avewright/proofwiki-math huggingface dataset:
Theorem: Sigma-Compact Space is Lindelöf

Every $\sigma$-compact space is a Lindelöf space.

Proof:
Let $T = \struct {S, \tau}$ be a $\sigma$-compact space.
{{Recall|Lindelöf Space|Lindelöf space}}
{{:Definition:Lindelöf Space}}
{{Recall|Sigma-Compact Space|$\sigma$-compact space}}
{{:Definition:Sigma-Compact Space}}
So by definition of $\sigma$-compact space:
:$T = \bigcup \HH$
where $\HH$ is a countable set of compact subspaces of $T$.
Let $\CC$ be an open cover of $T$.
Because they are compact, each $H \in \HH$ is covered by a finite number of elements of $\CC$.
Hence $T$ is covered by a countable union of a finite number of elements of $\CC$.
Hence $\CC$ has a countable subcover.
Hence the result.
{{qed}}
Theorem: Sequentially Compact Space is Countably Compact

A sequentially compact topological space is also countably compact.

Proof:
{{Recall|Countably Compact Space|countably compact space|index = 3}}
{{:Definition:Countably Compact Space/Definition 3}}
Let $T = \struct {S, \tau}$ be a sequentially compact topological space.
{{Recall|Sequentially Compact Space|sequentially compact space}}
{{:Definition:Sequentially Compact Space}}
The result follows from Limit of Sequence is Accumulation Point.
{{qed}}
{{ACC|Equivalence of Definitions of Countably Compact Space|3}}
Theorem: Sequentially Compact Space is Countably Compact

A sequentially compact topological space is also countably compact.

Proof:
{{Recall|Countably Compact Space|countably compact space|index = 1}}
{{:Definition:Countably Compact Space/Definition 1}}
Let $T = \struct {S, \tau}$ be a sequentially compact topological space.
{{Recall|Sequentially Compact Space|sequentially compact space}}
{{:Definition:Sequentially Compact Space}}
{{AimForCont}} $T$ is not countably compact.
Then there exists a countable open cover $\family {U_n}_{n \mathop \in \N}$ of $T$ which has no finite subcover.
Let:
:$\ds x_n \in S \setminus \bigcup_{i \mathop < n} U_i \ne \O$
for all $n \in \N$.
Then for arbitrary $p \in S$:
:$p \in U_n$
for at least some $n$.
Hence no infinite subsequence of $\sequence {x_n}$ converges to $p$.
This contradicts the fact that $T$ is sequentially compact.
Hence the result by Proof by Contradiction.
{{qed}}
Theorem: Countably Compact Space is Weakly Countably Compact

Every countably compact space is a weakly countably compact space.

Proof:
Let $T = \struct {S, \tau}$ be a countably compact space.
{{Recall|Weakly Countably Compact Space|weakly countably compact}}
{{:Definition:Weakly Countably Compact Space}}
{{Recall|Countably Compact Space|countably compact space|index = 1}}
{{:Definition:Countably Compact Space/Definition 1}}
Let $A \subseteq S$ be countably infinite.
{{AimForCont}} $A$ has no limit points.
By Set with no Limit Point is Closed, $A$ is closed.
By Open Neighborhood of Element of Set with no Limit Points, for every $a \in A$ there exists some open set $U_a \subseteq S$ such that:
:$U_a \cap A = \set a$
Then the set of all $U_a$ together with $S \setminus A$ forms a countable open cover $\UU$ of $S$.
By definition of countably compact space, $\UU$ has a finite subcover.
Thus $A$ may be covered by finitely many $U_a$.
This contradicts the fact that $A$ is (countably) infinite.
Hence by Proof by Contradiction $A$ must have a limit point.
Hence the result.
{{qed}}
Theorem: T1 Space is Weakly Countably Compact iff Countably Compact

Let $T = \struct {S, \tau}$ be a $T_1$ space.
Then $T$ is weakly countably compact {{iff}} $T$ is countably compact.

Proof:
=== Sufficient Condition ===
We have that a Countably Compact Space is Weakly Countably Compact whatever the separation properties.
{{qed|lemma}}
Theorem: Countably Compact Space is Pseudocompact

Let $T = \struct {S, \tau}$ be a countably compact space.
Then $T$ is a pseudocompact space.

Proof:
{{Recall|Pseudocompact Space|pseudocompact space}}
{{:Definition:Pseudocompact Space}}
Let $T = \struct {S, \tau}$ be a countably compact space.
{{Recall|Countably Compact Space|countably compact space}}
{{:Definition:Countably Compact Space/Definition 1}}
Let $f: S \to \R$ be an arbitrary continuous real-valued function.
For $n \in \N$, consider the set:
:$S_n = \set {x \in S: \size {\map f x} \le n}$
Let $\SS$ be the family of subsets of $S$ so defined:
:$\SS = \family {S_n}_{n \mathop \in \N}$
The continuity of $f$ guarantees that:
:$\forall s \in S: \map f s \in \R$
Hence it follows by the Axiom of Archimedes that:
:$\exists m \in \N: m \ge \size {\map f s}$
Let $n = \max \set {\size {\map f s}: s \in S}$
Then $\SS$, as defined above, forms a countable cover of $S$.
But because $T$ is a countably compact space, $S$ has a finite subcover $\CC$ of $\SS$.
Thus:
:$\exists N \in \N: N = \max \set {n \in \N: S_n \in \CC}$
That is:
:$\forall s \in S: \size {\map f s} \le N$
demonstrating that $f$ is a bounded real-valued function on $S$.
As $f$ is arbitrary, the result follows.
{{qed}}
Theorem: Normal Space is Pseudocompact iff Countably Compact

Let $T = \struct {S, \tau}$ be a normal space.
Then $T$ is pseudocompact {{iff}} $T$ is countably compact.

Proof:
Let $T = \struct {S, \tau}$ be a normal space.
By Countably Compact Space is Pseudocompact we already have that if $T$ is countably compact then $T$ is pseudocompact.
It remains to prove that if $T$ is pseudocompact then $T$ is countably compact.
{{AimForCont}} $T$ is not countably compact.
By definition of countably compact space, $S$ contains an infinite subset $H = \set {x_n}$ with no $\omega$-accumulation point in $T$.
Since, by definition of normal, $S$ is a $T_1$ space, $H$ is closed and discrete in the subspace topology.
{{Explain|$H$ is closed and discrete in the subspace topology.}}
By definition, $T$ is pseudocompact {{iff}} every continuous real-valued function on $S$ is bounded.
By definition of normal, we have that $S$ is a $T_4$ space.
Hence the Tietze Extension Theorem guarantees a continuous extension to $S$ of the unbounded continuous mapping $f: H \to \R$ defined by $\map f {x_n} = n$.
This contradicts our hypothesis that $S$ is pseudocompact.
It follows by Proof by Contradiction that $T$ is countably compact.
{{qed}}
Theorem: Countably Compact Lindelöf Space is Compact

Let $T = \struct {S, \tau}$ be a topological space.
Then:
:$T$ is a compact space
{{iff}}:
:$T$ is both:
::a Lindelöf space
::a countably compact space.

Proof:
=== Sufficient Condition ===
Let $T$ be a compact space.
Then from:
:Compact Space is Countably Compact
:Compact Space is Lindelöf
it follows that $T$ is both a Lindelöf space and a countably compact space.
{{qed|lemma}}
Theorem: Big Picard Theorem

Let:
:$U \subset \C$ be a domain
:$z_0 \in U$
:$f: U \setminus \set {z_0} \to \C$ be a holomorphic function with an essential singularity at $z_0$. 
Then $f$ omits at most one complex value $a \in \C$.

Proof:
{{MissingLinks}}
{{tidy|get rid of the digressional material}}
Montel's Theorem is a very close relative of Picard's theorem.
It states that a family of holomorphic functions that all omit the same two values is normal.
(Recall that a family $\FF$ of holomorphic functions $f: U \to \C$ is normal if every sequence in $\FF$ has a subsequence that converges locally uniformly to a holomorphic function or to infinity.)
{{WLOG}}, assume that $U = \Bbb D = \set {z \in \C: \cmod z < 1}$, and that $z_0 = 0$.
Suppose that $f: \Bbb D \setminus \set 0 \to \C$ is an analytic function that omits two different values $a, b \in \C$.
We need to show that $0$ is a pole or a removable singularity of $f$.
If $\map f z \to \infty$ as $z \to 0$, then $0$ is a pole.
So suppose that there is a sequence $z_n \to 0$ such that $\map f {z_n}$ converges to a value $w \ne \infty$.
We may assume that:
:$\cmod {z_n}$ is strictly decreasing
:$\cmod {z_n} < \dfrac 1 2$ for all $n$.
Let $\lambda_n := 2 z_n$.
Consider the sequence of functions:
:$f_n: \Bbb D \setminus \set 0 \to \C \setminus \set {a, b}$
such that:
:$\map {f_n} z := \map f {\lambda_n z}$
By Montel's Theorem (actually a stronger version of Montel's Theorem, also called the Fundamental Normality Test), this family is normal.
So we may assume, passing to a subsequence if necessary, that $f_n$ converges locally uniformly to a holomorphic function or to $\infty$.
Note that:
:$\map {f_n} {\dfrac 1 2} = \map f {z_n} \to w$
so $f_n$ does not converge to $\infty$.
Let us set:
:$\ds K_n := \max_{\cmod z \mathop = 1/2} \cmod {\map {f_n} z} = \max_{\cmod z \mathop = \cmod {z_n} } \cmod {\map f z}$
It follows from the above that
:$\ds K := \sup_{n \mathop \in \N} K_n < \infty$
But by the Maximum Principle:
:$\cmod {\map f z} \le \max \set {K_{n -1}, K_n} \le K$
whenever $\cmod {z_n} \le \cmod z \le \cmod {z_{n - 1} }$.
So $\cmod {\map f z} \le K$ for $\cmod z < \cmod {z_0}$, and hence $0$ is a removable singularity by the Riemann Removable Singularities Theorem.
Theorem: Compact Space is Weakly Locally Compact

Let $T = \struct {S, \tau}$ be a compact topological space.
Then $T$ is a weakly locally compact.

Proof:
Let $T$ be a compact space.
By definition, $T$ is weakly locally compact {{iff}} each point of $S$ has a compact neighborhood in $T$.
Let $x \in S$ be an arbitrary point of $S$.
By Topological Space is Neighborhood of all its Points, $S$ is a neighborhood of $x$.
Thus $x$ has a compact neighborhood in $T$.
Hence the result by definition of weakly locally compact.
{{qed}}
Theorem: Compact Space is Weakly Locally Compact

Let $T = \struct {S, \tau}$ be a compact topological space.
Then $T$ is a weakly locally compact.

Proof:
Follows directly from:
:Compact Space is Weakly $\sigma$-Locally Compact
:a weakly $\sigma$-locally compact space is weakly locally compact by definition.
{{qed}}
Theorem: Completed Riemann Zeta Function has Order One

The completed Riemann zeta function $\xi$ has order at most $1$.

Proof:
We are required to prove that:
:$\map \xi s = \dfrac 1 2 s \paren {s - 1} \pi^{-s/2} \map \Gamma {\dfrac s 2} \map \zeta s \ll \map \exp {\size s^\beta}$
for all $\beta > 1$, where $\ll$ is the order notation.
Note that by the Functional Equation for Riemann Zeta Function, it is sufficient to check this for $\map \Re s \ge \dfrac 1 2$.
We simply check this fact for each factor.
Evidently:
{{begin-eqn}}
{{eqn | l = \frac 1 2 s \paren {s - 1} \pi^{-s/2}
      | o = \ll
      | r = \map \exp {2 \log s - \frac s 2 \log \pi}
}}
{{eqn | o = \ll
      | r = \map \exp {c_1 \size s}
}}
{{end-eqn}}
for some $c_1 > 0$.
For the gamma factor, we have Stirling's Formula for Gamma Function:
:$\ds \log \map \Gamma s = \paren {s - \frac 1 2} \log s - s + \frac {\log 2 \pi} 2 + \sum_{n \mathop = 1}^{d-1} \frac {B_{2 n} } {2 n \paren {2 n - 1}s^{2 n - 1} } + \map \OO {s^{1 - 2 d} }$
This is valid only away from the poles of $\Gamma$ at $s = 0, -1, -2, \ldots$
However, it is assumed that $\map \Re s \ge \dfrac 1 2$, so this is not a problem.
The error term $\map \OO {s^{1 - 2 d} }$ is small for large $s$.
More generally, the largest contribution is the term $\paren {s - \dfrac 1 2} \log s$, so we have:
:$\log \map \Gamma {\dfrac s 2} \ll \size s \log \size s$
That is:
:$\map \Gamma {\dfrac s 2} \ll \map \exp {\size s \log \size s}$
Finally, from Integral Representation of Riemann Zeta Function in terms of Fractional Part, for $\map \Re s > \dfrac 1 2$:
:$\ds \map \zeta s = \frac s {s - 1} - s \int_1^\infty \fractpart x x^{-s - 1} \rd x$
It is seen that for $\map \Re s > \dfrac 1 2$, the integral is bounded, and therefore:
:$\paren {1 - s} \map \zeta s \ll \map \OO {\size s^2} \ll \map \exp {\size s}$
Combining these facts, and using that $\log s \ll s^\epsilon$ for all $\epsilon > 0$ (shown by Upper Bound of Natural Logarithm), we have:
:$\size {\map \xi s} \ll \map \exp {\size s^{1 + \epsilon} }$ 
for all $\epsilon > 0$, and the proof is complete.
{{qed}}
Category:Riemann Zeta Function
lyhe2xq7bhltr7xnf3w3twknshyuf9f
Theorem: Product Equation for Riemann Zeta Function

There exists a constant $B$ such that:
:$\ds \frac {\map {\zeta'} s} {\map \zeta s} = B - \frac 1 {s - 1} + \frac 1 2 \ln \pi - \frac 1 2 \frac {\map {\Gamma'} {s / 2 + 1} } {\map \Gamma {s / 2 + 1} } + \sum_\rho \paren {\frac 1 {s - \rho} + \frac 1 \rho}$
where:
:$\zeta$ is the Riemann zeta function
:$\rho$ runs over the nontrivial zeros of $\zeta$
:$\Gamma$ is the gamma function.

Proof:
Let $\xi$ be the completed Riemann zeta function:
{{begin-eqn}}
{{eqn | l = \map \xi s
      | r = \frac 1 2 s \paren {s - 1} \pi^{-s / 2} \map \Gamma {\frac s 2} \map \zeta s
      | c = {{Defof|Completed Riemann Zeta Function}}
}}
{{eqn | r = \paren {s - 1} \pi^{-s / 2} \map \Gamma {\frac s 2 + 1} \map \zeta s
      | c = Gamma Difference Equation: $\map \Gamma {\dfrac s 2 + 1} = \dfrac s 2 \map \Gamma {\dfrac s 2}$
}}
{{eqn | ll= \leadsto
      | l = \ln \map \xi s
      | r = \map \ln {s - 1} - \frac s 2 \ln \pi + \ln \map \Gamma {\frac s 2 + 1} + \ln \map \zeta s
      | c = taking the logarithm of both sides
}}
{{eqn | n = 1
      | ll= \leadsto
      | l = \quad \ds \frac {\map {\xi'} s} {\map \xi s}
      | r = \frac 1 {s - 1} - \frac 1 2 \ln \pi + \dfrac 1 2 \dfrac {\map {\Gamma'} {s / 2 + 1} } {\map \Gamma {s / 2 + 1} } + \frac {\map {\zeta'} s} {\map \zeta s}
      | c = taking the derivative of both sides
}}
{{end-eqn}}
We have that the Completed Riemann Zeta Function has Order One.
So, by the Hadamard Factorisation Theorem, there exist constants $A$, $B$ such that:
:$\ds \map \xi s = \map \exp {A + B s} : \prod_\rho \paren {1 - \frac s \rho} \map \exp {\frac s \rho}$
where $\rho$ runs over the zeros of $\xi$, that is, the nontrivial zeros of $\zeta$.
Therefore:
{{begin-eqn}}
{{eqn | l = \map \xi s
      | r = \map \exp {A + B s} : \prod_\rho \paren {1 - \frac s \rho} \map \exp {\frac s \rho}
      | c = Hadamard Factorisation Theorem
}}
{{eqn | ll= \leadsto
      | l = \ln \map \xi s
      | r = A + B s + \sum_\rho \paren {\map \ln {1 - \frac s \rho} + \frac s \rho}
      | c = taking the logarithm of both sides
}}
{{eqn | n = 2
      | ll= \leadsto
      | l = \quad \ds \frac {\map {\xi'} s} {\map \xi s}
      | r = B + \sum_\rho \paren {\frac 1 {s - \rho} + \frac 1 \rho}
      | c = taking the derivative of both sides
}}
{{end-eqn}}
Combining $(1)$ and $(2)$ we have:
{{begin-eqn}}
{{eqn | l = \frac 1 {s - 1} - \frac 1 2 \ln \pi + \dfrac 1 2 \dfrac {\map {\Gamma'} {s / 2 + 1} } {\map \Gamma {s / 2 + 1} } + \frac {\map {\zeta'} s} {\map \zeta s}
      | r = B + \sum_\rho \paren {\frac 1 {s - \rho} + \frac 1 \rho}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\map {\zeta'} s} {\map \zeta s}
      | r = B + \sum_\rho \paren {\frac 1 {s - \rho} + \frac 1 \rho} - \paren {\frac 1 {s - 1} - \frac 1 2 \ln \pi + \dfrac 1 2 \dfrac {\map {\Gamma'} {s / 2 + 1} } {\map \Gamma {s / 2 + 1} } }
      | c = 
}}
{{eqn | r = B - \frac 1 {s - 1} + \frac 1 2 \ln \pi - \frac 1 2 \frac {\map {\Gamma'} {s / 2 + 1} } {\map \Gamma {s / 2 + 1} } + \sum_\rho \paren {\frac 1 {s - \rho} + \frac 1 \rho}
      | c = rearranging terms
}}
{{end-eqn}}
{{qed}}
Theorem: Zeroes of Functions of Finite Order

Let $\map f z$ be an entire function which satisfies:
:$\map f 0 \ne 0$
:$\cmod {\map f z} \ll \map \exp {\map \alpha {\cmod z} }$
for all $z \in \C$ and some function $\alpha$, where $\ll$ is the order notation.
For $T \ge 1$, let:
:$\map N T = \# \set {\rho \in \C: \map f r = 0, \ \cmod \rho < T}$
where $\#$ denotes the cardinality of a set.
Then:
:$\map N T \ll \map \alpha {2 T}$

Proof:
Fix $T \ge 1$ and let $\rho_1, \rho_2, \ldots, \rho_n$ be an enumeration of the zeroes of $f$ with modulus less than $T$, counted with multiplicity.
By Jensen's Formula:
:$\ds \frac 1 {2 \pi} \int_0^{2 \pi} \ln \size {\map f {T e^{i \theta} } } \rd \theta = \ln \cmod {\map f 0} + \sum_{k \mathop = 1}^n \paren {\ln T - \ln \size {\rho_k} }$
Let $\rho_0 = 1$, $\rho_{n + 1} = T$, $r_k = \size {\rho_k}$.
Then:
{{begin-eqn}}
{{eqn | l = \int_0^T \map N t \frac {\d t} t
      | r = \sum_{k \mathop = 0}^n \int_{r_k}^{r_{k + 1} } \map N t \frac {\d t} t
}}
{{eqn | r = \sum_{k \mathop = 0}^n k \, \map \ln {\frac{r_{k + 1} } {r_k} }
      | c = as by the definition of $N$, it is constant value $k$ on each interval $\openint {\size {\rho_k} }  {\size {\rho_{k + 1} } }$
}}
{{eqn | r = \map \ln {\frac {T^n} {r_1 \dotsm r_n} }
}}
{{eqn | r = \sum_{k \mathop = 1}^n \paren {\ln T - \ln r_k}
}}
{{end-eqn}}
and
{{begin-eqn}}
{{eqn | l = \int_0^T \map N t \frac {\d t} t
      | r = \int_0^2 \map N {\frac {T \theta} 2} \frac {\d \theta} \theta
}}
{{eqn | o = \ge
      | r = \map N {\frac T 2} \int_1^2 \frac {\d \theta} \theta
      | c = Integration by Substitution
}}
{{eqn | r = \map N {\frac T 2} \ln 2
      | c = {{Defof|Logarithm}}
}}
{{end-eqn}}
Moreover, by hypothesis we have that:
:$\ds \frac 1 {2 \pi} \int_0^{2 \pi} \ln \size {\map f {T e^{i \theta} } } \rd \theta \le \sup_{\theta \mathop \in \closedint 0 {2 \pi} } \ln \size {\map f {T e^{i \theta} } } \ll \map \alpha T$
Putting these facts into Jensen's Formula we have:
:$\map N {\dfrac T 2} \ln 2 + \cmod {\map f 0} \ll \map \alpha T$
which implies:
:$\map N T \ll \map \alpha {2 T}$
{{qed}}
Category:Entire Functions
b34k189wsvs9qw3932p67acrair3zyw
Theorem: Strongly Locally Compact Space is Weakly Locally Compact

Let $T = \struct {S, \tau}$ be a strongly locally compact space.
Then $T$ is weakly locally compact.

Proof:
{{Recall|Weakly Locally Compact Space|weakly locally compact space}}
{{:Definition:Weakly Locally Compact Space}}
Let $T = \struct {S, \tau}$ be strongly locally compact.
{{Recall|Strongly Locally Compact Space|strongly locally compact space}}
{{:Definition:Strongly Locally Compact Space/Definition 1}}
Let $x \in S$.
By definition, there exists an open set $U_x$ of $T$ such that:
:$x \in U_x$
:${U_x}^-$ is compact
where ${U_x}^-$ denotes the closure of $U_x$.
From Set is Subset of its Topological Closure, $U_x \subseteq {U_x}^-$ and so $x \in {U_x}^-$.
Thus $x$ is contained in a compact neighborhood.
As this holds for all $x$, $T$ is a weakly locally compact.
{{qed}}
Theorem: Weakly Locally Compact Hausdorff Space is Strongly Locally Compact

Let $T = \struct {S, \tau}$ be a $T_2$ (Hausdorff) space.
Let $T$ be weakly locally compact.
Then $T$ is strongly locally compact.

Proof:
Let $x \in S$.
As $T$ is weakly locally compact, $x$ is contained in a compact neighborhood $N_x$.
As $T$ is a $T_2$ (Hausdorff) space, we can use the result Compact Subspace of Hausdorff Space is Closed.
Thus the interior of $N_x$ has a closure which is compact.
Hence the result, from definition of strongly locally compact space.
{{qed}}
Theorem: Weakly Sigma-Locally Compact iff Weakly Locally Compact and Lindelöf

Let $T = \struct {S, \tau}$ be a topological space.
{{TFAE}}
:$(1): \quad T$ is weakly $\sigma$-locally compact
:$(2): \quad T$ is weakly locally compact and Lindelöf

Proof:
=== 1 implies 2 ===
Let $T = \struct {S, \tau}$ be weakly $\sigma$-locally compact.
Then by definition:
:$T$ is $\sigma$-compact
:$T$ is weakly locally compact.
From $\sigma$-Compact Space is Lindelöf Space it follows directly that:
:$T$ is Lindelöf
:$T$ is weakly locally compact.
{{qed|lemma}}
Theorem: Second-Countable Space is Lindelöf

Let $T = \struct {S, \tau}$ be a topological space which is second-countable.
Then $T$ is also a Lindelöf space.

Proof:
{{Recall|Lindelöf Space|Lindelöf space}}
{{:Definition:Lindelöf Space}}
Let $T = \struct {S, \tau}$ be a second-countable space.
{{Recall|Second-Countable Space|second-countable space}}
{{:Definition:Second-Countable Space}}
Let $\BB$ be such a countable basis.
Let $\CC$ be an open cover of $T$.
An arbitrary $C \in \CC$ is the union of a subset $\BB_C \subseteq \BB$.
Let:
:$\ds \BB_0 = \bigcup_{C \mathop \in \CC} \BB_C \subseteq \BB$
which is itself countable. 
Now take for an arbitrary $B \in \BB_0$ a set $C_B \in \CC$ such that $B \subseteq C_B$.
Finally $\CC_0 = \set {C_B : B \in \BB_0}$ is a countable subcover of $\CC$. 
That is, $T$ is by definition Lindelöf.
{{qed}}
Theorem: Separable Space satisfies Countable Chain Condition

Let $T = \struct {S, \tau}$ be a separable topological space.
Then $T$ satisfies the countable chain condition.

Proof:
In order to demonstrate that $T$ satisfies the '''countable chain condition''', it is sufficient to demonstrate that every disjoint set of open sets of $T$ is countable.
Because $T$ is separable, there exists a subset $\set {y_n : n \in \N}$ of $S$ which is everywhere dense in $S$.
Now consider an indexed family $\family {U_j}_{j \mathop \in J}$ of non-empty open sets of $T$ such that:
:$\forall i, j \in J, i \ne j: U_i \cap U_j = \O$
Using Equivalence of Definitions of Everywhere Dense this implies that for every $j \in J$ there has to exist $n_j \in \N$ such that $y_{n_j} \in U_j$.
This gives rise to a well-defined mapping $f: J \to \N$ via $\map f j := n_j$.
In particular $f$ is injective:
{{AimForCont}} there were to exist $i, j \in J$, $i \ne j$ such that $n_i = n_j$.
Then:
:$y_{n_i} \in U_i \cap U_j$
But the latter is the empty set by assumption.
From this contradiction it follows that $J$ is countable by definition.
This concludes the proof. 
{{qed}}
Theorem: Logarithmic Derivative of Riemann Zeta Function

Let $\zeta$ be the Riemann zeta function:
:$\ds \forall s \in \C: \map \Re s > 1: \map \zeta s = \sum_{n \mathop \ge 1} n^{-s}$
Then for all $s \in \C$ with $\map \Re s > 1$:
:$\ds -\frac {\map {\zeta'} s} {\map \zeta s} = \sum_{n \mathop \ge 1} \map \Lambda n n^{-s}$
where $\Lambda$ is von Mangoldt's function.

Proof:
By Sum of Reciprocals of Powers as Euler Product:
:$\ds \map \zeta s = \prod_p \frac 1 {1 - p^{-s} }$
where $p$ ranges over the primes.
From Laws of Logarithms:
:$\ds \ln \map \zeta s = - \sum_p \map \ln {1 - p^{-s} }$
{{explain|Is the above equality really true for complex logarithm? I believe no. Probably, we need to check the identity on $\R_{>1}$, and later extend it to $\map \Re s > 1$  }}
Taking the derivative:
{{begin-eqn}}
{{eqn | l = \frac {\map {\zeta'} s} {\map \zeta s}
      | r = -\sum_p \frac {\paren {\ln p} p^{-s} } {1 - p^{-s} }
      | c = Logarithmic Derivative of Infinite Product of Analytic Functions and Derivative of Riemann Zeta Function
}}
{{eqn | r = -\sum_p \paren {\ln p} \paren {\frac 1 {1 - p^{-s} } - 1}
      | c = 
}}
{{eqn | r = -\sum_p \paren {\ln p} \sum_{n \mathop \ge 1} p^{-n s}
      | c = Sum of Infinite Geometric Sequence
}}
{{end-eqn}}
Also, by the definition of $\Lambda$:
:$\ds \sum_{n \mathop \ge 1} \map \Lambda n n^{-s} = \sum_p \paren {\ln p} \sum_{n \mathop \ge 1} p^{-n s}$
and the proof is complete.
{{qed}}
Category:Riemann Zeta Function
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