally Ordered Semigroup: : $r_{k-1} + 1 \le m \le r_k$ Therefore $m \in A_k$. {{qed}} {{improve|A better link instead of Sum with One is Immediate Successor in Naturally Ordered Semigroup}}
Let $r_0 = 0$, and let: :$\displaystyle \forall k \in \closedint 1 n: r_k = \sum_{j \mathop = 1}^k {p_j}$ Then: :$r_{k - 1} + p_k = r_k$ so: :$r_{k - 1} < r_k$. Thus by Isomorphism to Closed Interval, $\closedint {r_{k - 1} } {r_k}$ has $r_k - r_{k - 1} = p_k$ elements. As a consequence, there exists a bijection $\sigma_k: B_k \to \closedint {r_{k - 1} } {r_k}$ for each $k \in \closedint 1 n$. Let $\sigma: A \to \N$ be the mapping that satisfies: :$\forall x \in B_k: \forall k \in \N: \map \sigma x = \map {\sigma_k} x$ By Strictly Increasing Sequence on Ordered Set, $\sequence {r_k}_{0 \mathop \le k \mathop \le n}$ is a strictly increasing sequence of natural numbers. Thus by Strictly Increasing Sequence induces Partition, $\sigma: A \to \closedint 1 {r_n}$ is a bijection. By Isomorphism to Closed Interval, $\closedint 1 {r_n}$ has $r_n$ elements. Hence the result. {{Qed}}
Proof by induction: For all $n \in \N_{>0}$, let $\map P n$ be the proposition: :$\displaystyle n^2 = \sum_{j \mathop = 1}^n \paren {2 j - 1}$ === Basis for the Induction === $\map P 1$ is true, as this just says $1^2 = 1$. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\displaystyle k^2 = \sum_{j \mathop = 1}^k \paren {2 j - 1}$ Then we need to show: :$\displaystyle \paren {k + 1}^2 = \sum_{j \mathop = 1}^{k + 1} \paren {2 j - 1}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \paren {k + 1}^2       | r = k^2 + 2 k + 1       | c =  }} {{eqn | r = \sum_{j \mathop = 1}^k \paren {2 j - 1} + 2 k + 1       | c = Induction Hypothesis }} {{eqn | r = \sum_{j \mathop = 1}^k \paren {2 j - 1} + 2 \paren {k + 1} - 1       | c =  }} {{eqn | r = \sum_{j \mathop = 1}^{k + 1} \paren {2 j - 1}       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \N: n^2 = \sum_{j \mathop = 1}^n \paren {2 j - 1}$ {{qed}}