hcju/proofwikiqa huggingface dataset:
Because $f_1$ and $f_2$ are both bijections, it follows by definition that they are both surjections. Let $\left({t_1, t_2}\right) \in T_1 \times T_2$. Then: :as $f_1$ is a surjection, $\exists s_1 \in S_1: f_1 \left({s_1}\right) = t_1$ :as $f_2$ is a surjection, $\exists s_2 \in S_2: f_2 \left({s_2}\right) = t_2$ Thus: :$\exists \left({s_1, s_2}\right) \in S_1 \times S_2: f_1 \times f_2 \left({s_1, s_2}\right) = \left({t_1, t_2}\right)$ So $f_1 \times f_2$ is a surjection. Because $f_1$ and $f_2$ are both bijections, it follows by definition that they are both injections. Let: : $f_1 \times f_2 \left({a_1, a_2}\right) = \left({c_1, c_2}\right), f_1 \times f_2 \left({b_1, b_2}\right) = \left({d_1, d_2}\right)$ for some $\left({a_1, a_2}\right), \left({b_1, b_2}\right) \in S_1 \times S_2$. Suppose $\left({c_1, c_2}\right) = \left({d_1, d_2}\right)$. Then by Equality of Ordered Pairs: :$c_1 = d_1$ :$c_2 = d_2$ By definition of $f_1 \times f_2$: :$f_1 \left({a_1}\right) = c_1$ :$f_1 \left({b_1}\right) = d_1$ and: :$f_2 \left({a_2}\right) = c_2$ :$f_2 \left({b_2}\right) = d_2$ As $f_1$ is an injection: :$c_1 = d_1 \implies a_1 = b_1$ and as $f_2$ is an injection: :$c_2 = d_2 \implies a_2 = b_2$ Thus it follows that: :$\left({c_1, c_2}\right) = \left({d_1, d_2}\right) \implies \left({a_1, a_2}\right) = \left({b_1, b_2}\right)$ and so $f_1 \times f_2$ is an injection. So $f_1 \times f_2$ is a surjection and also an injection. Hence by definition, $f_1 \times f_2$ is a bijection. {{qed}} Category:Bijections Category:Cartesian Product a5gd6052hnmbkcqn4kezns4qy9nt7ci
Let $u = \map f y$ Then by the Chain Rule for Derivatives: :$\dfrac {\d u} {\d x} = \map {f'} y \dfrac {\d y} {\d x}$ Then: {{begin-eqn}} {{eqn | l = \dfrac {\d y} {\d x} - \dfrac {\map f y} {\map {f'} y} \map {\phi'} x       | r = \dfrac {\map \phi x \, \map {\phi'} x} {\map {f'} y}       | c =  }} {{eqn | ll= \leadsto       | l = \map {f'} y \dfrac {\d y} {\d x} - \map f y \, \map {\phi'} x       | r = \map \phi x \, \map {\phi'} x       | c = multiplying $(1)$ by $\map {f'} y$ }} {{eqn | ll= \leadsto       | l = \dfrac {\d u} {\d x} - u \, \map {\phi'} x       | r = \map \phi x \, \map {\phi'} x       | c = substituting $u$ for $\map f y$ }} {{end-eqn}} This is a linear first order ordinary differential equation in the form: :$\dfrac {\d u} {\d x} + \map P x u = \map Q x$ whose general solution from Solution to Linear First Order Ordinary Differential Equation is: :$\displaystyle u e^{\int P \rd x} = \int Q e^{\int P \rd x} \rd x + C$ In this instance: :$\map P x = -\map {\phi'} x$ and: :$\map Q x = \map \phi x \, \map {\phi'} x$ Thus: {{begin-eqn}} {{eqn | l = \int P \rd x       | r = -\int \map {\phi'} x \rd x       | c =  }} {{eqn | r = -\map \phi x + A       | c =  }} {{end-eqn}} Hence: :$\displaystyle u e^{-\map \phi x + A} = \int \map \phi x \, \map {\phi'} x e^{-\map \phi x + A} \rd x + C$ Let $v = \map \phi x$. Then by Integration by Substitution: {{begin-eqn}} {{eqn | l = \int \map \phi x \, \map {\phi'} x e^{-\map \phi x + A} \rd x       | r = \int v e^{-v + A} \rd v       | c = Integration by Substitution }} {{eqn | r = \int v e^{-v} e^A \rd v       | c =  }} {{eqn | r = e^A \int v e^{-v} \rd v       | c =  }} {{eqn | r = - e^A e^{-v} \paren {v + 1} + C'       | c = Primitive of $x e^{a x}$ }} {{end-eqn}} This leaves us with: :$u e^{-\map \phi x + A} = - e^A e^{-v} \paren {v + 1} + C$ subsuming $C'$ into $C$. Substituting back in, the general solution is seen to be: :$\map f y e^{-\map \phi x + A} = - e^{- \map \phi x + A} \paren {\map \phi x + 1} + C$ and so multiplying both sides by $e^{\map \phi x - A}$: {{begin-eqn}} {{eqn | l = \map f y        | r = \paren {- \map \phi x - 1} + C e^{\map \phi x - A} }} {{eqn | r = C e^{- A} e^{\map \phi x} - \map \phi x - 1  }} {{eqn | r = C e^{\map \phi x} - \map \phi x - 1        | c = By absorbing the constant factor $e^{- A}$ into $C$ }} {{end-eqn}} Hence the result. {{qed}}
Let $\card S = n$ and $\card T = m$. First suppose that $n = 0$, that is, that $S = \O$. From Cardinality of Set of All Mappings from Empty Set: :$\card {T^\O} = 1 = m^0$ and the result is seen to hold for $n = 0$. Next, suppose that $m = 0$, that is, that $T = \O$. From Cardinality of Set of All Mappings to Empty Set: :$\card {\O^S} = \begin{cases} 1 & : S = \O \\  0 & : S \ne \O \end{cases}$ So if $n > 0$: :$\card {\O^S} = 0 = 0^n$ and if $n = 0$: :$\card {T^S} = 1 = 0^0 = m^n$ and the result holds. This fits in with the preferred definition of the value of $0^0$. Finally, suppose $m > 0$ and $n > 0$. Let $\sigma: \N_n \to S$ and $\tau: T \to \N_n$ be bijections. Then the mapping $\Phi: T^S \to \paren {\N_m}^{\paren {\N_n} }$ defined as: :$\forall f \in T^S: \Phi \paren f = \tau \circ f \circ \sigma$ (where $\paren {\N_m}^{\paren {\N_n} }$ is the set of all mappings from $\N_n$ to $\N_m$) is also a bijection. So we need only consider the case where $S = \N_n$ and $T = \N_m$. Let $m \in \N_{>0}$. For each $n \in \N$, let $\Bbb T \paren {n, m}$ be the set of all mappings from $\N_n$ to $\N_m$. Let: :$\Bbb S = \set {n \in \N: \card {\Bbb T \tuple {n, m} } = m^n}$ We have seen that $0 \in \Bbb S$. Let $n \in \Bbb S$. Let $\rho: \Bbb T \tuple {n + 1, m} \to \Bbb T \paren {n, m}$ defined by: :$\forall f \in \Bbb T \tuple {n + 1, m}: \rho \paren f =$ the restriction of $f$ to $\N_n$ Given that $g \in \Bbb T \tuple {n, m}$, and $k \in \N_m$, let $g_k: \N_{n + 1} \to \N_m$ be defined by: :$\forall x \in \N_{n + 1}: g_k \paren x = \begin{cases} g \paren x & : x \in \N_n \\ k & : x = n \end{cases}$ Then: :$\rho^{-1} \paren {\set g} = \set {g_0, \ldots, g_{m-1} }$ Thus $\rho^{-1} \paren {\set g}$ has $m$ elements. So clearly: :$\set {\rho^{-1} \paren {\set g}: g \in \Bbb T \tuple {n, m} }$ is a partition of $\Bbb T \tuple {n + 1, m}$. Hence, as $n \in \Bbb S$, the set $\Bbb T \tuple {n + 1, m}$ has $m \cdot m^n = m^{n + 1}$ elements by Number of Elements in Partition. Thus $n + 1 \in \Bbb S$. By the Principle of Mathematical Induction: :$\Bbb S = \N$ and the proof is complete. {{qed}}
By Cardinality is Additive Function, we can apply the Inclusion-Exclusion Principle: If $f: \mathcal S \to \R$ is an additive function, then: {{begin-eqn}}      {{eqn | l = f \paren {\bigcup_{i \mathop = 1}^n S_i}       | r = \sum_{i \mathop = 1}^n f \paren {S_i}       | c = }} {{eqn | o = -       | r = \sum_{1 \mathop \le i \mathop < j \mathop \le n} f \paren {S_i \cap S_j}       | c = }} {{eqn | o = +       | r = \sum_{1 \mathop \le i \mathop < j \mathop < k \mathop \le n} f \paren {S_i \cap S_j \cap S_k}       | c = }} {{eqn | o = \cdots       | c = }} {{eqn | o = +       | r = \paren {-1}^{n - 1} f \paren {\bigcap_{i \mathop = 1}^n S_i}       | c = }} {{end-eqn}} {{qed}}
Let $a \in C$. {{begin-eqn}} {{eqn | l = x       | o = \in       | r = S       | c =  }} {{eqn | ll= \implies       | l = x       | r = x \circ \left({a \circ' a^{-1} }\right)       | c = Definition of Invertible Element }} {{eqn | ll= \implies       | l = x       | r = \left({x \circ a}\right) \circ' a^{-1}       | c = Definition of Associative }} {{eqn | ll= \implies       | l = x       | o = \in       | r = S \circ' C^{-1}       | c = Definition of Subset Product }} {{eqn | ll= \implies       | l = S       | o = \subseteq       | r = S \circ' C^{-1}       | c = Definition of Subset }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = y       | o = \in       | r = C       | c =  }} {{eqn | ll= \implies       | l = y^{-1}       | o = \in       | r = C^{-1}       | c = Definition of Inverse of Subset }} {{eqn | ll= \implies       | l = y^{-1}       | r = a \circ' a^{-1} \circ' y^{-1}       | c = Definition of Invertible Element }} {{eqn | ll= \implies       | l = y^{-1}       | r = a \circ' \left({y \circ' a}\right)^{-1}       | c = Inverse of Product }} {{eqn | ll= \implies       | l = y^{-1}       | r = a \circ' \left({y \circ a}\right)^{-1}       | c = $\circ'$ extends $\circ$ }} {{eqn | ll= \implies       | l = y^{-1}       | o = \in       | r = S \circ' C^{-1}       | c = Definition of Subset Product }} {{eqn | ll= \implies       | l = C^{-1}       | o = \subseteq       | r = S \circ' C^{-1}       | c = Definition of Subset }} {{end-eqn}} Thus, as: : $C^{-1} \subseteq S \circ' C^{-1}$ and: :$S \subseteq S \circ' C^{-1}$ by Union is Smallest Superset it follows that: :$S \cup C^{-1} \subseteq S \circ' C^{-1}$ From Subset Product defining Inverse Completion of Commutative Semigroup is Commutative Semigroup: :$S \circ' C^{-1}$ is a commutative semigroup. So $S \circ' C^{-1}$ is a semigroup which contains $S \cup C^{-1}$. By definition of generator of semigroup, it follows that: :$\left\langle{S \cup C^{-1} }\right\rangle \subseteq S \circ' C^{-1}$ Let $z = x \circ' y^{-1} \in S \circ' C^{-1}$. Then by definition: :$x \in S$ and: :$y^{-1} \in C^{-1}$ and so by definition of generator of semigroup: :$x \circ' y^{-1} \in \left\langle{S \cup C^{-1} }\right\rangle$ Thus: $S \circ' C^{-1} \subseteq \left\langle{S \cup C^{-1} }\right\rangle$ By definition of set equality: :$S \circ' C^{-1} = \left\langle{S \cup C^{-1} }\right\rangle$ and so by definition of the inverse completion: :$T = S \circ' C^{-1}$ {{qed}}
Let $e$ be the identity element of $\struct {S, \circ}$. Suppose $x \in S$ has two inverses: $y$ and $z$. Then: {{begin-eqn}} {{eqn | l = y       | r = y \circ e       | c = {{Defof|Identity Element}} }} {{eqn | r = y \circ \paren {x \circ z}       | c = {{Defof|Inverse Element}} }} {{eqn | r = \paren {y \circ x} \circ z       | c = {{Defof|Associative Operation}} }} {{eqn | r = e \circ z       | c = {{Defof|Inverse Element}} }} {{eqn | r = z       | c = {{Defof|Identity Element}} }} {{end-eqn}} Similarly: {{begin-eqn}} {{eqn | l = y       | r = e \circ y       | c = {{Defof|Identity Element}} }} {{eqn | r = \paren {z \circ x} \circ y       | c = {{Defof|Inverse Element}} }} {{eqn | r = z \circ \paren {x \circ y}       | c = {{Defof|Associative Operation}} }} {{eqn | r = z \circ e       | c = {{Defof|Inverse Element}} }} {{eqn | r = z       | c = {{Defof|Identity Element}} }} {{end-eqn}} So whichever way round you do it, $y = z$ and the inverse of $x$ is unique. {{qed}}
By Inverse in Monoid is Unique, any inverse of $x$ is unique, and can be denoted $x^{-1}$. From Inverse of Inverse in General Algebraic Structure: :$x^{-1}$ is invertible and its inverse is $x$. That is: :$\paren {x^{-1} }^{-1} = x$ {{qed}}
{{begin-eqn}} {{eqn | l = \left({a \circ b}\right) \circ \left({b^{-1} \circ a^{-1} }\right)       | r = \left({\left({a \circ b}\right) \circ b^{-1} }\right) \circ a^{-1}       | c = Associativity }} {{eqn | r = \left({a \circ \left({b \circ b^{-1} }\right)}\right) \circ a^{-1}       | c = Associativity }} {{eqn | r = \left({a \circ e}\right) \circ a^{-1}       | c = Behaviour of Inverse }} {{eqn | r = a \circ a^{-1}       | c = Behaviour of Identity }} {{eqn | r = e       | c = Behaviour of Inverse }} {{end-eqn}} Similarly for $\left({b^{-1} \circ a^{-1}}\right) \circ \left({a \circ b}\right)$. {{qed}}
By the definition of a right identity: :$\forall x \in S: x \circ e_R = x$ Thus in particular: :$e_R \circ e_R = e_R$ Therefore $e_R$ is idempotent under $\circ$. {{qed}}
By the definition of a left identity: :$\forall x \in S: e_L \circ x = x$ Thus in particular: : $e_L \circ e_L = e_L$ Therefore $e_L$ is idempotent under $\circ$. {{qed}}
Let $\left({S, \circ}\right)$ be an algebraic structure with more than one right identity. Take any two of these, and call them $e_{R_1}$ and $e_{R_2}$, where $e_{R_1} \ne e_{R_2}$. Suppose $\left({S, \circ}\right)$ has a left identity. Call this left identity $e_L$. Then, by the behaviour of $e_L$, $e_{R_1}$ and $e_{R_2}$: : $e_{R_1} = e_{R_1} \circ e_L = e_L$ : $e_{R_2} = e_{R_2} \circ e_L = e_L$ So $e_{R_1} = e_L = e_{R_2}$, which contradicts the supposition that $e_{R_1}$ and $e_{R_2}$ are different. Therefore, in an algebraic structure with more than one right identity, there can be no left identity. {{qed}}
From Inverse Completion is Commutative Semigroup: :$\left({T, \circ'}\right)$ is a commutative semigroup. By definition of inverse completion: :$\forall x \in C: \exists x^{-1} \in C: x \circ x^{-1} \in T$ Thus by definition of inverse element: :$e := x \circ x^{-1}$ is an identity element of $T$. By Identity is Unique, $e$ is the only such element of $T$. Thus $\left({T, \circ'}\right)$ is a semigroup with an identity element. So, by definition, $\left({T, \circ'}\right)$ is a commutative monoid. {{qed}} Category:Inverse Completions 92t9hf9kxhakzgw1zv7vnoqrizel2ts
From Subset Product defining Inverse Completion of Commutative Semigroup is Commutative Semigroup: :$S \circ' C^{-1}$ is a commutative semigroup. From Structure of Inverse Completion of Commutative Semigroup: :$T = S \circ' C^{-1}$ {{qed}}
=== Definition 1 implies Definition 2 === Let $\tau$ be a topology on $S$ by definition 1. $\text O 1$ is the same as $\text O 1'$, so $\text O 1'$ holds for $\tau$. From General Intersection Property of Topological Space: :if $\text O 2$ holds, then $\text O 2'$ holds. Thus $\tau$ is a topology on $S$ by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $\tau$ be a topology on $S$ by definition 2. $\text O 1'$ is the same as $\text O 1$, so $\text O 1$ holds for $\tau$. $\text O 2'$ states that the intersection of any finite subset of $\tau$ is an element of $\tau$. This applies when the subset of $\tau$ contains exactly $2$ sets. Thus $\text O 2$ is a direct consequence of $\text O 2'$. Also as a consequence of $\text O 2$, it follows that the intersection of an empty subset of $\tau$ is an element of $\tau$. From Intersection of Empty Set it follows that $S \in \tau$. So $\text O 3$ is a direct consequence of $\text O 2'$. Thus all the open set axioms hold. Thus $\tau$ is a topology on $S$ by definition 1. {{qed}}
Let $X$ be a $G_\delta$ set of $T$. Let $X = \displaystyle \bigcap \UU$ where $\UU$ is a countable intersection of open sets in $T$. Then from De Morgan's Laws: Difference with Intersection we have: :$\displaystyle S \setminus X = S \setminus \bigcap \UU = \bigcup_{U \mathop \in \UU} \paren {S \setminus U}$ By definition of closed set, each of the $S \setminus U$ are closed sets. So $\displaystyle \bigcup_{U \mathop \in \UU} \paren {S \setminus U}$ is a countable union of closed sets in $T$. Hence $S \setminus X$ is, by definition, an $F_\sigma$ set of $T$. {{qed}}
Let $f$ be piecewise continuous and bounded on $\closedint a b$. It is sufficient to prove that the improper integral $\displaystyle \int_{ {x_{i - 1} }^+}^{ {x_i}^-} \map f x \rd x$ exists for every $i \in \set {1, 2, \ldots, n}$. Let $i \in \set {1, 2, \ldots, n}$. Let $c$ be a point in $\openint {x_{i − 1} } {x_i}$. By definition, the improper integral $\displaystyle \int_{ {x_{i - 1} }^+}^{ {x_i}^-} \map f x \rd x$ exists {{iff}}: :$\displaystyle \lim_{\gamma \mathop \to {x_{i - 1} }^+} \int_\gamma^c \map f x \rd x$ and: : $\displaystyle \lim_{\gamma \mathop \to {x_i}^-} \int_c^\gamma \map f x \rd x$ both exist. By Bounded Piecewise Continuous Function is Darboux Integrable we know that $f$ is Darboux integrable on $\closedint a b$. Therefore, $f$ is Darboux integrable on every closed subinterval of $\closedint a b$.  Accordingly, the following Darboux integrals: :$\displaystyle \int_{x_{i - 1} }^c \map f x \rd x$ :$\displaystyle \int_{\gamma_1}^c \map f x \rd x$ :$\displaystyle \int_c^{\gamma_2} \map f x \rd x$ :$\displaystyle \int_c^{x_i} \map f x \rd x$ all exist where $\gamma_1 \in \openint {x_{i − 1} } c$ and $\gamma_2 \in \openint c {x_i}$. Note that $f$ is bounded on $\closedint a b$ as $f$ is piecewise continuous and bounded. Therefore, a bound $B$ exists for $f$ on $\closedint a b$. We have: {{begin-eqn}} {{eqn | l = \size {\int_{\gamma_1}^c \map f x \rd x - \int_{x_{i - 1} }^c \map f x \rd x}       | r = \size {\int_{x_{i - 1} }^{\gamma_1} \map f x \rd x} }} {{eqn | o = \le       | r = B \size {\gamma_1 - x_{i - 1} }       | c = Corollary to Upper and Lower Bounds of Integral }} {{end-eqn}} which approaches $0$ as $\gamma_1$ approaches $x_{i − 1}$. This shows that $\displaystyle \lim_{\gamma \mathop \to x_{i - 1}+} \int_\gamma^c \map f x \rd x$ equals $\displaystyle \int_{x_{i - 1} }^c \map f x \rd x$. From this we gather that $\displaystyle \lim_{\gamma \mathop \to  {x_{i - 1} }^+} \int_\gamma^c \map f x \rd x$ exists. Also: {{begin-eqn}} {{eqn | l = \size {\int_c^{x_i} \map f x \rd x - \int_c^{\gamma_2} \map f x \rd x}       | r = \size {\int_{\gamma_2}^{x_i} \map f x \rd x} }} {{eqn | o = \le       | r = B \size {x_i - \gamma_2}       | c = Corollary to Upper and Lower Bounds of Integral }} {{end-eqn}} which approaches $0$ as $\gamma_2$ approaches $x_i$. This shows that $\displaystyle \lim_{\gamma \mathop \to x_i-} \int_c^\gamma \map f x \rd x$ equals $\displaystyle \int_c^{x_i} \map f x \rd x$. From this we gather that $\displaystyle \lim_{\gamma \mathop \to {x_i}^-} \int_c^\gamma \map f x \rd x$ exists. Since $i$ is arbitrary, we have shown that $\displaystyle \lim_{\gamma \mathop \to {x_{i - 1} }^+} \int_\gamma^c \map f x \rd x$ and $\displaystyle \lim_{\gamma \mathop \to {x_i}^-} \int_c^\gamma \map f x \rd x$ exist for every $i \in \left\{{1, \ldots, n}\right\}$. Therefore, $\displaystyle \int_{ {x_{i - 1} }^+}^{ {x_i}^-} \map f x \rd x$ exists for every $i \in \set {1, 2, \ldots, n}$. That is, $f$ is a piecewise continuous function with improper integrals. {{qed}}
Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$. Let $X \subseteq S$ be an $F_\sigma$ set of $T$. From $F_\sigma$ and $G_\delta$ Subsets of Uncountable Finite Complement Space: :$X$ is neither open nor closed in $T$. Hence the result. {{qed}}
Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$. Let $V$ be a closed set of $T$. From Closed Set of Uncountable Finite Complement Topology is not $G_\delta$: :$V$ is not a $G_\delta$ set of $T$. Hence the result. {{qed}}
Let $U$ be an open set of $T$. As $S$ is uncountable, then so is $U$. By the definition of a finite complement topology, all closed sets of $T$ are finite. From Countable Union of Countable Sets is Countable, $U$ can not be expressed as the union of a countable number of closed sets. So by definition $U$ is not an $F_\sigma$ set. {{qed}}
Let $V$ be a closed set of $T$. {{AimForCont}} $V$ is $G_\delta$ set. Then by Complement of $G_\delta$ Set is $F_\sigma$ Set: :$S \setminus V$ is an $F_\sigma$ set. By definition of closed set, $S \setminus V$ is an open set of $T$. But by Open Set of Uncountable Finite Complement Topology is not $F_\sigma$: :$S \setminus V$ is not an $F_\sigma$ set. It follows by Proof by Contradiction that $V$ is not a $G_\delta$ set. {{qed}}
Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$. Let $X \subseteq S$ be a $G_\delta$ set of $T$. From $F_\sigma$ and $G_\delta$ Subsets of Uncountable Finite Complement Space: :$X$ is neither open nor closed in $T$. Hence the result. {{qed}}
Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$. Let $U$ be an open set of $T$. From Closed Set of Uncountable Finite Complement Topology is not $F_\sigma$: :$U$ is not an $F_\sigma$ set of $T$. Hence the result. {{qed}}
Let $T = \struct {S, \tau}$ be the discrete space on $S$. Let $x \in S$. Then by definition of discrete space: :$U = \set x$ is an open set of $T$. Consider the sequence $\sequence {x_n}$ defined as: :$\forall n \in \N: x_n = x$ That is: :$\sequence {x_n} = \tuple {x, x, x, \ldots}$ From Limit Point of Sequence in Discrete Space not always Limit Point of Open Set: :$x$ is not a limit point of $U$. But from Limit Point of Sequence is Adherent Point of Range: :$x$ is an adherent point of $\set x$. Hence the result. {{qed}}
Let $x \in S$. By definition of discrete space: :$U = \set x$ is an open set of $T$. Consider the sequence $\sequence {x_n}$ defined as: :$\forall n \in \N: x_n = x$ That is: :$\sequence {x_n} = \tuple {x, x, x, \ldots}$ Thus $x$ is the limit point of $\sequence {x_n}$. But: :$U \setminus \set x = \O$ and so $x$ is not a limit point of $U$. Hence the result. {{qed}}
Let $U$ be an open set of $T$ containing $\alpha$. By definition of accumulation point of $\sequence {x_n}$, $U$ contains infinitely many terms of $\sequence {x_n}$. As $\sequence {x_n}$ is a sequence of distinct terms: $U$ contains infinitely many elements of $\set {x_n: n \in \N}$. Thus by definition $\alpha$ is an $\omega$-accumulation point of $\set {x_n: n \in \N}$. {{qed}}
By definition, $\left\{ {x_n: n \in \N}\right\}$ contains all the terms of $\left \langle {x_n} \right \rangle$. In particular, $\left\{ {x_n: n \in \N}\right\}$ contains $\alpha$. Thus by definition $\alpha$ is an adherent point of $\left\{ {x_n: n \in \N}\right\}$. {{qed}} Category:Limit Points Category:Adherent Points ceqq0wwplri2wabadwh8wexrnj5lzps
Let $\alpha$ be an limit point of $\sequence {x_n}$. From Limit Point of Sequence is Accumulation Point, $\alpha$ is an accumulation point of $\sequence {x_n}$. From Accumulation Point of Sequence of Distinct Terms is Omega-Accumulation Point of Range, $\alpha$ is an $\omega$-accumulation point of $\set {x_n: n \in \N}$. {{qed}}
=== Definition 1 implies Definition 2 === Let $x$ be an isolated point of $H$ by definition 1. Then by definition: :$\exists U \in \tau: U \cap H = \set x$ Thus we have an open set in $T$ such that $x \in U$ contains no other point of $H$ than $x$. Thus, by definition, $x$ is not a limit point of $H$. Thus $x$ is an isolated point of $H$ by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $x$ be an isolated point of $H$ by definition 2. {{AimForCont}} $x$ is a limit point of $H$. Then by definition every open set $U \in \tau$ such that $x \in U$ contains some point of $H$ other than $x$. That is: :$\forall U \in \tau: x \in U \implies \exists y \in S, y \ne x: y \in U \cap H$ That is: :$\not \exists U \in \tau: U \cap H = \set x$ because all $U$ with $x$ in them are such that there is at least one point in $U \cap H$ apart from $x$. Thus by Proof by Contradiction $x$ is not a limit point of $H$. That is, $x$ is an isolated point of $H$ by definition 1. {{qed}}
Let $T = \struct {S, \tau}$ be a topological space. Let $H \subseteq S$. === Definition 1 implies Definition 2 === Let $H$ be a closed set of $T$ by definition 1. Let $H^\complement$ denote the relative complement of $H$ in $S$. By definition of closed set in $T$: :$H^\complement$ is open in $T$ From Set is Open iff Neighborhood of all its Points: :$\forall x \in S: x \notin H \implies H^\complement$ is a neighborhood of $x$. By definition of limit point: :$\forall x \in S: x \notin H \implies x$ is not a limit point of $H$ Thus $H$ is a closed set of $T$ by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $H$ be a closed set of $T$ by definition 2. Then by definition: $\forall x \in S: x \notin H \implies x$ is not a limit point of $H$ By definition of limit point: $\forall x \in S: x \notin H \implies H^\complement$ is a neighborhood of $x$ By Set is Open iff Neighborhood of all its Points: $H^\complement$ is open in $T$ Thus $H$ is a closed set of $T$ by definition 1. {{qed}}