hcju/proofwikiqa huggingface dataset:
=== Necessary Condition === Let $\displaystyle \lim_{x \mathop \to c} \map f x = l$. Let $\epsilon > 0$. Then by the definition of the limit of a function: :$\exists \delta > 0: \size {\map f x - l} < \epsilon$ provided $0 < \size {x - c} < \delta$. Now suppose that $\sequence {x_n}$ is a sequence of points of $\openint a b$ such that: :$\forall n \in \N_{>0}: x_n \ne c$ and: :$\displaystyle \lim_{n \mathop \to \infty} x_n = c$ Since $\delta > 0$, from the definition of the limit of a function: :$\exists N: \forall n > N: \size {x_n - c} < \delta$ But: :$\forall n \in \N_{>0}: x_n \ne c$ That means: :$0 < \size {x_n - c} < \delta$ But that implies: :$\size {\map f {x_n} - l} < \epsilon$ That is, given a value of $\epsilon > 0$, we have found a value of $N$ such that: :$\forall n > N: \size {\map f {x_n} - l} < \epsilon$ Thus: :$\displaystyle \lim_{n \mathop \to \infty} \map f {x_n} = l$ {{qed|lemma}} === Sufficient Condition === Suppose that for each sequence $\sequence {x_n}$ of points of $\openint a b$ such that $\forall n \in \N_{>0}: x_n \ne c$ and $\displaystyle \lim_{n \mathop \to \infty} x_n = c$, it is true that: :$\displaystyle \lim_{n \mathop \to \infty} \map f {x_n} = l$ {{AimForCont}} it is ''not'' true that: :$\displaystyle \lim_{x \mathop \to c} \map f x = l$ Thus: :$\exists \epsilon > 0: \forall \delta > 0: \exists x: 0 < \size {x - c} < \delta: \size {\map f {x_n} - l} \ge \epsilon$ In particular, if $\delta = \dfrac 1 n$, we can find an $x_n$ where $0 < \size {x - c} < \dfrac 1 n$ such that: :$\size {\map f {x_n} - l} \ge \epsilon$ But then $\sequence {x_n}$ is a sequence of points of $\openint a b$ such that: :$\forall n \in \N_{>0}: x_n \ne c$ and $\displaystyle \lim_{n \mathop \to \infty} x_n = c$ but for which it is ''not'' true that: :$\displaystyle \lim_{n \mathop \to \infty} \map f {x_n} = l$ The result follows by Proof by Contradiction. {{qed}}
=== Necessary Condition === Let $f$ be continuous at $a \in A_1$. Let $\displaystyle \lim_{n \mathop \to \infty} x_n = a$. Let $V$ be a neighborhood of $f \left({a}\right)$. Then by Metric Space Continuity by Inverse of Mapping between Neighborhoods $f^{-1} \left[{V}\right]$ is a neighborhood of $a$. By Limit of Sequence in Metric Space in Neighborhood: :$\exists N \in \N: n > N \implies x_n \in f^{-1} \left[{V}\right]$ Thus for each neighborhood $V$ of $f \left({a}\right)$ :$\exists N \in \N: n > N \implies f \left({x_n}\right) \in V$ By Limit of Sequence in Metric Space in Neighborhood: : $\displaystyle \lim_{n \mathop \to \infty} f \left({x_n}\right) = f \left({a}\right)$ {{qed|lemma}} === Sufficient Condition === Aiming for a contradiction, suppose $f$ is not continuous at $a \in A_1$. Then there is a neighborhood $V$ of $f \left({a}\right)$ such that for each neighborhood $U$ of $a$: :$f \left[{U}\right] \nsubseteq V$ In particular, for each open $\epsilon$-ball $B_\epsilon \left({a}\right)$ such that $\forall n \in \N_{>0}: \epsilon = \dfrac 1 n$: :$f \left({B_\epsilon \left({a}\right)}\right) \nsubseteq V$ Thus for each $n \in \N_{>0}$ there exists a point $x_n$ such that: :$x_n \subseteq B_\epsilon \left({a}\right)$ and: :$f \left({x_n}\right) \notin V$ Now: :$d_1 \left({a, x_n}\right) < \dfrac 1 n$ and therefore: :$\displaystyle \lim_{n \mathop \to \infty} x_n = a$ whereas $\displaystyle \lim_{n \mathop \to \infty} f \left({x_n}\right) = f \left({a}\right)$ is impossible, as: :$\forall n \in \N_{>0}: f \left({x_n}\right) \notin V$ {{qed}}
From Limit of Function by Convergent Sequences, we have: :$\displaystyle \lim_{x \mathop \to \xi} f \left({x}\right) = l$ {{iff}}: : for each sequence $\left \langle {x_n} \right \rangle$ of points of $\left({a \,.\,.\, b}\right)$ such that: :: $\forall n \in \N_{>0}: x_n \ne \xi$ :and: :: $\displaystyle \lim_{n \mathop \to \infty} x_n = \xi$ :it is true that: :: $\displaystyle \lim_{n \mathop \to \infty} f \left({x_n}\right) = l$ The result follows directly from this and the definition of continuity on an interval. {{qed}}
Note that $A$ is non-empty as the empty set does not admit an infimum (in $\R$). Suppose $\epsilon \in \R_{>0}$ such that: :$\forall x \in A: x - b \ge \epsilon$ Then: :$\forall x \in A: b + \epsilon \le x$ and so $b + \epsilon$ would be a lower bound of $A$ which is greater than $b$. But since $b$ is an infimum of $A$ there can be no such $b + \epsilon$. From that contradiction it follows that: :$\exists x \in A: x - b < \epsilon$ {{qed}}
From Infimum of Subset of Real Numbers is Arbitrarily Close: For $\epsilon = \dfrac 1 n$ there exists an $a_n \in A$ such that: :$a_n - b < \dfrac 1 n$ Since $b$ is an infimum of $A$: :$0 \le a_n - b$ Therefore: :$\displaystyle \lim_{n \mathop \to \infty} a_n = b$ {{qed}}
From Existence of Sequence in Set of Real Numbers whose Limit is Infimum: :$\displaystyle \lim_{n \mathop \to \infty} \map d {a, a_n} = b$ where $b$ is an infimum of $\map d {a, a_n}$. Hence the result by definition of distance to a subset of a metric space. {{qed}}
{{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} a_n       | r = c       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \lim_{n \mathop \to \infty} \map {d_\infty} {a_n, c}       | r = 0       | c = {{Defof|Convergent Sequence|subdef = Metric Space|index = 3}} }} {{eqn | ll= \leadstoandfrom       | l = \lim_{n \mathop \to \infty} \max_{i \mathop = 1}^k \set {\map {d_i} {a_{n i}, c_i} }       | r = 0       | c = {{Defof|Chebyshev Distance}} }} {{eqn | ll= \leadstoandfrom       | l = \max_{i \mathop = 1}^k \set {\lim_{n \mathop \to \infty} \map {d_i} {a_{n i}, c_i} }       | r = 0       | c = Limit of Image of Sequence }} {{eqn | ll= \leadstoandfrom       | lo= \forall i \in \set {1, 2, \ldots, k}:       | l = \lim_{n \mathop \to \infty} \map {d_i} {a_{n i}, c_i}       | r = 0       | c = {{Defof|Max Function}} }} {{end-eqn}} {{qed}}
Let $F_n$ be the $n$th Fibonacci number. By definition: :$F_0 = 0, F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3, \ldots$ Hence, $g_n$, the sum of the first $n$ Fibonacci numbers is :$g_0 = 0, g_1 = 1, g_2 = 2, g_3 = 4, g_4 = 7, \ldots$ Proof by induction: For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \sum_{j \mathop = 0}^n F_j = g_n = g_{n - 1} + g_{n - 2} + 1$ === Basis for the Induction === $P(2)$ is the case $g_2 = g_1 + g_0 + 1$, which holds: :$2 = 1 + 0 + 1$ This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is our induction hypothesis: :$\displaystyle \sum_{j \mathop = 0}^k F_j = \sum_{j \mathop = 0}^{k - 1} F_j + \sum_{j \mathop = 0}^{k - 2} F_j + 1$ Then we need to show: :$\displaystyle \sum_{j \mathop = 0}^{k + 1} F_j = \sum_{j \mathop = 0}^k F_j + \sum_{j \mathop = 0}^{k - 1} F_j + 1$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = g_{k + 1}       | r = \sum_{j \mathop = 0}^{k + 1} F_j       | c = Definition of $g_k$ }} {{eqn | r = \sum_{j \mathop = 0}^k F_j + F_{k + 1}       | c =  }} {{eqn | r = \sum_{j \mathop = 0}^{k - 1} F_j + \sum_{j \mathop = 0}^{k - 2} F_j + 1 + F_{k + 1}       | c = Induction Hypothesis }} {{eqn | r = \sum_{j \mathop = 0}^{k - 1} F_j + \sum_{j \mathop = 0}^{k - 2} F_j + 1 + (F_k + F_{k - 1})       | c = {{Defof|Fibonacci Numbers}}: $F_{k + 1} = F_k + F_{k - 1}$ }} {{eqn | r = \sum_{j \mathop = 0}^k F_j + \sum_{j \mathop = 0}^{k - 1} F_j + 1       | c = pulling terms into summations }} {{eqn | r = g_k + g_{k - 1} + 1       | c =  }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \ge 2: g_n = g_{n - 1} + g_{n - 2} + 1$ where $g_n$ is the sum of the first $n$ Fibonacci numbers. {{qed}} Category:Fibonacci Numbers dku20szhkbm8map6jy77dy7ntvtzh76
Let $\epsilon > 0$. Since $\displaystyle \lim_{n \mathop \to \infty} x_n = l$, it follows from the definition of limit that: :$\exists N: \forall n > N: d \left({x_n, l}\right) < \epsilon$ Now let $R = N$. Then from Strictly Increasing Sequence of Natural Numbers: : $\forall r > R: n_r \ge r$ Thus $n_r > N$ and so: :$d \left({x_n, l}\right) < \epsilon$ The result follows. {{qed}}
From Real Number Line is Metric Space, the set $\R$ under the usual metric is a metric space. By Convergent Sequence in Metric Space is Bounded it follows that: :$\exists M > 0: \forall n, m \in \N: \size {x_n - x_m} \le M$ Then for $n \in \N$, by the Triangle Inequality for Real Numbers: {{begin-eqn}} {{eqn | l = \size {x_n}       | r = \size {x_n - x_1 + x_1}       | c =  }} {{eqn | o = \le       | r = \size {x_n - x_1} + \size {x_1}       | c =  }} {{eqn | o = \le       | r = M + \size {x_1}       | c =  }} {{end-eqn}} Hence $\sequence {x_n}$ is bounded. {{qed}}
Let $\sequence {x_n}$ be a sequence in $\R$. Let $x_n \to l$ as $n \to \infty$. To show that $\sequence {x_n}$ is bounded sequence, we need to find $K$ such that: :$\forall n \in \N: \size {x_n} \le K$ Because $\sequence {x_n}$ converges: :$\forall \epsilon > 0: \exists N: n > N \implies \size {x_n - l} < \epsilon$ In particular, this is true when $\epsilon = 1$. That is: :$\exists N_1: \forall n > N_1: \size {x_n - l} < 1$ By Backwards Form of Triangle Inequality: :$\forall n > N_1: \size {x_n} - \size l \le \size {x_n - l} < 1$ That is: :$\size {x_n} < \size l + 1$ So we set: :$K = \max \set {\size {x_1}, \size {x_2}, \ldots, \size {x_{N_1} }, \size l + 1}$ and the result follows. {{qed}}
Let $\sequence {x_n}$ be decreasing and bounded below. Then $\sequence {-x_n}$ is increasing and bounded above. Thus the Monotone Convergence Theorem for Increasing Sequence applies and the proof follows. {{Qed}}
Suppose $\sequence {x_n}$ is increasing and bounded above. By the Continuum Property, it has a supremum, $B$. We need to show that $x_n \to B$ as $n \to \infty$. Let $\epsilon > 0$. By the definition of supremum, $B - \epsilon$ is not an upper bound. Thus: :$\exists x_N: x_N > B - \epsilon$ But $\sequence {x_n}$ is increasing. Hence: :$\forall n > N: x_n \ge x_N > B - \epsilon$ But $B$ ''is'' still an upper bound for $\sequence {x_n}$. Then: {{begin-eqn}} {{eqn | l = \forall n > N: B - \epsilon       | o = <       | r = x_n \le B       | c = }} {{eqn | ll= \leadsto       | l = \forall n > N: B - \epsilon       | o = <       | r = x_n < B + \epsilon       | c = Real Plus Epsilon }} {{eqn | ll= \leadsto       | l = \forall n > N: \size {x_n - B}       | o = <       | r = \epsilon       | c = Negative of Absolute Value: Corollary 1 }} {{end-eqn}} Hence the result. {{Qed}}
Let $S = \map {B_\epsilon} x$ be an open $\epsilon$-ball at $x$. Let $y = \tuple {y_1, y_2} \in \map {B_\epsilon} x$. Then: {{begin-eqn}} {{eqn | l = y       | o = \in       | m = \map {B_\epsilon} x       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \map d {y, x}       | o = <       | m = \epsilon       | c = {{Defof|Open Ball|Open $\epsilon$-Ball}} }} {{eqn | ll= \leadstoandfrom       | l = \sqrt {\paren {y_1 - x_1}^2 + \paren {y_2 - x_2}^2}       | o = <       | m = \epsilon       | c = {{Defof|Real Number Plane with Euclidean Metric}} }} {{eqn | ll= \leadstoandfrom       | l = \paren {y_1 - x_1}^2 + \paren {y_2 - x_2}^2       | o = <       | m = \epsilon^2 }} {{end-eqn}} But from Equation of Circle: :$\paren {y_1 - x_1}^2 + \paren {y_2 - x_2}^2 = \epsilon^2$ is the equation of a circle whose center is $\tuple {x_1, x_2}$ and whose radius is $\epsilon$. The result follows by definition of interior and Open Ball of Point Inside Open Ball. {{qed}}
Let $S = \map {B_\epsilon} x$ be an open $\epsilon$-ball at $x$. Let $y = \tuple {y_1, y_2, y_3} \in \map {B_\epsilon} x$. Then: {{begin-eqn}} {{eqn | l = y       | o = \in       | m = \map {B_\epsilon} x       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \map d {y, x}       | o = <       | m = \epsilon       | c = {{Defof|Open Ball|Open $\epsilon$-Ball}} }} {{eqn | ll= \leadstoandfrom       | l = \sqrt {\paren {y_1 - x_1}^2 + \paren {y_2 - x_2}^2 + \paren {y_3 - x_3}^2}       | o = <       | m = \epsilon       | c = {{Defof|Euclidean Metric|subdef = Ordinary Space}} }} {{eqn | ll= \leadstoandfrom       | l = \paren {y_1 - x_1}^2 + \paren {y_2 - x_2}^2 + \paren {y_3 - x_3}^2       | o = <       | m = \epsilon^2 }} {{end-eqn}} But from Equation of Sphere in Rectangular Coordinates: :$\paren {y_1 - x_1}^2 + \paren {y_2 - x_2}^2 + \paren {y_3 - x_3}^2 = \epsilon^2$ is the equation of a sphere whose center is $x$ and whose radius is $\epsilon$. The result follows by definition of interior and Open Ball of Point Inside Open Ball. {{qed}}
Let $a = \left({a_1, a_2}\right)$. From Open Ball in Cartesian Product under Chebyshev Distance: :$B_\epsilon \left({a; d_\infty}\right) = B_\epsilon \left({a_1; d}\right) \times B_\epsilon \left({a_2; d}\right)$ where $d$ is the usual (Euclidean) topology. From Open Ball in Real Number Line is Open Interval: :$B_\epsilon \left({a_1; d}\right) \times B_\epsilon \left({a_2; d}\right) = \left({a_1 - \epsilon \,.\,.\, a_1 + \epsilon}\right) \times \left({a_2 - \epsilon \,.\,.\, a_2 + \epsilon}\right)$ That is: :$x \in B_\epsilon \left({a; d_\infty}\right) \iff \left({a_2 - \epsilon < x_2 < a_2 + \epsilon}\right) \land \left({a_2 - \epsilon < x_2 < a_2 + \epsilon}\right)$ from which the result follows. {{qed}}
$\forall x, y \in X, \forall z \in A$, by the definition of the distance from $x$ to $A$, we have: :$\map d {x, A} \le \map d {x, z} \le \map d {x, y} + \map d {y, z}$ Taking infimum with respect to $z \in A$ to above inequality, we get: :$\map d {x, A} \le \map d {x, y} + \map d {y, A}$ Similarly, we have: :$\map d {y, A} \le \map d {x, y} + \map d {x, A}$ Therefore: :$\norm {\map d {x, A} - \map d {y, A} } \le \map d {x, y}$ We can conclude that $f$ is continuous. {{qed}}
=== Sufficient Condition === Let $x \in X$. Let every neighborhood of $x$ contain a point in $A$. Every open ball $\map {B_\epsilon} x$ centered at $x$ is seen to be a neighborhood of $x$ in $M$. But then this implies that for every $\epsilon \in \R_{\gt 0}$ there must exists a $y \in A$ such that: :$\map d {x, y} < \epsilon$ From this it is seen that $\map d {x, A} = 0$. {{qed|lemma}} === Necessary Condition === Let $\map d {x, A} = 0$ for $x \in X$. Let $S$ be a neighborhood of $x$ in $M$. From the definition of a neighborhood, there exists an open ball $\map {B_\epsilon} x$ centered at $x$ such that: :$\map {B_\epsilon} x \subseteq S$ We have that: :$\epsilon < 0$ and $\map d {x, A} = 0$ Hence there must exists a $y \in A$ such that: :$\map d {x, y} < \epsilon \implies \map {B_\epsilon} x \cap A \ne \O$ If otherwise then $\epsilon$ would be a lower bound for $\set {\map d {x, y} : y \in A}$ that is greater than $0$, which is a contradiction. Then from Set Intersection Preserves Subsets we have: :$\map {B_\epsilon} x \cap A \subseteq S \cap A$ and so: :$\map {B_\epsilon} x \cap A \ne \O \implies S \cap A \ne \O$ Hence the result. {{qed}}
From definition of continuous mapping: We just need to show that: :$\forall \tuple {a, b} \in A \times A: \forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall \tuple {x, y} \in A \times A: \map {d_\infty} {\tuple {x, y}, \tuple {a, b}} < \delta \implies \map d {\map {d'} {x, y}, \map {d'} {a, b}} < \epsilon$ Let $\tuple {a, b} \in A \times A$ and $\epsilon \in \R_{>0}$. Let $\tuple {x, y} \in A \times A$. Suppose $\map {d_\infty} {\tuple {x, y}, \tuple {a, b}} < \dfrac \epsilon 2$. Then: {{begin-eqn}} {{eqn | l = \map d {\map {d'} {x, y}, \map {d'} {a, b} }       | r = \size {\map {d'} {x, y} - \map {d'} {a, b} }       | c = {{Defof|Euclidean Metric on Real Number Line}} }} {{eqn | o = \le       | r = \size {\map {d'} {x, y} - \map {d'} {a, y} } + \size {\map {d'} {a, y} - \map {d'} {a, b} }       | c = Triangle Inequality }} {{eqn | o = \le       | r = \map {d'} {x, a} + \map {d'} {y, b}       | c = Reverse Triangle Inequality on $d'$ }} {{eqn | o = \le       | r = 2 \max \set {\map {d'} {x, a}, \map {d'} {y, b} }       | c = }} {{eqn | o = \le       | r = 2 \ \map {d_\infty} {\tuple {x, y}, \tuple {a, b} }       | c = {{Defof|Chebyshev Distance}} }} {{eqn | o =  <       | r = 2 \cdot \frac \epsilon 2 }} {{eqn | r = \epsilon }} {{end-eqn}} Therefore $d'$ is continuous. {{qed}}
=== Necessary Condition === Let $U$ be open in $M$. Let $a \in U$. Then by definition of open set: :$\exists \delta_a \in \R_{>0}: \map {B_{\delta_a} } {a, d} \subseteq U$ where $\map {B_{\delta_a} } {a, d}$ is the open $\delta_a$-ball of $a$ in $M$. Therefore: :$\displaystyle U = \bigcup_{a \mathop \in U} \map {B_{\delta_a} } {a, d}$ and so $U$ is a union of open balls. {{qed|lemma}} === Sufficient Condition === Let $U$ be a union of open balls of $M$. Let the centers of these open balls be the elements of an indexing set $I$. Then $U$ can be written as: :$\displaystyle U = \bigcup_{a \mathop \in I} \map {B_{\delta_a} } {a, d}$ where $\delta_a \in \R_{>0}$ is the radius of the open ball of $a$. Let $x \in U$. Then by definition of union: :$\exists a \in I: x \in \map {B_{\delta_a} } {a, d}$ From Open Ball is Neighborhood of all Points Inside, $\map {B_{\delta_a} } {a, d}$ is a neighborhood of $x$. By Set is Subset of Union: :$\map {B_{\delta_a} } {a, d} \subseteq U$ From Superset of Neighborhood in Metric Space is Neighborhood, it follows that $U$ is a neighborhood of $x$. Since $x$ is arbitrary, it follows that $U$ is a neighborhood of each of its points. Hence by definition, $U$ is open in $M$. {{qed}}
From Empty Set is Open in Metric Space, $\varnothing$ is open in $M$. From Empty Set is Closed in Metric Space, $\varnothing$ is closed in $M$. {{qed}}
From Metric Space is Open in Itself, $A$ is open in $M$. But: :$\varnothing = \relcomp A A$ where $\complement_A$ denotes the set complement relative to $A$. The result follows by definition of closed set. {{qed}}
From Empty Set is Open in Metric Space, $\O$ is open in $M$. But: :$A = \relcomp A \O$ where $\complement_A$ denotes the set complement relative to $A$. The result follows by definition of closed set. {{qed}}
From Metric Space is Open in Itself, $A$ is open in $M$. From Metric Space is Closed in Itself, $A$ is closed in $M$. {{qed}}
=== Necessary Condition === Let $H$ be closed in $M$. Suppose that: :$\displaystyle \lim_{n \mathop \to \infty} a_n = a$ and: :$\forall n \in \N_{>0}: a_n \in H$ If the set $\left\{ {a_1, a_2, \ldots}\right\}$ is infinite then every neighborhood of $a$ contains infinitely many points of $H$. Thus $a$ is a limit point of $H$. So by definition of closed set, $a \in H$. On the other hand, if $\left\{ {a_1, a_2, \ldots}\right\}$ is finite, then for some $N \in \N$: :$n, m > N \implies a_n = a_m$ Since: :$\displaystyle \lim_{n \mathop \to \infty} a_n = a$ Then: :$\forall n > N: d \left({a_n, a}\right) = 0$ Thus: :$a_n = a$ and so: :$a \in H$ {{qed|lemma}} === Sufficient Condition === Let $H$ be a set such that: :for each sequence $\left\langle{a_n}\right\rangle$ such that $\displaystyle \lim_{n \mathop \to \infty} a_n = a$, it follows that $a \in H$. Let $b$ be a limit point of $H$. Then $b$ is the limit of a convergent sequence of points of $H$. By hypothesis, $b \in H$. Thus $H$ is a closed set by definition. {{Qed}}
=== Necessary Condition === Let $H$ be closed in $M$. Let $x \in A: \map d {x, H} = 0 \implies x \in H$. By Existence of Sequence in Subset of Metric Space whose Limit is Infimum there exists a sequence $\sequence {a_n}$ of points of $H$ such that: :$\displaystyle \lim_{n \mathop \to \infty} \map d {x, a_n} = 0$ So every neighborhood of $x$ contains points of $H$. If some $a_n \in H$ then $x \in H$. Otherwise each $a_n$ is different from $x$ Then $x$ is the limit of $\sequence {a_n}$ and hence of $H$. Thus $x \in H$ by definition of closed set. {{qed|lemma}} === Sufficient Condition === Suppose $H$ is such that: :$\forall x \in A: \map d {x, H} = 0 \implies x \in H$ Let $x$ be a limit point of $H$. Then $\map d {x, H} = 0$. {{explain|There should be a result proving the statement: $x$ is a limit point of $H$ implies $\map d {x, H} {{=}} 0$. I thought it was already there but I can't find it.}} Thus $H$ contains all its limit points and so is by definition a closed set. {{qed}}
First the following is established. Let $W \in A_2$. We note that: : $f^{-1} \left[{A_2}\right] = A_1$ Hence, from Preimage of Set Difference under Mapping: :$f^{-1} \left[{A_2 \setminus W}\right] = A_1 \setminus f^{-1} \left[{W}\right]$ === Necessary Condition === Suppose that: :for every $V \subseteq A_2$ which is closed in $M_2$, $f^{-1} \left({V}\right)$ is closed in $M_1$. Let $U$ be open in $M_2$. Then by definition $A_2 \setminus U$ is closed in $M_2$. By hypothesis, $f^{-1} \left[{A_2 \setminus U}\right] = A_1 \setminus f^{-1} \left[{U}\right]$ is closed in $M_1$. So $f^{-1} \left[{U}\right]$ is open in $M_1$. This is true for any $U \in A_2$ as $U$ is arbitrary. It follows by definition that $f$ is continuous. {{qed|lemma}} === Sufficient Condition === Now let $f$ be continuous. Let $V$ be closed in $M_2$. Then $A_2 \setminus V$ is open in $M_2$. As $f$ is continuous, $f^{-1} \left[{A_2 \setminus V}\right] = A_1 \setminus f^{-1} \left[{V}\right]$ is open in $M_1$. So $f^{-1} \left[{V}\right]$ is closed in $M_1$. {{qed}}
From Metric Space is Closed in Itself, $C1$ holds. {{qed|lemma}} From Empty Set is Closed in Metric Space, $C2$ holds. {{qed|lemma}} Let $\displaystyle \bigcup_{i \mathop = 1}^n V_i$ be the union of a finite number of closed sets of $M$. Then from De Morgan's laws: :$\displaystyle A \setminus \bigcup_{i \mathop = 1}^n V_i = \bigcap_{i \mathop = 1}^n \paren {A \setminus V_i}$ By definition of closed set, each of the $A \setminus V_i$ is by definition open in $M$. We have that $\displaystyle \bigcap_{i \mathop = 1}^n \paren {A \setminus V_i}$ is the intersection of a finite number of open sets of $M$. Therefore, by Finite Intersection of Open Sets of Metric Space is Open, $\displaystyle \bigcap_{i \mathop = 1}^n \paren {A \setminus V_i} = A \setminus \bigcup_{i \mathop = 1}^n V_i$ is likewise open in $M$. Then by definition of closed set, $\displaystyle \bigcup_{i \mathop = 1}^n V_i$ is closed in $M$. Thus $C3$ holds. {{qed|lemma}} Let $I$ be an indexing set (either finite or infinite). Let $\displaystyle \bigcap_{i \mathop \in I} V_i$ be the intersection of a indexed family of closed sets of $M$ indexed by $I$. Then from De Morgan's laws: :$\displaystyle A \setminus \bigcap_{i \mathop \in I} V_i = \bigcup_{i \mathop \in I} \paren {A \setminus V_i}$ By definition of closed set, each of $A \setminus V_i$ are by definition open in $M$. We have that $\displaystyle \bigcup_{i \mathop \in I} \paren {A \setminus V_i}$ is the union of a family of open sets of $M$ indexed by $I$. Therefore, by definition of a topology, $\displaystyle \bigcup_{i \mathop \in I} \paren {A \setminus V_i} = A \setminus \bigcap_{i \mathop \in I} V_i$ is likewise open in $M$. Then by definition of closed set, $\displaystyle \bigcap_{i \mathop \in I} V_i$ is closed in $M$. Thus $C4$ holds. {{qed}}
Consider the closed real interval: :$\closedint {\dfrac 1 n} 1 \subseteq \R$ From Closed Real Interval is Closed Set, $\closedint {\dfrac 1 n} 1$  is closed in $\R$ for all $n \in \N$. But: :$\displaystyle \bigcup_{n \mathop \in \N} \closedint {\dfrac 1 n} 1 = \hointl 0 1$ The result follows from Half-Open Real Interval is neither Open nor Closed. {{qed}}
Let $U$ be an open set of $M$. Then by definition: :$\forall y \in U: \exists \epsilon \in \R_{>0}: B_\epsilon \left({y}\right) \subseteq U$ Thus: :$\displaystyle U = \bigcup_{y \mathop \in U} B_\epsilon \left({y}\right)$ {{qed}}
From Infimum of Bounded Below Set of Reals is in Closure: :$\map \inf A \in \map \cl A$ From Set is Closed iff Equals Topological Closure: :$A = \map \cl A$ Therefore $\map \inf A \in A$. {{qed}}
Let $a \in H_i$. Then by definition of isolated point: :$\exists \epsilon \in \R_{>0}: \left\{{x \in H: d \left({x, a}\right) < \epsilon}\right\} = \left\{{a}\right\}$ But by metric space axiom $M1$: :$d \left({a, a}\right) = 0$ and so: :$\left\{{x \in H: 0 < d \left({x, a}\right) < \epsilon}\right\} = \varnothing$ So by definition $a$ is not a limit point of $H$. That is: :$a \notin H'$ or: :$a \in \complement_A \left({H'}\right)$ It follows from Intersection with Complement is Empty iff Subset that: :$H' \cap H^i = \varnothing$ {{qed}}