hcju/proofwikiqa huggingface dataset:
By definition, $T$ is '''scattered''' {{iff}} every non-empty subset $H$ of $S$ contains at least one point which is isolated in $H$. Let $H$ be a non-empty subset of $S$. Let $x \in H$. Let $x \ne 0$. By definition of either-or space, $\set x$ is open in $T$. So $\set x$ is an open set of $x$ containing only $x$. Thus by definition $x$ is isolated. Thus if $H$ is a non-empty subset of $S$ which contains any element other than $0$, $H$ contains at least one isolated point. The other case is that $H$ contains no $x$ such that $x \ne 0$. That is: $H = \set 0$. But from Singleton Point is Isolated, $0$ is an isolated point in $H$. Thus in all cases $H$ contains at least one point which is isolated in $H$. So by definition $T$ is a scattered space. {{qed}}
By definition, we have that $\O \in \tau$. We also have that $S \in \tau$ as $\relcomp S S = \O$ which is trivially finite. {{qed|lemma}} Let $A, B \in \tau$. Let $H = A \cap B$. Then: {{begin-eqn}} {{eqn | l = H       | r = A \cap B       | c = }} {{eqn | ll= \leadsto       | l = \relcomp S H       | r = \relcomp S {A \cap B}       | c = }} {{eqn | r = \relcomp S A \cup \relcomp S B       | c = De Morgan's laws: Complement of Intersection }} {{end-eqn}} But as $A, B \in \tau$ it follows that $\relcomp S A$ and $\relcomp S B$ are both finite. Hence their union is also finite. Thus $\relcomp S H$ is finite. So $H = A \cap B \in \tau$ as its complement is finite. {{qed|lemma}} Let $\UU \subseteq \tau$. From De Morgan's laws: Complement of Union: :$\displaystyle \relcomp S {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \relcomp S U$ But as: :$\forall U \in \UU: \relcomp S U \in \tau$ each of the $\relcomp S U$ is finite. Hence so is their intersection. So $\displaystyle \relcomp S {\bigcup \UU}$ is finite. So by definition: :$\displaystyle \bigcup \UU \in \tau$ So $\tau$ is a topology on $T$. {{qed}}
Let $U \in \tau$ be any open set of $T$. From Infinite Subset of Finite Complement Space Intersects Open Sets, we have that $U \cap H$ is infinite {{iff}} $H$ is infinite. Let $x \in S$. Then every open set $U$ in $T$ such that $x \in U$ also contains an infinite number of points of $H$ other than $x$. Thus, by definition, $x$ is a limit point of $H$. {{qed}}
Let $H$ be a countably infinite subset of $S$. From Closure of Infinite Subset of Finite Complement Space, the closure of $H$ is $S$. So by definition $H$ is everywhere dense in $T$. Hence the result by definition of separable space. {{qed}}
Let $H$ be an infinite subset of $S$. From Limit Points of Infinite Subset of Finite Complement Space, every point of $S$ is a limit point of $H$. Hence the result from the definition of closure. {{qed}}
Let $T_H = \struct {H, \tau_H}$ be a subspace of $T$. Let $\CC$ be an open cover of $T_H$. Let $U \in \CC$ be any set in $C$. $U$ covers all but a finite number of points of $T_H$. So for each of those points we pick an element of $\CC$ which covers each of those points. Hence we have a finite subcover of $T_H$. So by definition $T_H$ is a compact space. {{qed}}
Recall the definition of a perfectly $T_4$ space :Every closed set in $T$ can be written as a countable intersection of open sets of $T$. Let $V$ be a closed set in $T$. From Closed Set of Uncountable Finite Complement Topology is not $G_\delta$: :$V$ is not a $G_\delta$ set. The result follows by definition of perfectly $T_4$ space. {{qed}}
Let $\alpha \in \R$. Let $e: \R \to \C$ be the mapping defined as: :$\map e \alpha := \map \exp {2 \pi i \alpha}$ Let $\zeta_q$ be a primitive $q$th root of unity. Let: :$\displaystyle \map {\eta_q} n := \sum_{1 \mathop \le a \mathop \le q} \map e {\frac {a n} q}$ By Complex Roots of Unity in Exponential Form this is the sum of all $q$th roots of unity. Therefore: :$\displaystyle \map {\eta_q} n = \sum_{d \mathop \divides q} \map {c_d} n$ By the Möbius Inversion Formula, this gives: :$\displaystyle \map {c_q} n = \sum_{d \mathop \divides q} \map {\eta_d} n \map \mu {\frac q d}$ Now by Sum of Roots of Unity, we have: :$\displaystyle \map {c_q} n = \sum_{d \mathop \divides q} d \map \mu {\frac q d}$ as required. {{questionable|This demonstrates only the sum over $d \divides q$, not over $d \divides \gcd \set {q, n}$.}} {{qed}} {{Namedfor|Jan Cornelis Kluyver|cat = Kluyver}} Category:Analytic Number Theory kcc0luy1raznt4yqkvk73ebjz1alvnx
Let $U_n = \set {\map \exp {\dfrac {2 \pi i k} n}: 0 \le k \le n - 1}$. Let $V = \set {1, \dotsc, {\alpha_k}^{n - 1} }$. By Complex Roots of Unity in Exponential Form it is sufficient to show that $U_n = V$ {{iff}} $\gcd \set {n, k} = 1$. Let $\gcd \set {n, k} = d > 1$. Then there are $n', k' \in \N$ such that: :$n' = d n$ and: :$k' = d k$ Then we have: :$\alpha_k = \map \exp {\dfrac {2 \pi i k'} {n'} }$ and: :$\alpha_k^{n'} = \map \exp {2 \pi i k'} = 1$ Therefore: :$V = \set {1, \dotsc, \alpha^{n' - 1} }$ such that $n' < n$. So: :$\cmod V = n' < n = \cmod {U_n}$ and $U_n \ne V$. Let $\gcd \set {n, k} = 1$. Let: :$\map \exp {\dfrac {2 \pi i k} n}^d = \map \exp {\dfrac {2 \pi i k} n} = 1$ Then it must be the case that $\dfrac {k d} n \in \Z$. Since $\gcd \set {n, k}\ = 1$ it follows that: :$n \divides d$ and so: :$d \ge n$ Therefore $\set {1, \dotsc, \alpha^{n - 1} }$ are distinct Hence $\card V = \card {U_n}$. Moreover each element of $V$ can be written in the form: :$\map \exp {\dfrac {2 \pi i k} n}$ with $0 \le k \le n - 1$. It follows that $V = U_n$. {{qed}}
{{ProofWanted|The material on this page was originally raised by User:Yohji kinomoto.}}
Let $U$ be a countably infinite subset of $S$. Then it can be written as: :$\displaystyle U = \bigcup_{x \mathop \in U} \set x$ But as $\set x$ is a finite subset of $S$, it is by definition the complement of an open set in $T$. So by definition $\set x$ is closed in $T$. So: :$\displaystyle U = \bigcup_{x \mathop \in U} \left\{{x}\right\}$ is a countable union of closed sets of $T$. That is, by definition, $U$ is an $F_\sigma$ set. As $U$ is countably infinite, its complement is uncountable, not finite, and so $U$ is not open in $T$. But as $U$ is also not finite, it is not the complement of an open set and so is not closed in $T$ either. From Complement of $F_\sigma$ Set is $G_\delta$ Set it follows that the complement of $U$ is a $G_\delta$ set, and similarly neither open nor closed. {{qed}}
{{AimForCont}} some $x \in S$ has a countable local basis. That means: :there exists a countable set $\BB_x \subseteq \tau$ such that: :$\forall B \in \BB_x: x \in B$ and such that: :every open neighborhood of $x$ contains some $B \in \BB_x$. Let $y \in S$ with $y \ne x$. Because its complement relative to $S$ is finite, we have that $S \setminus \set y$ is an open set containing $x$. Because  $\BB_x$ is a neighborhood basis at $x$, there exists some $B \in \BB_x$ such that: :$B \subseteq S \setminus \set y$ whence $y \notin B$. So: :$y \notin \bigcap \BB_x$ So: {{begin-eqn}} {{eqn | l = \bigcap \BB_x       | r = \set x       | c =  }} {{eqn | ll= \leadsto       | l = S \setminus \set x       | r = S \setminus \bigcap \BB_x       | c =  }} {{eqn | r = \bigcup_{B \mathop \in \BB_x} \paren {S \setminus B}       | c = De Morgan's Laws: Difference with Intersection }} {{end-eqn}} By definition, each of $S \setminus B$ is finite. From Countable Union of Countable Sets is Countable it follows that $\displaystyle \bigcup_{B \mathop \in \BB_x} \paren {S \setminus B}$ is countable. So $S \setminus \set x$ and therefore $S$ is also countable. From this contradiction (as we have specified that $S$ is uncountable) it follows that our assumption that $x \in S$ has a countable local basis must be false. Hence by definition $T$ can not be first-countable. {{qed}}
Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$. We have that a Finite Complement Topology is Separable. But we also have that an Uncountable Finite Complement Space is not First-Countable. Hence the result, by Proof by Counterexample. {{qed}}
Let $q, r \in \N$ such that: :$\gcd \set {q, r} = 1$ where $\gcd \set {q, r}$ denotes the greatest common divisor of $q$ and $r$. Then: {{begin-eqn}} {{eqn | l = \map {c_q} n \, \map {c_r} n       | r = \sum_{d_1 \mathop \divides \gcd \set {n, q} } d_1 \, \map \mu {\frac q {d_1} } \sum_{d_2 \mathop \divides \gcd \set {n, r} } d_2 \, \map \mu {\frac r {d_2} }       | c = Kluyver's Formula for Ramanujan's Sum }} {{eqn | r = \sum_{\substack {d_1 \mathop \divides \gcd \set {n, q} \\ d_2 \mathop \divides \gcd \set {n, r} } } d_1 d_2 \, \map \mu {\frac q {d_1} } \, \map \mu {\frac r {d_2} }       | c =  }} {{eqn | r = \sum_{d \mathop \divides \gcd \set {n, q} \gcd \set {n, r} } d \, \map \mu {\frac {q r} d}       | c = Möbius Function is Multiplicative, and setting $d = d_1 d_2$ }} {{eqn | r = \sum_{d \mathop \divides \gcd \set {n, q r} } d \, \map \mu {\frac {q r} d}       | c = GCD with One Fixed Argument is Multiplicative Function }} {{eqn | r = \map {c_{q r} } n       | c = Kluyver's Formula for Ramanujan's Sum }} {{end-eqn}} This completes the proof. {{qed}} Category:Analytic Number Theory 92a5u5ragspb52bp1qa1wgrtfjbbata
It is clear that $\phi \left({n}\right) < n$ for all $n$, so it is sufficient to prove that: :$\displaystyle \lim_{n \to \infty} \frac {n^{1 - \epsilon}}{\phi \left({n}\right)} = 0$ By Multiplicative Function that Converges to Zero on Prime Powers it is sufficient to prove that: :$\displaystyle \lim_{p^k \to \infty} \frac{p^{k(1-\epsilon)}}{\phi(p^k)} = 0$ as $p^k$ ranges through all prime powers. By Euler Phi Function of Prime Power we have: : $\phi \left({p^k}\right) = p^k - p^{k-1}$ for a prime power $p^k$. Therefore: {{begin-eqn}} {{eqn|l=\frac{p^{k \left({1 - \epsilon}\right)} } {\phi \left({p^k}\right)}      |r=\frac{p^{k \left({1 - \epsilon}\right)} } {p^k - p^{k-1} }      |c= }} {{eqn|l=      |r=\frac p {p-1} \frac{p^{k \left({1 - \epsilon}\right)} }{p^k}      |c= }} {{eqn|l=      |r=\frac p {p-1} \frac 1 {p^{k \epsilon} }      |c= }} {{eqn|l=      |o=\le      |r=\frac 2 {p^{k \epsilon} }      |c=since $p / \left({p - 1}\right) \le 2$ for all primes $p$ }} {{end-eqn}} Therefore: :$\displaystyle \lim_{p^k \to \infty} \frac{p^{k \left({1 - \epsilon}\right)}} {\phi \left({p^k}\right)} \le \lim_{p^k \to \infty} \frac 2 {p^{k \epsilon} } = 0$ {{qed}} Category:Analytic Number Theory Category:Euler Phi Function f8cck1beowzt7p4l1eapw1am9ggntt1
Let $x, y \in S$. Then: :$S \setminus \set x$ is an open set of $T$ containing $y$ but not $x$. :$S \setminus \set y$ is an open set of $T$ containing $x$ but not $y$. Hence the result from definition of $T_1$ space. {{qed}}
Let $U_1, U_2 \in \tau$ be non-empty open sets of $T$. We have that $U_1$ and $U_2$ are both infinite if $S$ is. From Infinite Subset of Finite Complement Space Intersects Open Sets, they intersect each other. Hence the result from definition of irreducible space. {{qed}}
Let $U \in \tau$ be any non-empty open set of $T$. Then $\relcomp S U$ is finite. We have that: :$H = H \cap \paren {U \cup \relcomp S U} = \paren {H \cap U} \cup \paren {H \cap \relcomp S U}$ {{AimForCont}} $H \cap U$ is finite. Since $H \cap \relcomp S U \subseteq \relcomp S U$, $H \cap \relcomp S U$ is also finite. $H = \paren {H \cap U} \cup \paren {H \cap \relcomp S U}$ is the union of two finite sets, and so it is finite. It is a contradiction that $H$ is infinite and finite at the same time, so $H \cap U$ must be infinite. {{qed}}
Suppose $T$ has at least two points $x, y \in S$. Because $T$ is irreducible, there are no two disjoint open sets such that $x$ is in one and $y$ is in the other. This contradicts the fact that $T$ is Hausdorff. Thus $T$ has only one point. {{qed}} Category:Irreducible Spaces Category:Hausdorff Spaces 77maej8b2acclyq5geemvmi4h4g2vae
We have: :Finite Complement Space is Irreducible :Irreducible Hausdorff Space is Singleton {{qed}}
We have that a Finite Complement Space is a $T_1$ space. From $T_1$ Space is $T_0$ Space, $T$ is a $T_0$ space. We then have that a Finite Complement Space is not $T_2$. From Regular Space is $T_2$ Space, $T$ is not a regular space. By definition, a regular space is a space that is both a $T_0$ space and a $T_3$ space. But $T$ is a $T_0$ space and not a regular space. So it follows that $T$ cannot be a $T_3$ space. Next we have that a Normal Space is a $T_3$ Space. But as $T$ is not a $T_3$ space, $T$ cannot be a normal space. By definition, a normal space is a space that is both a $T_1$ space and a $T_4$ space. But $T$ is a $T_1$ space and not a normal space. So it follows that $T$ cannot be a $T_4$ space. Finally we have that a $T_5$ Space is a $T_4$ Space. But as $T$ is not a $T_4$ space, $T$ cannot be a $T_5$ space. {{qed}}
The Finite Complement Topology is Compact. Also, a Finite Topological Space is Compact. So $D$, defined as the indiscrete space on a doubleton, is also compact. The result follows directly from Tychonoff's Theorem. {{qed}}
From Separation Axioms on Double Pointed Topology, we have that: :$T \times D$ is not a $T_0$ space. :$T \times D$ is not a $T_1$ space. :$T \times D$ is not a $T_2$ space. Also from Separation Axioms on Double Pointed Topology, we have that: :$T \times D$ is a $T_3$ space {{iff}} $T$ is a $T_3$ space :$T \times D$ is a $T_4$ space {{iff}} $T$ is a $T_4$ space :$T \times D$ is a $T_5$ space {{iff}} $T$ is a $T_5$ space But from Finite Complement Space is not $T_3$, $T_4$ or $T_5$: :$T$ is not a $T_3$ space :$T$ is not a $T_4$ space :$T$ is not a $T_5$ space Hence the result. {{qed}}
By Sum Over Divisors of von Mangoldt is Logarithm: :$\displaystyle \ln n = \sum_{d \mathop \divides n} \map \Lambda d$ Hence: {{begin-eqn}} {{eqn | l = \map \Lambda n       | r = \sum_{d \mathop \divides n} \map \mu d \, \map \ln {\frac n d}       | c = Möbius Inversion Formula }} {{eqn | r = \sum_{\substack {d \mathop \divides n \\ d \mathop \le y} } \map \mu d \, \map \ln {\frac n d} + \sum_{\substack {d \mathop \divides n \\ d \mathop > y} } \map \mu d \, \map \ln {\frac n d}       | c =  }} {{end-eqn}} Taking the second summation in that last line: {{begin-eqn}} {{eqn | l = \sum_{\substack {d \mathop \divides n \\ d \mathop > y} } \map \mu d \, \map \ln {\frac n d}       | r = \sum_{\substack {d \mathop \divides n \\ d \mathop > y} } \map \mu d \sum_{c \mathop \divides n / d} \map \Lambda c       | c =  }} {{eqn | r = \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop > y} } \map \mu d \, \map \Lambda c       | c =  }} {{eqn | r = \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop > y, \, c \mathop > z} } \map \mu d \, \map \Lambda c + \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop > y, \, c \mathop \le z} } \map \mu d \, \map \Lambda c       | c =  }} {{end-eqn}} Again, taking the second summation in that last line: :$\displaystyle \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop > y, \, c \mathop \le z} } \map \mu d \, \map \Lambda c = \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ c \mathop \le z} } \map \mu d \, \map \Lambda c - \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop \le y, \, c \mathop \le z} } \map \mu d \, \map \Lambda c$ Putting this together: :$\displaystyle \map \Lambda n = \sum_{\substack {d \mathop \divides n \\ d \mathop \le y} } \map \mu d \, \map \ln {\frac n d} + \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop > y, c \mathop > z} } \map \mu d \, \map \Lambda c + \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ c \mathop \le z} } \map \mu d \, \map \Lambda c - \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop \le y, c \mathop \le z} } \map \mu d \, \map \Lambda c$ It remains to be shown that: :$\displaystyle \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ c \mathop \le z} } \map \mu d \, \map \Lambda c = 0$ The summation is expressed as: :$\displaystyle \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ c \mathop \le z} } \map \mu d \, \map \Lambda c = \sum_{\substack {c \mathop \le z \\ c \mathop \divides n} } \map \Lambda c \sum_{d \mathop \divides \frac n c} \map \mu d$ Now we have $c \le z < n$, so: :$\dfrac n c > 1$ Therefore, by the lemma to Sum of Möbius Function over Divisors: :for each $c$ the inner sum vanishes. This shows that: :$\displaystyle \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ c \mathop \le z} } \map \mu d \, \map \Lambda c = 0$ as required. {{qed}} {{Namedfor|Robert Charles Vaughan|cat = Vaughan}} Category:Analytic Number Theory 6ci23vilevkgb4exsd9dt9uypq89ciz
Let $n \geq 1$, and by the Fundamental Theorem of Arithmetic write $n = p_1^{e_1}\cdots p_k^{e_k}$ with $p_1,\ldots,p_k$ distinct primes and $e_1,\ldots,e_k > 0$. Now $d | n$ if any only if $d = p_1^{f_1}\cdots p_k^{f_k}$ with $0 \leq f_i \leq e_i$ for $i = 1,\ldots, k$. By the definition of $\Lambda$, for such $d$ we have $\Lambda(d) \neq 0$ if and only if there is exactly one $i \in \{1,\ldots, k\}$ such that $f_i > 0$. If this is the case, say $d = p_i^{f_i}$ then $\Lambda(d) = \log p_i$. Therefore, :$\displaystyle \sum_{d | n}\Lambda(d) = \sum_{i = 1}^k e_i \log p_i$ Also, we have {{begin-eqn}} {{eqn|l=\log n      |r=\log(p_1^{e_1} \cdots p_k^{e_k})      |c= }} {{eqn|l=      |r= \sum_{i = 1}^k e_i \log p_i      |c=By Sum of Logarithms }} {{end-eqn}} Thus we indeed have :$\displaystyle \displaystyle \sum_{d | n}\Lambda(d) = \log n$ {{qed}} Category:Analytic Number Theory 48twvhkcbcsk2ja85z9y4r3rv9l2yyz
By hypothesis, there exist only finitely many prime powers $p^k$ such that $\size {\map f {p^k} } > 1$. Let $\displaystyle A = \prod_{\size {\map f {p^k} } \mathop > 1} \size {\map f {p^k} }$. Thus $A \ge 1$. Let $0 < \dfrac \epsilon A$.  There exist only finitely many prime powers $p^k$ such that $\size {\map f {p^k} } > \dfrac \epsilon A$. Therefore there are only finitely many integers $n$ such that: :$\size {\map f {p^k} } > \dfrac \epsilon A$ for every prime power $p^k$ that divides $n$. Therefore if $n$ is sufficiently large there exists a prime power $p^k$ that divides $n$ such that: :$\size {\map f {p^k} } < \dfrac \epsilon A$ Therefore $n$ can be written as: :$\displaystyle n = \prod_{i \mathop = 1}^ r p_i^{k_i} \prod_{i \mathop = r + 1}^{r + s} p_i^{k_i} \prod_{i \mathop = r + s + 1}^{r + s + t} p_i^{k_i}$ where $t \ge 1$ and: {{begin-eqn}} {{eqn | l = 1       | o = \le       | m = \size {\map f {p_i^{k_i} } }       | mo=        | r =        | c = for $i = 1, \ldots, r$ }} {{eqn | l = \frac \epsilon A       | o = \le       | m = \size {\map f {p_i^{k_i} } }       | mo= <       | r = 1       | c = for $i = r + 1, \ldots, r + s$ }} {{eqn | o =        | m = \size {\map f {p_i^{k_i} } }       | mo= <       | r = \frac \epsilon A       | c = for $i = r + s + 1, \ldots, r + s + t$ }} {{end-eqn}} Therefore: {{begin-eqn}} {{eqn | l = \size {\map f n}       | r = \prod_{i \mathop = 1}^r \map f {p_i^{k_i} } \prod_{i \mathop = r + 1}^{r + s} \map f {p_i^{k_i} } \prod_{i \mathop = r + s + 1}^{r + s + t} \map f {p_i^{k_i} }       | c = because $f$ is multiplicative }} {{eqn | o = <       | r = A \paren {\frac \epsilon A}^t       | c =  }} {{eqn | o = <       | r = \epsilon       | c = because $t \ge 1$ }} {{end-eqn}} This shows that $\map f n$ can be made arbitrarily small for sufficiently large $n$. {{qed}} Category:Analytic Number Theory Category:Multiplicative Functions sl3n43xpxm69xs440y1uacxtzcykkja
We have from Finite Complement Topology is Minimal $T_1$ Topology that $\tau$ space is an expansion of a finite complement space. But from the definition, any such finite complement space on a finite set is discrete. But as the Discrete Topology is Finest Topology, $\tau$ must itself be the discrete topology on $S$. {{qed}}
Let $T = \left({S, \tau}\right)$ be a $T_1$ space on a finite set $S$. Let $U \subseteq S$ be any subset of $S$.  Let $H = \complement_S \left({U}\right)$ be the complement of $U$ relative to $S$. Then by Relative Complement of Relative Complement we have that $U = \complement_S \left({H}\right)$. We can write $H$ as: :$\displaystyle H = \bigcup_{x \mathop \in H} \left\{{x}\right\}$ From Equivalence of Definitions of $T_1$ Space, $\forall x \in H: \left\{{x}\right\}$ is closed in $T$. As $S$ is a finite set, it follows that $H$ is a finite union of closed sets of $T$. By Topology Defined by Closed Sets, $H$ is therefore closed in $T$. By definition of closed set, $U = \complement_S \left({H}\right)$ is open in $T$. As $U$ is arbitrary, it follows that: :$\forall U \subseteq S: S \in \tau$ by definition of a topology. Hence the result, by definition of the discrete topology. {{qed}}
Let $T = \struct {S, \tau}$ be the finite complement space on $S$. Let $U \in \tau$ be any open set of $T$. Let $H = \relcomp S U$ be the complement of $U$ relative to $S$. By definition of finite complement topology, $H \subseteq S$ is a finite subset of $S$. Let $T' = \struct {S, \tau'}$ be any arbitrary $T_1$ (Fréchet) space on $S$. From Equivalence of Definitions of $T_1$ Space: :$\forall x \in S: \set x$ is closed in $T'$. We can write $H$ as: :$\displaystyle H = \bigcup_{x \mathop \in H} \set x$ As $H$ is finite, it follows that $H$ is a finite union of closed sets of $T'$. By Topology Defined by Closed Sets, $H$ is therefore closed in $T'$. By Relative Complement of Relative Complement we have that $U = \relcomp S H$. So, by definition of closed set, we have that $U$ is open in $T'$. So we have shown that for any arbitrary $U \in \tau$ it follows that $U \in \tau'$. So $\tau \subseteq \tau'$ and so by definition $\tau$ is coarser than $\tau'$. {{qed}}
We have that a Finite Complement Space is Irreducible. The result follows from Irreducible Space is Connected. {{qed}}
We have that a Finite Complement Space is Irreducible. The result follows from Irreducible Space is Locally Connected. {{qed}}
Let $\mathcal R_f \subseteq S \times S$ be the relation induced by $f$: :$\tuple {s_1, s_2} \in \mathcal R_f \iff \map f {s_1} = \map f {s_2}$ Then from Relation Induced by Mapping is Equivalence Relation, $\mathcal R_f$ is an equivalence relation. The result follows from Relation Partitions Set iff Equivalence. {{qed}} Category:Mapping Theory mn1p432i9cwga3h01701aau75sdxeus
{{AimForCont}} $f: \closedint 0 1 \to S$ is a path on $T$. Then $f$ is by definition continuous. By definition of the finite complement topology, for all $x \in S$, the set $\set x$ is closed. Now consider the set: :$F = \set {\map {f^{-1} } x: x \in S}$ From Continuity Defined from Closed Sets, each of the elements of $F$ is closed. Also, from Mapping Induces Partition on Domain, the elements of $F$ are pairwise disjoint. As $T$ is a countable finite complement topology, its underlying set $S$ is countably infinite by definition. Hence $F$ is also countably infinite by nature of $f$ being a mapping. Furthermore, we have $\displaystyle \bigcup F = \closedint 0 1$. Hence $F$ is a countably infinite set of pairwise disjoint closed sets whose union is $\closedint 0 1$. From Closed Unit Interval is not Countably Infinite Union of Disjoint Closed Sets, this is impossible. From this contradiction, $f$ cannot be continuous, and so cannot be a path on $T$. So $T$ cannot be path-connected. {{qed}}
Let $\BB$ be a basis for $T$. Let $B \in \BB$. {{AimForCont}} $f: \closedint 0 1 \to B$ is a path on $T$. Then $f$ is by definition continuous. Now consider the set: :$F = \set {\map {f^{-1} } x: x \in S}$ From Continuity Defined from Closed Sets, each of the elements of $F$ is closed. Also, from Mapping Induces Partition on Domain, the elements of $F$ are pairwise disjoint. From Basis of Countable Finite Complement Topology consists of Countably Infinite Sets, $B$ is a countably infinite set. Hence $F$ is also countably infinite by nature of $f$ being a mapping. Furthermore, we have $\displaystyle \bigcup F = \closedint 0 1$. Hence $F$ is a countably infinite set of pairwise disjoint closed sets whose union is $\closedint 0 1$. From Closed Unit Interval is not Countably Infinite Union of Disjoint Closed Sets, this is impossible. From this contradiction, $f$ cannot be continuous, and so cannot be a path on $B$. So $B$ cannot be path-connected. So, by definition, $T$ is not locally path-connected. {{qed}}