hcju/proofwikiqa huggingface dataset:
Follows directly from: :Particular Point Topology is Closed Extension Topology of Discrete Topology :Limit Points in Subset of Closed Extension Space {{qed}}
Every open set of $T = \struct {S, \tau_p}$ except $\O$ contains the point $p$ by definition. So every open set $U \in \tau_p$ such that $x \in U$ contains $p$. So by definition of the limit point of a set, $x$ is a limit point of $U$. {{qed}}
Let $\epsilon$ be a positive real number. Since $f$ is absolutely continuous, there exists real $\delta_1 > 0$ such that for all sets of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with:  :$\displaystyle \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta_1$ we have:  :$\displaystyle \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \frac \epsilon 2$ Similarly, since $g$ is absolutely continuous, there exists real $\delta_2 > 0$ such that whenever:  :$\displaystyle \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta_2$ we have:  :$\displaystyle \sum_{i \mathop = 1}^n \size {\map g {b_i} - \map g {a_i} } < \frac \epsilon 2$ Let:  :$\delta = \map \min {\delta_1, \delta_2}$ Then, for all sets of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with:  :$\displaystyle \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$ we have:  :$\displaystyle \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \frac \epsilon 2$ and: :$\displaystyle \sum_{i \mathop = 1}^n \size {\map g {b_i} - \map g {a_i} } < \frac \epsilon 2$ We then have: {{begin-eqn}} {{eqn	| l = \sum_{i \mathop = 1}^n \size {\map {\paren {f + g} } {b_i} - \map {\paren {f + g} } {a_i} } 	| r = \sum_{i \mathop = 1}^n \size {\paren {\map f {b_i} - \map f {a_i} } + \paren {\map g {b_i} - \map g {a_i} } } }} {{eqn	| o = \le 	| r = \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } + \sum_{i \mathop = 1}^n \size {\map g {b_i} - \map g {a_i} } 	| c = Triangle Inequality for Real Numbers }} {{eqn	| o = < 	| r = \frac \epsilon 2 + \frac \epsilon 2 }} {{eqn	| r = \epsilon }} {{end-eqn}} whenever:  :$\displaystyle \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$ Since $\epsilon$ was arbitrary: :$f + g$ is absolutely continuous. {{qed}} Category:Absolutely Continuous Functions nqfoj663e9mkgdcghrotrjnsi82vvuw
{{begin-eqn}} {{eqn | l = \int_0^\infty p \lambda^{p - 1} \int_{E_\lambda} \size f^r \rd \mu \rd \lambda       | r = \int_0^\infty \int_{E_\lambda} p \lambda^{p - 1} \size f^r \rd \mu \rd \lambda }} {{eqn | r = \int_X \int_0^{\size {\map f x} } p \lambda^{p - 1} \size f^r \rd \lambda \rd \mu       | c = by Tonelli's Theorem }} {{eqn | r = \int_X \size f^r \int_0^{\size {\map f x} } p \lambda^{p - 1} \rd \lambda \rd \mu }} {{eqn | r = \int_X \size f^r \size f^p \rd \mu       | c = by Integral of Power }} {{eqn | r = \int_X \size f^{p + r} \rd \mu }} {{end-eqn}} We have that for any measurable $A \in \Sigma$: :$\map \mu A = \displaystyle \int_A 1 \rd \mu$ Therefore, for $\lambda > 0$: :$\map \mu {E_\lambda} = \displaystyle \int_{E_\lambda} 1 \rd \mu$ which can also be written: :$\map \mu {E_\lambda} = \displaystyle \int_{E_\lambda} \size f^0 \rd \mu$ Therefore, taking $r = 0$ in the above, we obtain: {{begin-eqn}} {{eqn | l = \int_0^\infty p \lambda^{p - 1} \map m \lambda \rd \lambda       | r = \int_0^\infty p \lambda^{p - 1} \int_{E_\lambda} \size f^0 \rd \mu \rd \lambda       | c =  }} {{eqn | r = \int_X \size f^p \rd \mu       | c =  }} {{end-eqn}} {{qed}} Category:Measure Theory e0w1gsd8i9iaj9udgcw28lqo0zizs99
Every open set of $T^*_p = \struct {S^*_p, \tau^*_{\bar p} }$ except $S^*_p$ does not contain the point $p$ by definition. So every open set $U \in \tau^*_{\bar p}$ such that $p \in U$ (there is only the one such open set) contains $x$. So by definition of the limit point of a set, $p$ is a limit point of $U$. {{qed}}
Since the derivative of $f$ is bounded, there exists some $M \in \R_{> 0}$ such that:  :$\size {\map {f'} x} \le M$ for all $x \in \openint a b$. Let $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq \closedint a b$ be a collection of disjoint closed real intervals. Note that for each $i \in \set {1, 2, \ldots, n}$: :$f$ is continuous on $\closedint {a_i} {b_i}$ and differentiable on $\openint {a_i} {b_i}$. So, by the Mean Value Theorem, for each $i$ there exists some $\xi_i \in \openint {a_i} {b_i}$ such that: :$\map {f'} {\xi_i} = \dfrac {\map f {b_i} - \map f {a_i} } {b_i - a_i}$ We then have: {{begin-eqn}} {{eqn	| l = \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } 	| r = \sum_{i \mathop = 1}^n \size {\map {f'} {\xi_i} } \paren {b_i - a_i} }} {{eqn	| o = \le 	| r = M \sum_{i \mathop = 1}^n \paren {b_i - a_i} 	| c = since $\xi_i \in \openint a b$, we have $\size {\map {f'} {\xi_i} } \le M$ }} {{end-eqn}} Let $\epsilon$ be a positive real number.  Then for all collections of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq \closedint a b$ with: :$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \frac \epsilon M$ we have: :$\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \epsilon$ Since $\epsilon$ was arbitrary: :$f$ is absolutely continuous. {{qed}} Category:Absolutely Continuous Functions 0fg9ljkpexbqj5lldwzjnp4rnf0t3it
Any open cover $\CC$ of $T$ must contain an open set of $T$ which contains $0$. So $\openint {-1} 1$ will always be covered by one set in $\CC$, leaving just $-1$ and $1$ possibly needing to be included in at most two other sets. So $\CC$ has a subcover containing at most three sets. Hence $T$ is a compact space by definition. {{qed}}
We have: :Compact Space is Sigma-Compact :Sigma-Compact Space is Lindelöf Hence the result. {{qed}} Category:Compact Spaces Category:Lindelöf Spaces rq8ho9t43qo43ua1sdqq3ndsddg55j6
{{questionable|circular reasoning}} Let $x_0$ be an irrational number. We seek $a_0, a_1, \ldots \in \Z$ such that $x_0 = \left[{a_0, a_1, \ldots}\right]$. From Bound for Difference of Limit of Simple Infinite Continued Fraction with Convergent: :$x_1$ lies strictly between any successive pair of its convergents. So, for a start, it has to lie between $C_0 = a_0$ and $C_1 = a_0 + \dfrac 1 {a_1}$. In particular, as $a_1 \ge 1$, we know that $a_0 < x_0 < a_0 + 1$. So $a_0 = \left \lfloor {x_0} \right \rfloor$ where $\left \lfloor {x_0} \right \rfloor$ is the floor function of $x_0$. We note, in particular, that $a_0$ is therefore determined by $x_0$ alone. Now we write: :$x_0 = \left \lfloor {x_0} \right \rfloor + \left\{{x_0}\right\}$ where $\left\{{x_0}\right\}$ is the fractional part of $x_0$. Then: :$x_0 = a_0 + \cfrac 1 {a_1 + \cfrac 1 {a_2 + \cfrac 1 {\ddots}}} = \left \lfloor {x_0} \right \rfloor + \cfrac 1 {a_1 + \cfrac 1 {a_2 + \cfrac 1 {\ddots}}} = \left \lfloor {x_0} \right \rfloor + \dfrac 1 {\left[{a_1, a_2, a_3, \ldots}\right]}$ From Real Number minus Floor: : $0 \le \left\{{x_0}\right\} < 1$ But because $x_0$ is irrational, $\left\{{x_0}\right\} \ne 0$. So $0 < \left\{{x_0}\right\} < 1$. So: $\left[{a_1, a_2, a_3, \ldots}\right] = \dfrac 1 {x_0 - \left \lfloor {x_0} \right \rfloor} = \dfrac 1 {\left\{{x_0}\right\}}$. Now we write $x_1 = \dfrac 1 {\left\{{x_0}\right\}}$. Then $x_1 = \left[{a_1, a_2, a_3, \ldots}\right]$. As $\left\{{x_0}\right\} < 1$ we have that $x_1 = \dfrac 1 {\left\{{x_0}\right\}} > 1$. So $x_1$ is an irrational number greater than $a_1$ which is positive. Repeating the argument leads to $a_1 = \left \lfloor {x_1}\right \rfloor$ and so $a_1$ is determined uniquely from $x_1$ and hence from $x_0$. In the same way, $x_2 = \dfrac 1 {\left\{{x_1}\right\}}$ and so $x_2 = \left[{a_2, a_3, a_4, \ldots}\right]$. And so on. {{qed}}
From Absolutely Continuous Real Function is Continuous: :$f$ and $g$ are continuous. From Closed Real Interval is Compact in Metric Space: :$\closedint a b$ is compact. Therefore, by Continuous Function on Compact Subspace of Euclidean Space is Bounded: :$f$ and $g$ are bounded. That is, there exists $M_f, M_g \in \R_{> 0}$ such that:  :$\size {\map f x} \le M_f$ :$\size {\map g x} \le M_g$ for all $x \in \closedint a b$. Let $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq \closedint a b$ be a set of disjoint closed real intervals. Then: {{begin-eqn}} {{eqn	| l = \sum_{i \mathop = 1}^n \size {\map {\paren {f \times g} } {b_i} - \map {\paren {f \times g} } {a_i} } 	| r = \sum_{i \mathop = 1}^n \size {\map f {b_i} \map g {b_i} - \map f {a_i} \map g{a_i} + \map f {a_i} \map g {b_i} - \map f {a_i} \map g {b_i} } }} {{eqn	| r = \sum_{i \mathop = 1}^n \size {\map g {b_i} \paren {\map f {b_i} - \map f {a_i} } + \map f {a_i} \paren {\map g {b_i} - \map g {a_i} } } }} {{eqn	| o = \le 	| r = \sum_{i \mathop = 1}^n \size {\map g {b_i} } \size {\map f {b_i} - \map f {a_i} } + \sum_{i \mathop = 1}^n \size {\map f {a_i} } \size {\map g {b_i} - \map g {a_i} } 	| c = Triangle Inequality for Real Numbers }} {{eqn	| o = \le 	| r = M_g \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } + M_f \sum_{i \mathop = 1}^n \size {\map g {b_i} - \map g {a_i} } 	| c = since $a_i, b_i \in \closedint a b$ }} {{end-eqn}} Let $\epsilon$ be a positive real number. Since $f$ is absolutely continuous, there exists $\delta_1 > 0$ such that for all sets of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq \closedint a b$ with:  :$\displaystyle \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta_1$ we have:  :$\displaystyle \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \frac \epsilon {2 M_g}$ Similarly, since $g$ is absolutely continuous, there exists $\delta_2 > 0$ such that whenever: :$\displaystyle \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta_2$ we have: :$\displaystyle \sum_{i \mathop = 1}^n \size {\map g {b_i} - \map g {a_i} } < \frac \epsilon {2 M_f}$ Let: :$\delta = \map \min {\delta_1, \delta_2}$ Then, whenever: :$\displaystyle \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$ We have both: :$\displaystyle \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \frac \epsilon {2 M_g}$ :$\displaystyle \sum_{i \mathop = 1}^n \size {\map g {b_i} - \map g {a_i} } < \frac \epsilon {2 M_f}$ and hence:  {{begin-eqn}} {{eqn	| l = \sum_{i \mathop = 1}^n \size {\map {\paren {f \times g} } {b_i} - \map {\paren {f \times g} } {a_i} } 	| o = < 	| r = M_g \times \frac \epsilon {2 M_g} + M_f \times \frac \epsilon {2 M_f} }} {{eqn	| r = \epsilon }} {{end-eqn}} Since $\epsilon$ was arbitrary, we have: :$f \times g$ is absolutely continuous. {{qed}} Category:Absolutely Continuous Functions 1dzalwwopk9m4yyu8d8dex5ma42gck3
Proof by Counterexample: Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology. Let $\sequence {a_n}$ be the sequence defined as: {{begin-eqn}} {{eqn | l = \sequence {a_n}       | r = \begin {cases} 1 & : \text {$n$ odd} \\ n / 2 & : \text {$n$ even} \end {cases}       | c =  }} {{eqn | r = \sequence {1, 1, 1, 2, 1, 3, 1, 4, \dotsc}       | c =  }} {{end-eqn}} Then $\sequence {a_n}$ has exactly one accumulation point, that is $1$. However, $1$ is not a limit point of $\sequence {a_n}$, as $\sequence {a_n}$ has no limit point. Hence the result. {{qed}}
From Closure of Intersection is Subset of Intersection of Closures, it is seen that it is always the case that: :$\paren {H_1 \cap H_2}^- \subseteq {H_1}^- \cap {H_2}^-$ It remains to be shown that it does not always happen that: :$\paren {H_1 \cap H_2}^- = {H_1}^- \cap {H_2}^-$ The result is demonstrated by Proof by Counterexample.
Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number. Then the open interval $\openint {-\epsilon} \epsilon$ contains $0$ and all elements $a_m$ of $S$ such that $0 < \dfrac 1 m < \epsilon$. Hence $\dfrac 1 m \in \openint {-\epsilon} \epsilon$. Hence the result by definition of limit point of $S$. {{qed}}
Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number. Then the open interval $\openint {-\epsilon} \epsilon$ contains $0$ and all elements $a_m$ of $S$ such that $0 < \dfrac 1 m < \epsilon$. We have that: :$\forall n \in \N: n \ge m$ have the property that $0 < \dfrac 1 n < \epsilon$. Hence there are a countably infinite number of terms of $\sequence {a_n}$ such that $a_n \in \openint {-\epsilon} \epsilon$. Hence the result by definition of $\omega$-accumulation point of $S$. {{qed}}
Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number. Then the open interval $\openint {-\epsilon} \epsilon$ contains $0$ and all elements $a_m$ of $S$ such that $0 < \dfrac 1 m < \epsilon$. We have that: :$\forall n \in \N: n \ge m$ have the property that $0 < \dfrac 1 n < \epsilon$. Hence there are a countably infinite number of terms of $\sequence {a_n}$ such that $a_n \in \openint {-\epsilon} \epsilon$. Hence the result by definition of accumulation point of $\sequence {a_n}$. {{qed}}
The open interval $\openint 1 1$ contains $0$, and also contains all terms of $\sequence {a_n}$ with odd indices greater than $1$. However, all terms of $\sequence {a_n}$ with even indices are outside $\openint {-\dfrac 1 2} {\dfrac 1 2}$. Hence $0$ cannot be a limit point of $\sequence {a_n}$. {{qed}}
Proof by induction on $n$: === Basis for the Induction === When $n = 1$, we have: :$\map f x = a x + b$ for some $a, b \in \Z_p$ and $a \ne 0$ Suppose $x_1, x_2 \in \Z_p$ are two roots of $\map f x$. Then: {{begin-eqn}} {{eqn | l = a x_1 + b       | o = \equiv       | r = a x_2 + b       | rr = \equiv 0       | rrr = \pmod p }} {{eqn | ll = \leadsto       | l = a x_1       | o = \equiv       | r = a x_2       | rrr = \pmod p }} {{eqn | ll = \leadsto       | l = x_1       | o = \equiv       | r = x_2       | rrr = \pmod p       | c = since $a \perp p$ }} {{end-eqn}} Hence these two roots must be the same, implying that there is at most $1$ root. This is our base case. === Induction Hypothesis === This is our induction hypothesis: :Any polynomial in one variable of degree $k$ has at most $k$ roots in $\Z_p$. It is to be demonstrated that: :Any polynomial in one variable of degree $k + 1$ has at most $k + 1$ roots in $\Z_p$. === Induction Step === This is our induction step: Consider $n = k + 1$, and let $f$ be a polynomial in one variable of degree $k + 1$. If $f$ does not have a root in $\Z_p$, our claim is satisfied. Hence suppose $f$ does have a root $x_0$. From Ring of Integers Modulo Prime is Field, $\Z_p$ is a field. Applying the Polynomial Factor Theorem, since $\map f {x_0} = 0$: :$\map f x = \paren {x - x_0} \map Q x$ where $Q$ is a polynomial of degree $k$. By Euclid's Lemma for Prime Divisors: :$\map f x = 0 \iff x - x_0 = 0$ or $\map Q x = 0$ By induction hypothesis, $Q$ has at most $k$ roots. Hence $f$ has at most $k + 1$ roots. By the Principle of Mathematical Induction, the theorem is true for any $n$. {{qed}} {{Namedfor|Joseph Louis Lagrange}} Category:Number Theory Category:Polynomial Theory Category:Proofs by Induction 6lpsyjudkseo2swix5o3u7a4nui5874
It suffices to prove the result for prime values of $b$. For $n = 0$, it is clear that $1 + 2^0 = 2$ is not a perfect power. For $n > 0$, $1 + 2^n$ is odd. Hence for the equation to hold $a$ must be odd as well. Writing $a = 2 m + 1$ we have: {{begin-eqn}} {{eqn | l = 1 + 2^n       | r = \paren {2 m + 1}^b }} {{eqn | r = \sum_{i \mathop = 0}^b \binom b i \paren {2 m}^i \paren 1^{b - i}       | c = Binomial Theorem }} {{eqn | r = \sum_{i \mathop = 0}^b \binom b i \paren {2 m}^i }} {{eqn | r = 1 + \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^i       | c = Binomial Coefficient with Zero }} {{eqn | l = 2^n       | r = \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^i }} {{eqn | r = 2 m \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1}       | c = $m \ne 0$ }} {{end-eqn}} Since all factors of $2^n$ are powers of $2$: :$\ds \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1}$ is a power of $2$. But since each summand is non-negative: :$\ds \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1} \ge 2$ we must have $\ds \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1}$ is even. We have: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1}       | r = \binom b 1 + \sum_{i \mathop = 2}^b \binom b i \paren {2 m}^{i - 1} }} {{eqn | r = b + 2 m \sum_{i \mathop = 2}^b \binom b i \paren {2 m}^{i - 2} }} {{eqn | o = \equiv       | r = b       | rr = \pmod 2 }} {{end-eqn}} Therefore we must have $b = 2$, the only even prime. In that case: {{begin-eqn}} {{eqn | l = 2^n       | r = \paren {2 m + 1}^2 - 1 }} {{eqn | r = 4 m^2 + 4 m + 1 - 1 }} {{eqn | r = 4 m \paren {m + 1} }} {{end-eqn}} So $m$ and $m + 1$ are powers of $2$. The only $m$ satisfying this is $1$, giving the solution: {{begin-eqn}} {{eqn | l = a       | r = 2 m + 1       | c =  }} {{eqn | r = 3       | c =  }} {{eqn | l = 2^n       | r = 3^2 - 1       | c =  }} {{eqn | r = 8       | c =  }} {{eqn | ll= \leadsto       | l = n       | r = 3       | c =  }} {{end-eqn}} {{qed}}
Each Fermat number is in the form of $1 + 2^n$ for some $n \in \Z$. This $n$ must also be a power of $2$. From 1 plus Power of 2 is not Perfect Power except 9 we have: :$1 + 2^n = a^b$ has only one solution $\tuple {n, a, b} = \tuple {3, 3, 2}$. But $3$ is not a power of $2$. Hence no Fermat numbers are perfect powers. {{qed}} Category:Fermat Numbers blocil9c4mkehf14is8l2emcyx5njxl
We have: :Closed Ordinal Space is Compact :Compact Space is Countably Compact {{qed}}
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$. From Uncountable Closed Ordinal Space is Countably Compact, $\closedint 0 \Omega$ is a countably compact space. So every sequence in $\hointr 0 \Omega$ has an accumulation point in $\closedint 0 \Omega$. {{LinkWanted|sequence in $\hointr 0 \Omega$ has an accumulation point in $\closedint 0 \Omega$}} But $\Omega$ cannot be an accumulation point of any sequence in $\closedint 0 \Omega$. {{LinkWanted|$\Omega$ cannot be an accumulation point of any sequence in $\closedint 0 \Omega$}} So every sequence in $\hointr 0 \Omega$ has an accumulation point in $\hointr 0 \Omega$. This means that $\hointr 0 \Omega$ is countably compact. {{qed}}
{{AimForCont}} $\hointr 0 \Omega$ is a metacompact space. From Open Ordinal Space is not Compact in Closed Ordinal Space we have that $\hointr 0 \Omega$ is a countably compact space. From Metacompact Countably Compact Space is Compact it follows that $\hointr 0 \Omega$ is a compact space. But from Open Ordinal Space is not Compact in Closed Ordinal Space this contradicts the fact that $\hointr 0 \Omega$ is not a compact space. Hence the result by Proof by Contradiction. {{qed}}
{{AimForCont}} $\hointr 0 \Omega$ is a paracompact space. From Paracompact Space is Metacompact, it follows that $\hointr 0 \Omega$ is a metacompact space. But from Uncountable Open Ordinal Space is not Metacompact this contradicts the fact that $\hointr 0 \Omega$ is not a metacompact space. Hence the result by Proof by Contradiction. {{qed}}
{{AimForCont}} $\hointr 0 \Omega$ is a Lindelöf space. From Ordinal Space is Completely Normal, $\hointr 0 \Omega$ is a completely normal. From Sequence of Implications of Separation Axioms, $\hointr 0 \Omega$ is a $T_3$ space. From Lindelöf $T_3$ Space is Paracompact, it follows that $\hointr 0 \Omega$ is a paracompact space. But this contradicts the fact that from Uncountable Open Ordinal Space is not Paracompact, $\hointr 0 \Omega$ is not a paracompact space. Hence the result by Proof by Contradiction. {{qed}}
{{AimForCont}} $\hointr 0 \Omega$ is a $\sigma$-compact space. From Sigma-Compact Space is Lindelöf, $\hointr 0 \Omega$ is a Lindelöf space. But this contradicts the fact that from Uncountable Open Ordinal Space is not Lindelöf, $\hointr 0 \Omega$ is not a Lindelöf space. Hence the result by Proof by Contradiction. {{qed}}
We have: :Closed Ordinal Space is Compact :Compact Space is Lindelöf {{qed}}
We have: :Closed Ordinal Space is Compact :Compact Space is $\sigma$-Compact {{qed}}
We have that: :Uncountable Open Ordinal Space is First-Countable :Uncountable Open Ordinal Space is Countably Compact The result follows from First-Countable Space is Sequentially Compact iff Countably Compact. {{qed}}
From Sum of Euler Numbers by Binomial Coefficients Vanishes we have: $\forall n \in \Z_{>0}: \displaystyle \sum_{k \mathop = 0}^n \binom {2 n} {2 k} E_{2 k} = 0$ If: :$\forall n \in \Z_{>0}: \displaystyle \sum_{k \mathop = 0}^n \binom {2 n} {2 k} E_{2 k} = 0$ then: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k} + E_{2 n}       | r = 0 }} {{eqn | l = E_{2 n}       | r = - \sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k} }} {{end-eqn}} {{qed}} Category:Euler Numbers 2qn8jp12mmcxnp5wqbpnk8phl58hd20
From Euler's Theorem, we have $a^{\map \phi n} \equiv 1 \pmod n$. Applying Integer to Power of Multiple of Order we see that $\map \phi n$ is a multiple of $c$. {{qed}} Category:Integer to Power of Multiple of Order Category:Number Theory kwzul7e0s2q95w09jdl9y42jzr6ycl5
It is sufficient to prove the result for prime divisors. The general argument for all divisors follows from the argument: :$\paren {a \, 2^c + 1} \paren {b \, 2^c + 1} = a b \, 2^{2 c} + \paren {a + b} \, 2^c + 1 = \paren {a b \, 2^c + a + b} \, 2^c + 1$ So the product of two factors of the form preserves that form. Let $n \in \N$. Let $p$ be a prime divisor of $F_n = 2^{2^n} + 1$. Then we have: :$2^{2^n} \equiv -1 \pmod p$ Squaring both sides: :$2^{2^{n + 1}} \equiv 1 \pmod p$ From Integer to Power of Multiple of Order, the order of $2$ modulo $p$ divides $2^{n + 1}$ but not $2^n$. Therefore it must be $2^{n + 1}$. Hence: {{begin-eqn}} {{eqn | ll = \exists k \in \Z:       | l = \map \phi p       | r = k \, 2^{n + 1}       | c = Corollary to Integer to Power of Multiple of Order }} {{eqn | l = p - 1       | r = k \, 2^{n + 1}       | c = Euler Phi Function of Prime }} {{eqn | l = p       | r = k \, 2^{n + 1} + 1 }} {{end-eqn}} {{qed}}
It is sufficient to prove the result for prime divisors. The general argument for all divisors follows from the argument: :$\paren {a \, 2^c + 1} \paren {b \, 2^c + 1} = a b \, 2^{2 c} + \paren {a + b} \, 2^c + 1 = \paren {a b \, 2^c + a + b} \, 2^c + 1$ So the product of two factors of the form preserves that form. Let $p$ be a prime divisor of $F_n = 2^{2^n} + 1$. From Euler's Result: :$\exists q \in \Z: p = q \, 2^{n + 1} + 1$ Since $n \ge 2$, $q \, 2^{n + 1}$ is divisible by $2^{2 + 1} = 8$. Hence: :$p \equiv 1 \pmod 8$ By Second Supplement to Law of Quadratic Reciprocity: :$\paren {\dfrac 2 p} = 1$ so $2$ is a quadratic residue modulo $p$. Hence: :$\exists x \in \Z: x^2 = 2 \pmod p$ We have shown $2^{2^n} \equiv -1 \pmod p$ and $2^{2^{n + 1} } \equiv 1 \pmod p$. By Congruence of Powers: :$x^{2^{n + 1} } \equiv 2^{2^n} \equiv -1 \pmod p$ :$x^{2^{n + 2} } \equiv 2^{2^{n + 1}} \equiv 1 \pmod p$ From Integer to Power of Multiple of Order, the order of $x$ modulo $p$ divides $2^{n + 2}$ but not $2^{n + 1}$. Therefore it must be $2^{n + 2}$. Hence: {{begin-eqn}} {{eqn | ll= \exists k \in \Z:       | l = \map \phi p       | r = k \, 2^{n + 2}       | c = Corollary to Integer to Power of Multiple of Order }} {{eqn | l = p - 1       | r = k \, 2^{n + 2}       | c = Euler Phi Function of Prime }} {{eqn | l = p       | r = k \, 2^{n + 2} + 1 }} {{end-eqn}} {{qed}}