hcju/proofwikiqa huggingface dataset:
By definition of the standard discrete metric: :$d \left({x, y}\right) = \begin{cases} 0 & : x = y  \\  1 & : x \ne y \end{cases}$ Let $\epsilon \in \R_{>0}$ such that $\epsilon < 1$. Then: {{begin-eqn}} {{eqn | l = B_\epsilon \left({a}\right)       | r = \left\{ {x \in A: d \left({x, a}\right) < \epsilon}\right\}       | c = Definition of Open $\epsilon$-Ball of $a$ }} {{eqn | ll= \implies       | l = B_\epsilon \left({a}\right)       | r = \left\{ {a}\right\}       | c =  }} {{end-eqn}} So by definition $\left\{ {a}\right\}$ is a neighborhood of $a$. Let $\mathcal N_a$ be the system of neighborhoods of $a$. Let $N \in \mathcal N_a$ be a neighborhood of $a$. Then: {{begin-eqn}} {{eqn | l = a       | o = \in       | r = N       | c = Point in Metric Space is Element of its Neighborhood }} {{eqn | ll= \implies       | l = \left\{ {a}\right\}        | o = \subseteq       | r = N       | c =  }} {{eqn | ll= \implies       | l = B_\epsilon \left({a}\right)       | o = \subseteq       | r = N       | c = for all $\epsilon \in \R_{>0}$ such that $\epsilon < 1$ }} {{end-eqn}} Hence the result by definition of basis of a system of neighborhoods. {{qed}}
Let $S \subseteq A$. Let $a \in S$. From Neighborhoods in Standard Discrete Metric Space, $\left\{{a}\right\}$ is a neighborhood of $a$. As $a \in S$ it follows from Singleton of Element is Subset that $\left\{{a}\right\} \subseteq S$. The result follows from Superset of Neighborhood in Metric Space is Neighborhood. {{qed}}
Let $x \in \R: x \ne c$. Let $\epsilon \in \R_{>0}$. Let $\delta < \left\vert{x - c}\right\vert$. Then by definition of the Heaviside step function: :$\forall y \in \left[{x - \delta \,.\,.\, x + \delta}\right]: \mu_c \left({x}\right) = \begin{cases} 0 & : x < c \\ 1 & : x > c \end{cases}$ Thus: :$\forall y \in \left[{x - \delta \,.\,.\, x + \delta}\right]: \mu_c \left({y}\right) = \mu_c \left({x}\right)$ Thus: :$\left\vert{y - x}\right\vert \implies : \mu_c \left({y}\right) - \mu_c \left({x}\right) < \epsilon$ and so $\mu_c$ is continuous at $\R$ except at $x$. {{qed|lemma}} Now suppose $x = c$. Let $\epsilon = \dfrac 1 3$. Let $\delta \in \R_{>0}$. Let $\alpha \in \R_{>0}: \alpha < \delta$. Let $y = x + \alpha$. Let $y \in \R: \left\vert{y - x}\right\vert < \delta$ and $y \ne x$. Suppose $\mu_c \left({y}\right) - \mu_c \left({x}\right) < \epsilon$. Then: {{begin-eqn}} {{eqn | l = y       | r = x + \alpha       | c =  }} {{eqn | ll= \implies       | l = \mu_c \left({y}\right)       | r = 1       | c =  }} {{end-eqn}} while: {{begin-eqn}} {{eqn | l = y       | r = x - \alpha       | c =  }} {{eqn | ll= \implies       | l = \mu_c \left({y}\right)       | r = 0       | c =  }} {{end-eqn}} So whatever the value of $f \left({x}\right)$, either: :$f \left({x + \alpha}\right) - f \left({x}\right) > \epsilon$ or: :$f \left({x - \alpha}\right) - f \left({x}\right) > \epsilon$ It follows that $\mu_c$ is not continuous at $c$. {{qed}}
By definition, $\BB_{\map f a}$ be a basis for the neighborhood system at $\map f a$ {{iff}}: :$\forall N_a \subseteq M_2: \exists N \in \BB_{\map f a}: N \subseteq N_a$ where $N_a$ denotes a neighborhood of $\map f a$ in $M_2$. === Necessary Condition === Let $f$ be continuous at $a$ with respect to the metrics $d_1$ and $d_2$. Then by Metric Space Continuity by Inverse of Mapping between Neighborhoods: :for each neighborhood $N$ of $\map f a$ in $M_2$, $f^{-1} \\sqbrk N$ is a neighborhood of $a$. In particular, let $N \in \BB_{\map f a}$. Then $f^{-1} \sqbrk N$ is a neighborhood of $a$. {{qed|lemma}} === Sufficient Condition === Let $f$ be such that: :for each neighborhood $N$ in $\BB_{\map f a}$, $f^{-1} \sqbrk N$ is a neighborhood of $a$. Let $N'$ be any neighborhood of $\map f a$. Then by definition: :$\exists B \in \BB_{\map f a}: B \subseteq N'$ From Image of Subset under Relation is Subset of Image: Corollary 3: :$f^{-1} \sqbrk B \subseteq f^{-1} \sqbrk {N'}$ By hypothesis $f^{-1} \sqbrk B$ is a neighborhood of $a$. From Superset of Neighborhood in Metric Space is Neighborhood it follows that $f^{-1} \sqbrk {N'}$ is a neighborhood of $a$. As $N'$ is arbitrary, the result follows. {{qed}}
Let $N$ be a neighborhood of $a$ in $M$. Then by definition: :$\exists \epsilon' \in \R_{>0}: \map {B_{\epsilon'} } a \subseteq N$ where $\map {B_{\epsilon'} } a$ is the open $\epsilon'$-ball at $a$. From Open Ball in Real Number Line is Open Interval: :$\map {B_{\epsilon'} } a = \openint {a - \epsilon'} {a + \epsilon'}$ Let $\epsilon \in \R_{>0}: \epsilon < \epsilon'$. Then by definition of closed interval and open interval: :$\closedint {a - \epsilon} {a + \epsilon} \subseteq \openint {a - \epsilon'} {a + \epsilon'}$ From Subset Relation is Transitive: :$\closedint {a - \epsilon} {a + \epsilon} \subseteq N$ Also by definition of closed interval and open interval: :$\openint {a - \epsilon} {a + \epsilon} \subseteq \closedint {a - \epsilon} {a + \epsilon}$ From Open Real Interval is Open Ball, $\openint {a - \epsilon} {a + \epsilon}$ is the open $\epsilon$-ball at $a$. Thus by definition, $\closedint {a - \epsilon} {a + \epsilon}$ is a neighborhood of $a$ in $M$. Hence there exists a neighborhood $\closedint {a - \epsilon} {a + \epsilon}$ of $a$ which is a subset of $N$. Hence the result by definition of basis for the neighborhood system of $a$. {{qed}}
Let $N$ be a neighborhood of $a$ in $M$. Then by definition: :$\exists \epsilon' \in \R_{>0}: \map {B_{\epsilon'} } a \subseteq N$ where $\map {B_{\epsilon'} } a$ is the open $\epsilon'$-ball at $a$. From Open Ball in Real Number Line is Open Interval: :$\map {B_{\epsilon'} } a = \openint {a - \epsilon'} {a + \epsilon'}$ From Between two Real Numbers exists Rational Number: :$\exists \epsilon \in \Q: 0 < \epsilon < \epsilon'$ and so: :$\openint {a - \epsilon} {a + \epsilon} \subseteq \openint {a - \epsilon'} {a + \epsilon'}$ From Open Real Interval is Open Ball :$\map {B_\epsilon} a = \openint {a - \epsilon} {a + \epsilon}$ is the open $\epsilon$-ball at $a$. From Subset Relation is Transitive: :$\openint {a - \epsilon} {a + \epsilon} \subseteq N$ From Open Ball is Neighborhood of all Points Inside, $\openint {a - \epsilon} {a + \epsilon}$ is a neighborhood of $a$ in $M$. Hence there exists a neighborhood $\openint {a - \epsilon} {a + \epsilon}$ of $a$ which is a subset of $N$. Hence the result by definition of basis for the neighborhood system of $a$. {{qed}}
Let $N$ be a neighborhood of $a$ in $M$. Then by definition: :$\exists \epsilon' \in \R_{>0}: \map {B_{\epsilon'} } a \subseteq N$ where $\map {B_{\epsilon'} } a$ is the open $\epsilon'$-ball at $a$. From Open Ball in Real Number Line is Open Interval: :$\map {B_{\epsilon'} } a = \openint {a - \epsilon'} {a + \epsilon'}$ From Between two Real Numbers exists Rational Number: :$\exists \epsilon'' \in \Q: 0 < \epsilon'' < \epsilon'$ Let $\epsilon''$ be expressed in canonical form: :$\epsilon'' = \dfrac p q$ Let $\epsilon = \dfrac 1 q$ Then $\epsilon \le \epsilon'' < \epsilon'$ and so: :$\openint {a - \epsilon} {a + \epsilon} \subseteq \openint {a - \epsilon'} {a + \epsilon'}$ From Open Real Interval is Open Ball :$\map {B_\epsilon} a = \openint {a - \epsilon} {a + \epsilon}$ is the open $\epsilon$-ball at $a$. From Subset Relation is Transitive: :$\openint {a - \epsilon} {a + \epsilon} \subseteq N$ From Open Ball is Neighborhood of all Points Inside, $\openint {a - \epsilon} {a + \epsilon}$ is a neighborhood of $a$ in $M$. Hence there exists a neighborhood $\openint {a - \epsilon} {a + \epsilon}$ of $a$ which is a subset of $N$. Hence the result by definition of basis for the neighborhood system of $a$. {{qed}}
Let $N$ be a neighborhood of $a$ in $M$. Then by definition: :$\exists \epsilon' \in \R_{>0}: \map {B_{\epsilon'} } a \subseteq N$ where $\map {B_{\epsilon'} } a$ is the open $\epsilon'$-ball at $a$. From Open Ball in Real Number Line is Open Interval: :$\map {B_{\epsilon'} } a = \openint {a - \epsilon'} {a + \epsilon'}$ From Between two Real Numbers exists Rational Number: :$\exists \epsilon'' \in \Q: 0 < \epsilon'' < \epsilon'$ Let $\epsilon''$ be expressed in canonical form: :$\epsilon'' = \dfrac p q$ Let $\epsilon''' = \dfrac 1 q$ Then $\epsilon''' \le \epsilon'' < \epsilon'$ If $q \le k$, let $\epsilon = \dfrac 1 {k + 1}$ Otherwise, let $\epsilon = \epsilon'''$. Then: :$\openint {a - \epsilon} {a + \epsilon} \subseteq \openint {a - \epsilon'} {a + \epsilon'}$ From Open Real Interval is Open Ball :$\map {B_\epsilon} a = \openint {a - \epsilon} {a + \epsilon}$ is the open $\epsilon$-ball at $a$. By its method of construction: :$\map {B_\epsilon} a \in \set {\map {B_\epsilon} a: \epsilon \in \set {\dfrac 1 n: n \in \N, n > k} }$ From Subset Relation is Transitive: :$\openint {a - \epsilon} {a + \epsilon} \subseteq N$ From Open Ball is Neighborhood of all Points Inside, $\openint {a - \epsilon} {a + \epsilon}$ is a neighborhood of $a$ in $M$. Hence the result by definition of basis for the neighborhood system of $a$. {{qed}}
Suppose $\BB_a$ be finite. Let the elements of $\BB_a$ be enumerated as $N_1, N_2, \ldots, N_n$. For each $N_k \in \BB_a$, let $B_{\epsilon_k} \left({a}\right)$ be the open $\epsilon_k$-ball of $a$ for some $\epsilon_k \in \R_{>0}$. Let $\alpha = \min \set {\epsilon_k: k \in \set {1, 2, \ldots, n} }$. Consider the open interval: :$I = \openint {a - \beta} {a + \beta}$ where $\beta = \dfrac \alpha 2$ By the method of construction of $I$, it follows that $\map {B_{\epsilon_k} } a \nsubseteq I$ for any $k$. Thus $I$ is a neighborhood of $a$ which does not have an element of $\BB_a$ as a subset. So $\BB_a$ cannot be a basis for the neighborhood system of $a$. Hence the result. {{qed}}
Let $x, y \in A: x \ne y$. From Distinct Points in Metric Space have Disjoint Open Balls, there exist disjoint open $\epsilon$-balls $B_\epsilon \left({x}\right)$ and $B_\epsilon \left({y}\right)$ containing $x$ and $y$ respectively. From Open Ball is Neighborhood of all Points Inside it follows that $B_\epsilon \left({x}\right)$ and $B_\epsilon \left({y}\right)$ are neighborhoods of $x$ and $y$ respectively. Hence the result. {{qed}}
Let $\epsilon \in \R_{>0}$. Let $x = \tuple {x_1, x_2, \ldots, x_n} \in \AA$. Then: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = \map {B_\epsilon} {a; d_\infty}       | c =  }} {{eqn | ll= \leadsto       | l = \map {d_\infty} {x, a}       | o = <       | r = \epsilon       | c = {{Defof|Open Ball|Open $\epsilon$-Ball}} }} {{eqn | ll= \leadsto       | l = \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, a_i} }       | o = <       | r = \epsilon       | c = {{Defof|Chebyshev Distance}} }} {{eqn | ll= \leadsto       | lo= \forall i \in \set {1, 2, \ldots, n}       | l = \map {d_i} {x_i, a_i}       | o = <       | r = \epsilon       | c = {{Defof|Maximum Element}} }} {{eqn | ll= \leadsto       | lo= \forall i \in \set {1, 2, \ldots, n}       | l = x_i       | o = \in       | r = \map {B_\epsilon} {a_i; d_i}       | c = {{Defof|Open Ball|Open $\epsilon$-Ball}} }} {{eqn | ll= \leadsto       | l = x       | o = \in       | r = \prod_{i \mathop = 1}^n \map {B_\epsilon} {a_i; d_i}       | c = {{Defof|Finite Cartesian Product}} }} {{eqn | ll= \leadsto       | l = \map {B_\epsilon} {a; d_\infty}       | o = \subseteq       | r = \prod_{i \mathop = 1}^n \map {B_\epsilon} {a_i; d_i}       | c = {{Defof|Subset}} }} {{end-eqn}} And then: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = \prod_{i \mathop = 1}^n \map {B_\epsilon} {a_i; d_i}       | c =  }} {{eqn | ll= \leadsto       | lo= \forall i \in \set {1, 2, \ldots, n}       | l = x_i       | o = \in       | r = \map {B_\epsilon} {a_i; d_i}       | c = {{Defof|Finite Cartesian Product}} }} {{eqn | ll= \leadsto       | lo= \forall i \in \set {1, 2, \ldots, n}       | l = \map {d_i} {x_i, a_i}       | o = <       | r = \epsilon       | c = {{Defof|Open Ball|Open $\epsilon$-Ball}} }} {{eqn | ll= \leadsto       | l = \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, a_i} }       | o = <       | r = \epsilon       | c = {{Defof|Maximum Element}} }} {{eqn | ll= \leadsto       | l = \map {d_\infty} {x, a}       | o = <       | r = \epsilon       | c = {{Defof|Chebyshev Distance}} }} {{eqn | ll= \leadsto       | l = x       | o = \in       | r = \map {B_\epsilon} {a; d_\infty}       | c = {{Defof|Open Ball|Open $\epsilon$-Ball}} }} {{eqn | ll= \leadsto       | l = \prod_{i \mathop = 1}^n \map {B_\epsilon} {a_i; d_i}       | o = \subseteq       | r = \map {B_\epsilon} {a; d_\infty}       | c = {{Defof|Subset}} }} {{end-eqn}} The result follows by definition of set equality. {{qed}}
Consider the countable set: :$\mathcal B_a := \left\{ {B_\epsilon \left({a}\right): \exists n \in \Z_{>0}: \epsilon = \dfrac 1 n}\right\}$ That is, let $\mathcal B_a$ be the set of all open $\epsilon$-balls of $a$ such that $\epsilon$ is of the form $\dfrac 1 n$ for (strictly) positive integral $n$. Let $N$ be a neighborhood of $a$. Then by definition of neighborhood: :$\exists \epsilon' \in \R_{>0}: B_{\epsilon'} \left({a}\right) \subseteq N$ From Between two Real Numbers exists Rational Number: :$\exists \epsilon'' \in \Q: 0 < \epsilon'' < \epsilon'$ Let $\epsilon''$ be expressed in canonical form as: :$\epsilon'' = \dfrac p q$ where $p$ and $q$ are coprime integers and $q > 0$. Then: :$\epsilon := \dfrac 1 q \le \dfrac p q$ Thus $0 < \epsilon < \epsilon'$. So: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = B_\epsilon \left({a}\right)       | c =  }} {{eqn | ll= \implies       | l = d \left({x, a}\right)       | o = <        | r = \epsilon       | c = Definition of Open $\epsilon$-Ball }} {{eqn | ll= \implies       | l = d \left({x, a}\right)       | o = <        | r = \epsilon'       | c = as $\epsilon < \epsilon'$ }} {{eqn | ll= \implies       | l = x       | o = \in        | r = B_{\epsilon'} \left({a}\right)       | c = Definition of Open $\epsilon$-Ball }} {{eqn | ll= \implies       | l = B_\epsilon \left({a}\right)       | o = \subseteq        | r = B_{\epsilon'} \left({a}\right)       | c = Definition of Subset }} {{eqn | ll= \implies       | l = B_\epsilon \left({a}\right)       | o = \subseteq        | r = N       | c = Subset Relation is Transitive }} {{end-eqn}} From Open Ball is Neighborhood of all Points Inside, $B_\epsilon \left({a}\right)$ is a neighborhood of $a$. But as $\epsilon = \dfrac 1 q$ where $q \in \Z_{>0}$ it follows that: : $B_\epsilon \left({a}\right) \in \left\{ {B_\epsilon \left({a}\right): \exists n \in \Z_{>0}: \epsilon = \dfrac 1 n}\right\}$ Hence the result. {{qed}}
Let $N$ be a neighborhood of $a$ in $M$. Then by definition of neighborhood: :$\exists \epsilon \in \R_{>0}: B_\epsilon \left({a, d_\infty}\right) \subseteq N$ where $B_\epsilon \left({a}\right)$ is the open $\epsilon$-ball of $a$ under $d_\infty$. From Open Ball in Cartesian Product under Chebyshev Distance: :$\displaystyle B_\epsilon \left({a; d_\infty}\right) = \prod_{i \mathop = 1}^n B_\epsilon \left({a_i; d_i}\right)$ From Open Ball is Neighborhood of all Points Inside, each $B_\epsilon \left({a_i; d_i}\right)$ is a neighborhood of $a_i$. By definition of basis for system of neighborhoods: :$\exists N_i \in \mathcal B_{a_i}: N_i \subseteq B_\epsilon \left({a_i; d_i}\right)$ From Cartesian Product of Subsets: :$\displaystyle \prod_{i \mathop = 1}^n N_i \subseteq \prod_{i \mathop = 1}^n B_\epsilon \left({a_i; d_i}\right) \subseteq N$ But $\displaystyle \prod_{i \mathop = 1}^n N_i \in \mathcal B_a = \prod_{i \mathop = 1}^n \mathcal B_{a_i}$ Hence the result by definition of basis for system of neighborhoods. {{qed}}
Let $\epsilon \in \R_{>0}$. Let $a = \tuple {a_1, a_2, \ldots, a_n} \in \AA$. Let $\map {B_\epsilon} {a_i; d_i}$ be the open $\epsilon$-ball of $a_i$ in $M_i$. From Open Ball in Cartesian Product under Chebyshev Distance: :$\displaystyle \map {B_\epsilon} {a; d_\infty} = \prod_{i \mathop = 1}^n \map {B_\epsilon} {a_i; d_i}$ By definition of $i$th projection: :$\forall x \in \map {B_\epsilon} {a; d_\infty}: \map {\pr_i} x \in \map {B_\epsilon} {a_i; d_i}$ Thus by definition of image of subset of $\AA$ under $\pr_i$: :$\pr_i \sqbrk {\map {B_\epsilon} {a; d_\infty} } \subseteq  \map {B_\epsilon} {a_i; d_i}$ Thus by definition of continuity at a point, $\pr_i$ is continuous at $a$. As $a$ is arbitrary, it follows that $\pr_i$ is continuous at all points of $\AA$. Hence, by definition, $\pr_i$ is continuous from $\AA$ to $A_i$. As $i$ is arbitrary, it follows that the result holds for all $i \in \set {1, 2, \ldots, n}$. {{qed}}
{{WLOG}}, let $i \in \left\{ {1, 2, \ldots, n}\right\}$ be arbitrary. === Necessary Condition === Let $f$ be continuous. From Projection from Cartesian Product under Chebyshev Distance is Continuous, $\operatorname{pr}_i: \mathcal A \to A_i$ is continuous on $\mathcal A$. From Composite of Continuous Mappings between Metric Spaces is Continuous it follows that $\operatorname{pr}_1 \circ f$ is continuous on $\mathcal A$. {{qed|lemma}} === Sufficient Condition === Let $\operatorname{pr}_i \circ f$ be continuous on $X$. Let $a \in X$. By definition of continuity at a point: :$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \operatorname{pr}_i \circ f \left[{B_\delta \left({a; d'}\right)}\right] \subseteq B_\epsilon \left({\operatorname{pr}_i \circ f \left({a}\right); d_i}\right)$ where $B_\epsilon \left({\operatorname{pr}_i \circ f \left({a}\right); d_i}\right)$ denotes the open $\epsilon$-ball of $\operatorname{pr}_i \circ f \left({a}\right)$ with respect to the metric $d_i$, and similarly for $B_\delta \left({a; d'}\right)$. From Projection from Cartesian Product under Chebyshev Distance is Continuous, we have that: :$\forall \epsilon \in \R_{>0}: \operatorname{pr}_i \left[{B_\epsilon \left({f \left({a}\right); d_\infty}\right)}\right] \subseteq B_\epsilon \left({\operatorname{pr}_i \left({f \left({a}\right)}\right); d_i}\right)$ From Open Ball in Cartesian Product under Chebyshev Distance: :$\displaystyle B_\epsilon \left({f \left({a}\right); d_\infty}\right) = \prod_{i \mathop = 1}^n B_\epsilon \left({\operatorname{pr}_i \circ f \left({a}\right); d_i}\right)$ From Metric Space Continuity by Inverse of Mapping between Open Balls: :$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: B_\delta \left({a; d'}\right) \subseteq \left({\operatorname{pr}_i \circ f}\right)^{-1} \left[{B_\epsilon \left({\operatorname{pr}_i \circ f \left({a}\right); d_i}\right)}\right]$ where $B_\epsilon \left({\operatorname{pr}_i \circ f \left({a}\right); d_i}\right)$ denotes the open $\epsilon$-ball of $\operatorname{pr}_i \circ f \left({a}\right)$ with respect to the metric $d_i$, and similarly for $B_\delta \left({a; d'}\right)$. {{finish|not going anywhere, need to rethink}}
Let $f \left({a}\right) = l$ where $l > 0$. As $f$ is continuous: :$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \left\vert{y - x}\right\vert < \delta \implies \left\vert{f \left({y}\right) - f \left({x}\right)}\right\vert < \epsilon$ Let $\epsilon = \dfrac l 2 = k$. Then: :$\exists \delta' \in \R_{>0}: \forall y \in \R: \left\vert{y - x}\right\vert < \delta' \implies \left\vert{f \left({y}\right) - f \left({a}\right)}\right\vert < \dfrac l 2$ Thus: :$\forall x \in \left({a - \delta' \,.\,.\, a + \delta'}\right): f \left({x}\right) > \dfrac l 2$ Let $\delta = \dfrac {\delta'} 2$ Then $a - \delta \in \left({a - \delta' \,.\,.\, a + \delta'}\right)$ and $a + \delta \in \left({a - \delta' \,.\,.\, a + \delta'}\right)$. Thus: :$\left[{a - \delta \,.\,.\, a + \delta}\right] \subseteq \left({a - \delta' \,.\,.\, a + \delta'}\right)$ and hence the result. {{qed}}
=== Necessary Condition === Let $\displaystyle \lim_{n \mathop \to \infty} a_n = a$. Let $V$ be a neighborhood of $a$. By definition of neighborhood: :$\exists \epsilon \in \R_{>0}: B_\epsilon \left({a}\right) \subseteq V$ where $B_\epsilon \left({a}\right)$ denotes the open $\epsilon$-ball of $a$ in $M$. By definition of limit: :$\exists N \in \N: n > N \implies d \left({a, a_n}\right) < \epsilon$ Hence $a_n \in V$. {{qed|lemma}} === Sufficient Condition === Let $\left\langle{a_n}\right\rangle$ be such that for each neighborhood $V$ of $a$: :$\exists N \in \N: n > N \implies a_n \in V$ Let $\epsilon \in \R_{>0}$. Then by Open Ball is Neighborhood of all Points Inside, $B_\epsilon \left({a}\right)$ is a neighborhood of $a$. Let $N \in \N$ be such that: :$\forall n > N: a_n \in B_\epsilon \left({a}\right)$ Then: : $d \left({a, a_n}\right) < \epsilon$ and so by definition of limit: :$\displaystyle \lim_{n \mathop \to \infty} a_n = a$ {{qed}}
There are two cases to consider: : $(1): \quad x = 0$ : $(2): \quad x \ne 0$ Let $x = 0$. Then: :$x^2 = 0$ and thus :$0 \le x^2$ {{qed|lemma}} Let $x \ne 0$. From Square of Non-Zero Real Number is Strictly Positive it follows that: :$0 < x^2$ and so by definition: :$0 \le x^2$ {{Qed}}
We are given that $0 < x < 1$. By direct application of Real Number Ordering is Compatible with Multiplication, it follows that: : $0 \times x < x \times x < 1 \times x$ and the result follows. {{Qed}}
We have that Real Numbers form Ordered Integral Domain. Thus Square of Element Less than Unity in Ordered Integral Domain applies directly. {{qed}}
As $x > 1$ it follows that $x > 0$. Thus by Real Number Ordering is Compatible with Multiplication: : $x \times x > 1 \times x$ and the result follows. {{Qed}}
=== Definition 1 implies Definition 2 === Let $f: \R \to \R$ be the function defined as: :$\forall x \in \R: \map f x = \begin{cases} x & : x > 0 \\ 0 & : x = 0 \\ -x & : x < 0 \end{cases}$ Let $x > 0$. We have that: {{begin-eqn}} {{eqn | l = \paren {-x}^2       | r = \paren {-1}^2 x^2       | c =  }} {{eqn | r = x^2       | c =  }} {{eqn | ll= \leadsto       | l = +\sqrt {\paren {-x}^2}       | r = x       | c =  }} {{eqn | r = +\sqrt {x^2}       | c =  }} {{end-eqn}} Thus $\map f x = +\sqrt {x^2}$ {{qed|lemma}} === Definition 2 implies Definition 1 === Let $f: \R \to \R$ be the function defined as: :$\forall x \in \R: \map f x = +\sqrt {x^2}$ Let $x > 0$. Then by definition of square and square root: :$+\sqrt {x^2} = x$ Let $x = 0$. Then $+\sqrt {0^2} = 0$. Let $x < 0$. Thus let $x = -y$. Then: :$+\sqrt {\paren {-y}^2} = y = -x$ {{qed}}
{{AimForCont}} $S$ were bounded above. Then $S$ has a supremum $B$. As $x > 1$, it follows that $\dfrac B x < B$ and so therefore $\dfrac B x$ can not be an upper bound. Therefore: :$\exists n \in \N: x^n > \dfrac B x \implies x^{n + 1} > B$ So $B$ can not be an upper bound. From that contradiction it can be concluded that $S$ can not have an upper bound. So by Proof by Contradiction it follows that $S$ is unbounded above. {{qed}}
When $n = 0$ it follows that $x^n = 1$ and so $\sup S \ge 1$. Let $x^k \in S$. As $x < 1$, it follows from Real Number Ordering is Compatible with Multiplication that: :$x^{k + 1} < x^k$ So: :$\forall x \in S: x \le 1$ Hence it follows that $\sup S = 1$. As $x > 0$, it follows by Real Number Ordering is Compatible with Multiplication that: :$\forall x \in S: x \ge 0$ Therefore $x$ is a lower bound of $S$. Now suppose $h > 0$ is also a lower bound of $S$. Then: :$\forall n \in \N: x^n \ge h$ and: :$\forall n \in \N: \paren {\dfrac 1 x}^n \le \dfrac 1 h$ But as $0 < x < 1$ it follows that: :$\dfrac 1 x > 1$ Thus $\dfrac 1 h$ is an upper bound for $\set {\paren {\dfrac 1 x}^n: n \in \N}$. From Power of Real Number greater than One is Unbounded Above, this has been shown to be unbounded above. Therefore there can be no such lower bound $h > 0$ of $S$. Hence $\inf S = 0$. {{qed}}