hcju/proofwikiqa huggingface dataset:
Let $T$ be the Sorgenfrey line. From Sorgenfrey Line is Lindelöf, $T$ is a Lindelöf space. From Sorgenfrey Line is not Second-Countable, $T$ is not a second-countable space. Hence the result. {{qed}}
Let $T$ be the Sorgenfrey line. From Sorgenfrey Line is Separable, $T$ is a separable space. From Sorgenfrey Line is not Second-Countable, $T$ is not a second-countable space. Hence the result. {{qed}}
Let $T$ be a countable complement topology on an uncountable set. From Countable Complement Space Satisfies Countable Chain Condition, $T$ satisfies the countable chain condition. From Countable Complement Space is not Separable, $T$ is not a separable space. Hence the result. {{qed}}
=== 1 implies 2 === Follows immediately from Topological Closure is Closed. {{qed|lemma}} === 2 implies 1 === Let $x \in S$ Let $U$ be an open neighborhood of $x$. Let $K \subseteq S$ be a closed compact subspace with $U \subseteq K$. By Topological Closure is Closed $\overline U$ is closed in $T$. By Set Closure as Intersection of Closed Sets, the closure $\overline U$ of $U$ in $T$ satisfies $\overline U \subseteq K$. By Closed Set in Closed Subspace, $\overline U$ is closed in $K$. By Closed Subspace of Compact Space is Compact, $\overline U$ is compact. {{qed}} Category:Locally Compact Spaces sqw7k5g1hdld7o5j88i7ie2xga2e8w5
=== Existence === By induction on the length of $w$. ==== Basis for the induction ==== Let $w$ have length $0$. Then $w$ is the empty word. By Empty Group Word is Reduced, $w$ is reduced. ==== Induction step ==== Let $w$ have length $n \geq 1$. If $w$ is reduced, we are done. Otherwise, let $k \in \{1, \ldots, n-1\}$ be such that $w_{k+1} = w_k^{-1}$. Then $w$ has an elementary reduction of length $n-2$. By the induction hypothesis, $w$ has a reduction. The result follows by induction. {{qed|lemma}} === Uniqueness === By induction on the length of $w$. ==== Basis for the induction ==== Let $w$ have length $0$. By Length of Reduced Form of Group Word is at most Length of Word, a reduced form of $w$ must have length $0$. Thus a reduced form of $w$ can only be $w$ itself. {{qed|lemma}} ==== Induction step ==== Let $w$ have length $n \geq 1$. Let $u$ and $v$ be reduced forms of $w$. Let $u_1$ and $v_1$ be elementary reductions of $w$ with reduced forms $u$ and $v$. By Distinct Elementary Reductions of Group Word have Equal Elementary Reduction, there exists a word $t$ which is an elementary reduction of $u_1$ and $v_1$. Let $s$ be a reduced form of $t$. Then $s$ is a reduced form of $u_1$ and $v_1$. By the induction hypothesis, $s=u=v$. The result follows by induction. {{qed}} Category:Group Words Category:Proofs by Induction g30odvifjgviqb49p320x3p2rdd1lqo
:$\xymatrix { X \ar[d]_\iota \ar[rd]^{\forall f} & \\ F_X \ar[d]_\pi \ar[r]^{\exists ! g} & H \\ F_X^{\mathrm {ab} } \ar[ru]_{\exists ! h} }$ Lef $H$ be any abelian group and $f:X \to H$ be any mapping. By Universal Property of Free Group on Set, there exists a unique group homomorphism $g:F_X \to H$ such that $g \circ \iota = f$. By Universal Property of Abelianization of Group, there exists a unique group homomorphism $h:F_X^{\mathrm {ab} } \to H$ such that $h \circ \pi = g$. Therefore, $h \circ (\pi \circ \iota) = f$. {{Handwaving|Actually we have to show that $\pi \circ \iota$ is the natural inclusion from $X$ to $F_X^{\mathrm {ab} }$...}} Hence, $F_X^{\mathrm {ab} }$ satisfies the Universal Property of Free Abelian Group on Set. Hence $F_X^{\mathrm {ab} }$ is a Free Abelian Group on $X$. {{qed}} Category:Group Theory hq65lur9bypni916vpjsxgl3gmdoo43
Let $T = \struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology. From Real Number Line is Paracompact, $T$ is a paracompact space. From Real Number Line is not Countably Compact, $T$ is not a countably compact space. From Compact Space is Countably Compact, it follows that  $T$ is not a compact space. Hence the result. {{qed}}
Let $T$ be the Dieudonné plank. From Dieudonné Plank is Metacompact, $T$ is a metacompact space. From Dieudonné Plank is not Paracompact, $T$ is not a paracompact space. Hence the result. {{qed}}
From the definition of paracompact space, a paracompact space is also a metacompact space. The result follows from Metacompact Countably Compact Space is Compact. {{qed}}
Let $T = \struct {S, \tau}$ be the canonical minimal Hausdorff non-compact space. This space has been so named on {{ProofWiki}} in order to allow reference to it without needing to describe it whenever it is mentioned. By Canonical Minimal Hausdorff Non-Compact Space is Minimal Hausdorff, $\tau$ is the minimal subset of the power set $\powerset S$ such that $T$ is a Hausdorff space. By Canonical Minimal Hausdorff Non-Compact Space is not Compact, $T$ is not a compact topological space. Hence the result. {{qed}}
Let $T = \struct {S, \tau}$ be the canonical maximal compact non-Hausdorff space. This space has been so named on {{ProofWiki}} in order to allow reference to it without needing to describe it whenever it is mentioned. By Canonical Maximal Compact Non-Hausdorff Space is Maximal Compact, $\tau$ is the maximal subset of the power set $\powerset S$ such that $T$ is compact. By Canonical Maximal Compact Non-Hausdorff Space is not Hausdorff, $T$ is not a  Hausdorff space. Hence the result. {{qed}}
Let $T = \struct {\Z_{\ge 0}, \tau}$ be the topological space formed by the discrete topology on the set of positive integers. Let $T' = \struct {\displaystyle \prod_{\alpha \mathop \in \Z_{\ge 0} } \struct {\Z_{\ge 0}, \tau}_\alpha, \tau'}$ be the countable Cartesian product of $\struct {\Z_{\ge 0}, \tau}$ indexed by $\Z_{\ge 0}$ with the Tychonoff topology $\tau'$. From Countable Discrete Space is Sigma-Compact, $T$ is $\sigma$-compact. From Countable Product of Countable Discrete Spaces is not Sigma-Compact, $T'$ is not $\sigma$-compact. Hence the result. {{qed}}
Let $\mathbb I$ denote the closed unit interval. Let $T = \struct {\mathbb I, \tau}$ be the topological space consisting of $\mathbb I$ under the usual (Euclidean) topology. Let $T' = \mathbb I^{\mathbb I} = \struct {\displaystyle \prod_{\alpha \mathop \in \mathbb I} \struct {\mathbb I, \tau}_\alpha, \tau'}$ be the uncountable Cartesian product of $\struct{\mathbb I, \tau}$ indexed by $\mathbb I$ with the Tychonoff topology $\tau'$. From Closed Real Interval is Sequentially Compact, $T$ is  sequentially compact. {{explain|The above has not been proven yet, but I think it is: $\mathbb I$ is a metric space, and a compact metric space is sequentially compact.}} But from Uncountable Cartesian Product of Closed Unit Interval is not Sequentially Compact, $T$ is not a sequentially compact space. Hence the result. {{qed}}
Let $T$ denote the Novak space. Let $T \times T$ denote the Cartesian product of the Novak space with itself under the product topology. From Novak Space is Countably Compact, $T$ is a countably compact space. But from Cartesian Product of Novak Spaces is not Countably Compact, $T \times T$ is not a countably compact space. Hence the result. {{qed}}
Let $T = \struct {\Z_{\ge 0}, \tau}$ be the topological space formed by the discrete topology on the set of positive integers. Let $T' = \struct {\displaystyle \prod_{\alpha \mathop \in \Z_{\ge 0} } \struct {\Z_{\ge 0}, \tau}_\alpha, \tau'}$ be the countable Cartesian product of $\struct {\Z_{\ge 0}, \tau}$ indexed by $\Z_{\ge 0}$ with the Tychonoff topology $\tau'$. From Discrete Space is Strongly Locally Compact, $T$ is strongly locally compact. From Strongly Locally Compact Space is Weakly Locally Compact, $T$ is weakly locally compact. From Countable Product of Countable Discrete Spaces is not  Weakly Locally Compact, $T'$ is not weakly locally compact. Hence the result. {{qed}}
Let $T$ be the Sorgenfrey line. Let $T' = T \times T$ be Sorgenfrey's half-open square topology. From Sorgenfrey Line is Lindelöf, $T$ is a Lindelöf space. From Sorgenfrey's Half-Open Square Topology is Not Lindelöf, $T'$ is not a Lindelöf space. Hence the result. {{qed}}
Let $T = \struct {\Z_{\ge 0}, \tau}$ denote the topological space consisting of the set of positive integers $\Z_{\ge 0}$ under the discrete topology. Let $I$ be an indexing set with uncountable cardinality. Let $T' = \struct {\displaystyle \prod_{\alpha \mathop \in \mathbb I} \struct {\Z_{\ge 0}, \tau}_\alpha, \tau'}$ be the uncountable Cartesian product of $\struct {\Z_{\ge 0}, \tau}$ indexed by $I$ with the Tychonoff topology $\tau'$. From Discrete Space is First-Countable, $T$ is a first-countable space. But from Uncountable Cartesian Product of Discrete Topology on Positive Integers is not First-Countable, $T$ is not a first-countable space. Hence the result. {{qed}}
Let $T = \struct {\Z_{\ge 0}, \tau}$ denote the topological space consisting of the set of positive integers $\Z_{\ge 0}$ under the discrete topology. Let $I$ be an indexing set with uncountable cardinality. Let $T' = \struct {\displaystyle \prod_{\alpha \mathop \in I} \struct {\Z_{\ge 0}, \tau}_\alpha, \tau'}$ be the uncountable Cartesian product of $\struct {\Z_{\ge 0}, \tau}$ indexed by $I$ with the Tychonoff topology $\tau'$. From Countable Discrete Space is Second-Countable, $T$ is a second-countable space. But from Uncountable Cartesian Product of Discrete Topology on Positive Integers is not Second-Countable, $T$ is not a second-countable space. Hence the result. {{qed}}
Let $T = \struct {\Z_{\ge 0}, \tau}$ denote the topological space consisting of the set of positive integers $\Z_{\ge 0}$ under the discrete topology. Let $I$ be an indexing set with uncountable cardinality. Let $T' = \struct {\displaystyle \prod_{\alpha \mathop \in I} \struct {\Z_{\ge 0}, \tau}_\alpha, \tau'}$ be the uncountable Cartesian product of $\struct {\Z_{\ge 0}, \tau}$ indexed by $I$ with the Tychonoff topology $\tau'$. From Countable Discrete Space is Separable, $T$ is a separable space. But from Uncountable Cartesian Product of Discrete Topology on Positive Integers is not Separable, $T$ is not a separable space. Hence the result. {{qed}}
Let $T$ be the Sorgenfrey line. Let $T' = T \times T$ be Sorgenfrey's half-open square topology. From Sorgenfrey Line is Paracompact Space, $T$ is a paracompact space. From Sorgenfrey's Half-Open Square Topology is Not Paracompact Space, $T'$ is not a paracompact space. Hence the result. {{qed}}
Let $T$ be the Sorgenfrey line. Let $T' = T \times T$ be Sorgenfrey's half-open square topology. From Sorgenfrey Line is Metacompact Space, $T$ is a metacompact space. From Sorgenfrey's Half-Open Square Topology is Not Metacompact Space, $T'$ is not a metacompact space. Hence the result. {{qed}}
We have Open Continuous Image of Paracompact Space is not always Countably Metacompact. We also have: :Paracompact Space is Metacompact :Metacompact Space is Countably Metacompact Hence the result. {{qed}}
We have that: :$H^- \cup \left({S \setminus H}\right)^- \subseteq S$ by definition of $S$. From Union with Relative Complement: :$H \cup \left({S \setminus H}\right) = S$ From Set is Subset of its Topological Closure: {{begin-eqn}} {{eqn | l = H       | o = \subseteq       | r = H^- }} {{eqn | l = \left({S \setminus H}\right)       | o = \subseteq       | r = \left({S \setminus H}\right)^- }} {{end-eqn}} From Set Union Preserves Subsets: :$H \subseteq H^-, \left({S \setminus H}\right) \subseteq \left({S \setminus H}\right)^- \implies H \cup \left({S \setminus H}\right) \subseteq H^- \cup \left({S \setminus H}\right)^-$ which means: :$S \subseteq H^- \cup \left({S \setminus H}\right)^-$ The result follows by definition of equality of sets. {{qed}} Category:Set Closures Category:Set Union Category:Relative Complement 0z90hn0o58uhcg3xuvrxbt5mkyy90y1
Note that from Clopen Set contains Components of All its Points: :$\map {\operatorname {Comp}_x} T \subseteq K_x$ It remains to be demonstrated that it is not always the case that $K_x \subseteq \map {\operatorname {Comp}_x} T$. Let $T$ be the nested rectangle space in the Euclidean plane. Let $L_1$ and $L_2$ be the boundary lines of $T$. Let $x \in L_1$. From Boundary Line in Nested Rectangle Space is Component $L_1$ is a component of $T$. That is: :$L_1 = \map {\operatorname {Comp}_x} T$  From Union of Boundary Lines in Nested Rectangle Space is Quasicomponent, $L_1 \cup L_2$ is a quasicomponent of $T$. By Quasicomponent is Intersection of Clopen Sets: :$L_1 \cup L_2 = \displaystyle \bigcap_{x \mathop \in K} K$ clopen in $T$ Thus, while $\map {\operatorname {Comp}_x} T \subseteq L_1 \cup L_2$, it is not the case that $L_1 \cup L_2 \subseteq \map {\operatorname {Comp}_x} T$. Hence the result. {{qed}}
By definition of clopen, $H$ is open in $T$. By definition of closed set, $\relcomp S H$ is closed in $T$. By definition of clopen, $H$ is closed in $T$.  By definition of closed set, $\relcomp S H$ is open in $T$. Thus $\relcomp S H$ is both open in $T$ and closed in $T$. Hence the result, by definition of clopen set. {{qed}} Category:Relative Complement Category:Clopen Sets hqzjrlvf07mzscg6air2zsnp7pojz8q
By Set with Relative Complement forms Partition, $H$ and $\relcomp S H$ form a partition of $S$. By Complement of Clopen Set is Clopen, $\relcomp S H$ is also a clopen set of $T$. By definition of clopen set, both $H$ and $\relcomp S H$ are open in $T$. Thus $H$ and $\relcomp S H$ are a pair of open sets in $T$ forming a partition of $S$. Hence the result, by definition of separation. {{qed}} Category:Clopen Sets Category:Separations mtqtkd0juvgm6ca51szmsyuesx1qhac
Let $T = \struct {S, \tau_p}$ be a finite particular point space. From Particular Point Space is Path-Connected, $T$ is path-connected. Therefore $S$ is a path component in $T$. But from Particular Point Space is not Arc-Connected, $T$ is not arc-connected. Therefore $S$ is not an arc component in $T$. Hence the result. {{qed}}
Let $C$ be the closed topologist's sine curve embedded in the real Euclidean plane. From Closed Topologist's Sine Curve is Connected, $C$ is connected in $T$ Therefore $C$ is a component in the subspace of $T$ induced by $C$. From Closed Topologist's Sine Curve is not Path-Connected, $C$ is not path-connected. Therefore $C$ is not a path component in the subspace of $T$ induced by $C$. Hence the result. {{qed}}
From Component of Point is not always Intersection of its Clopen Sets, the set intersection of the clopen sets containing a point $x$ may not always be contained in the component of $x$. The result follows from Quasicomponent is Intersection of Clopen Sets. {{qed}}
Let $(a_0, a_1, \ldots)$ be its continued fraction expansion. Let $(p_n)_{n\geq 0}$ and $(q_n)_{n\geq 0}$ be its numerators and denominators. Then $C_n = p_n/q_n$ is the $n$th convergent. By Accuracy of Convergents of Continued Fraction Expansion of Irrational Number, for $n \ge 2$: :$\left|{x - \dfrac {p_n} {q_n}}\right| < \dfrac 1 {q_n q_{n + 1} }$ By Lower Bounds for Denominators of Simple Continued Fraction: :$q_nq_{n+1} \geq n$ for $n \geq 5$. So from Basic Null Sequences and the Squeeze Theorem: : $\dfrac 1 {q_n q_{n+1} } \to 0$ as $n \to \infty$. Thus $C_n = p_n/q_n$ converges to $x$. That is, $(a_0, a_1, \ldots)$ converges to $x$. {{qed}}
By Simple Infinite Continued Fraction Converges to Irrational Number, the value of $(a_n)_{n\geq 0}$ exists and is irrational. Let $(b_n)_{n\geq 0}$ be its continued fraction expansion. By Continued Fraction Expansion of Irrational Number Converges to Number Itself, $(a_n)_{n\geq 0}$ and $(b_n)_{n\geq 0}$ have the same value. The result will be achieved by the Second Principle of Mathematical Induction. First we note that if $\left[{a_0, a_1, a_2, \ldots}\right] = \left[{b_0, b_1, b_2, \ldots}\right]$ then $a_0 = b_0$ since both are equal to the integer part of the common value. {{explain|a result proving the above}} This is our basis for the induction. Now suppose that for some $k \ge 1$, we have: :$a_0 = b_0, a_1 = b_1, \ldots, a_k = b_k$. Then all need to do is show that $a_{k+1} = b_{k+1}$. Now: :$\left[{a_0, a_1, a_2, \ldots}\right] = \left[{a_0, a_1, \ldots, a_k, \left[{a_{k+1}, a_{k+2}, \ldots}\right]}\right]$ and similarly :$\left[{b_0, b_1, b_2, \ldots}\right] = \left[{b_0, b_1, \ldots, b_k, \left[{b_{k+1}, b_{k+2}, \ldots}\right]}\right]$. {{explain|this needs to be proved}} As these have the same value and have the same first $k$ partial quotients, it follows that: :$\left[{a_{k+1}, a_{k+2}, \ldots,}\right] = \left[{b_{k+1}, b_{k+2}, \ldots}\right]$. But now $a_{k+1} = b_{k+1}$ as each is equal to the integer part of the value of this simple infinite continued fraction. Hence the result. {{qed}}
Note that indeed a Simple Infinite Continued Fraction Converges to Irrational Number. The result follows from: : Continued Fraction Expansion of Irrational Number Converges to Number Itself : Continued Fraction Expansion of Limit of Simple Infinite Continued Fraction equals Expansion Itself {{qed}}
Let $T = \struct {S, \tau}$ be an irreducible space by Definition 3. That is, let every non-empty open set in $T$ have a non-empty intersection with every other non-empty  open set. Let $U \in \tau$ be open in $T$ such that $U \ne \O$. If $U = S$ then $U^- = S$ trivially. So, let $U \ne S$. Let $x \in S$ such that $x \ne U$. Let $V$ be any non-empty open set of $T$ such that $x \in V$. By hypothesis, $U \cap V \ne \O$. Therefore $V$ contains some point of $U$ distinct from $x$. $V$ is arbitrary, so it follows that every open set $V$ of $T$ such that $x \in V$ contains some point of $U$ distinct from $x$. That is, $x$ is a limit point of $U$. So every point of $T$ is a limit point of $U$. Thus, by definition, every point of $T$ is in the closure of $U$. The argument reverses directly.
Let $T = \struct {S, \tau}$ be a topological space which is ultraconnected. From Ultraconnected Space is Path-Connected, $T$ is path-connected. The result follows from Path-Connected Space is Connected. {{qed}}
Let $T$ be a countable finite complement space. From Finite Complement Space is Irreducible, $T$ is an irreducible space. From Countable Finite Complement Space is not Path-Connected, $T$ is not path-connected. Hence the result. {{qed}}
Let $T$ be an excluded point space. From Excluded Point Space is Ultraconnected, $T$ is an ultraconnected space. From Excluded Point Space is not Arc-Connected, $T$ is not arc-connected. Hence the result. {{qed}}
Note that indeed Finite Simple Continued Fraction has Rational Value. The result follows from: * Value of Continued Fraction Expansion of Rational Number equals Number Itself * Continued Fraction Expansion of Value of Simple Finite Continued Fraction equals Expansion Itself {{qed}}
Let $T$ be the Arens-Fort space. From Arens-Fort Space is not Locally Connected, $T$ is not a locally connected space. The result follows from Components and Quasicomponents of Arens-Fort Space are Equal. {{qed}}
By Power Set of Finite Set is Finite, the power set $\mathcal P \left({S}\right)$ is finite. By Subset of Finite Set is Finite, $\tau \subseteq \mathcal P \left({S}\right)$ is finite. The result follows from Irreducible Space with Finitely Many Open Sets is Path-Connected. {{qed}}
Let $T$ be a discrete topological space with more than $1$ point. From Discrete Space is Locally Connected, $T$ is a locally connected space. From Non-Trivial Discrete Space is not Connected, $T$ is not a connected space. Hence the result. {{qed}}
Let $C$ be the closed topologist's sine curve embedded in the real Euclidean plane. From Closed Topologist's Sine Curve is Connected, $C$ is connected in $T$ From Closed Topologist's Sine Curve is not Locally Connected, $C$ is not locally connected. Hence the result. {{qed}}
Follows immediately from: * Irreducible Space with Finitely Many Open Sets has Generic Point * Topological Space with Generic Point is Path-Connected {{qed}}
By Path-Connectedness is Equivalence Relation, it suffices to prove that every point is path-connected with $g$. Let $x \in S$. Define a path $\gamma: \closedint 0 1 \to S$ by: :$\map \gamma t = \begin{cases} x & : t \le \dfrac 1 2 \\ g & : t > \dfrac 1 2 \end{cases}$ We show that $\gamma$ is indeed continuous. Let $U \subseteq S$ be open and non-empty. Because $g$ is a generic point, $g \in U$. If $x \in U$, then its preimage $\gamma^{-1} \sqbrk U = \closedint 0 1$ is open. If $x \notin U$, then its preimage $\gamma^{-1} \sqbrk U = \hointl {\dfrac 1 2} 1$ is also open. Thus $x$ and $g$ are path-connected. {{qed}} Category:Path-Connected Spaces 15rjj65wcrgh4ibiu4mwgszydj02m6b
==== Length 0 ==== If $n = 0$, we have $x = \sqbrk {a_0, \ldots, a_n} = a_0$ by definition of value. By Real Number is Integer iff equals Floor: :$\floor x = \floor {a_0} = a_0$ ==== Length 1 ==== If $n = 1$, then: :$x = \sqbrk {a_0, a_1} = a_0 + \dfrac 1 {a_1}$ If $a_1 > 1$, then: :$a_0 < a_0 + \dfrac 1 {a_1} < a_0 + 1$ By definition of floor function: :$\floor x = a_0$ If $a_1 = 1$, then $x = a_0 + 1$. By Real Number is Integer iff equals Floor: :$\floor x = a_0 + 1$ ==== Length at least 2 ==== By Value of Finite Continued Fraction of Real Numbers is at Least First Term: :$\sqbrk {a_0, \ldots, a_n} > a_0$ :$\sqbrk {a_1, \ldots, a_n} > a_1 \ge 1$ Thus: :$a_0 < \sqbrk {a_0, \ldots, a_n} = a_0 + \dfrac 1 {\sqbrk {a_1, \ldots, a_n} } < a_0 + 1$ By definition of floor function: :$\floor x = a_0$ {{qed}}
If $n=0$, we have $x = [a_0] = a_0$ by definition of value. Let $n>0$. By definition of value: :$[a_0, a_1, \ldots, a_n] = a_0 + \dfrac 1 {[a_1, a_2, \ldots, a_n]}$ By Value of Finite Continued Fraction of Strictly Positive Real Numbers is Strictly Positive: :$[a_1, a_2, \ldots, a_n] > 0$. Thus  :$[a_0, a_1, \ldots, a_n] = a_0 + \dfrac 1 {[a_1, a_2, \ldots, a_n]} > a_0$ {{qed}}
This will be proved by induction on the number of partial quotients. For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition that the continued fraction $\sqbrk {a_0, a_1, \ldots, a_n}$ has a rational value. === Basis for the Induction === $P(0)$ is true, as $\sqbrk {a_0}$ is an integer and therefore rational. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: The continued fraction $\sqbrk {a_0, a_1, \ldots, a_k}$ has a rational value. Then we need to show that the continued fraction $\sqbrk {a_0, a_1, \ldots, a_k, a_{k + 1} }$ also has a rational value. === Induction Step === Consider the continued fraction $\sqbrk {a_0, a_1, \ldots, a_k, a_{k + 1} }$. By definition of value of a finite continued fraction: :$\sqbrk {a_0, a_1, \ldots, a_k, a_{k + 1} } = a_0 + \dfrac 1 {\sqbrk {a_1, a_2, \ldots, a_k, a_{k + 1} } }$ $a_0$ is an integer, as we have seen. By the induction hypothesis, $\sqbrk {a_1, a_2, \ldots, a_k, a_{k + 1} }$ is a rational number, and so is its reciprocal. Hence as $a_0 + \dfrac 1 {\sqbrk {a_1, a_2, \ldots, a_k, a_{k + 1} } }$ is the sum of two rational numbers, it is itself rational. So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore for all $n \in \N^*$, the continued fraction $\sqbrk {a_0, a_1, \ldots, a_n}$ has a rational value. {{qed}}
We show that either: :$x \in \left[ C_n..C_{n+1}\right]$ or :$x \in \left[ C_{n+1}..C_n\right]$ so that the result follows from: :Difference between Adjacent Convergents of Simple Continued Fraction :Distance between Point of Real Interval and Endpoint is at most Length ==== Odd case ==== Let $n \geq 1$ be odd. By Limit of Subsequence equals Limit of Sequence: :$x = \displaystyle\lim_{k \to\infty} C_{2k}$ For all $2k \geq n+1$, by: :Even Convergents of Simple Continued Fraction are Strictly Increasing :Even Convergent of Simple Continued Fraction is Strictly Smaller than Odd Convergent we have $C_{n+1} \leq C_{2k} < C_n$. By Lower and Upper Bounds for Sequences, $x \in \left[ C_{n+1}..C_n\right]$. ==== Even case ==== Let $n \geq 0$ be even. By Limit of Subsequence equals Limit of Sequence: :$x = \displaystyle\lim_{k \to\infty} C_{2k+1}$ For all $2k+1 \geq n+1$, by: :Odd Convergents of Simple Continued Fraction are Strictly Decreasing :Even Convergent of Simple Continued Fraction is Strictly Smaller than Odd Convergent we have $C_{n} < C_{2k+1} \leq C_{n+1}$. By Lower and Upper Bounds for Sequences, $x \in \left[ C_n..C_{n+1}\right]$. {{qed}}