avewright/proofwiki-math huggingface dataset:
Theorem: Upper Sum of Refinement

Let $\closedint a b$ be a closed interval.
Let $P$ be a finite subdivision of $\closedint a b$. 
Let $Q$ be a refinement of $P$. 
Then: 
:$\map U {f, P} \le \map U {f, Q}$
where $\map U {f, P}$ and $\map U {f, Q}$ denote the upper Darboux sum of $f$ with respect to $P$ and $Q$ respectively.

Proof:
Write: 
:$P = \set {x_0, x_1, \ldots, x_k}$
and:
:$Q = \set {y_0, y_1, \ldots, y_l}$
where:
:$a = x_0 < x_1 < \ldots < x_k = b$
and:
:$a = y_0 < y_1 < \ldots < y_l = b$
Since $P \subseteq Q$, we have $k \le l$ from Cardinality of Subset of Finite Set.
Set: 
:$M_i = \sup \set {\map f x : x \in \closedint {x_{i - 1} } {x_i} }$
for each $1 \le i \le k$.
Also set: 
:${\tilde M}_j = \sup \set {\map f x : x \in \closedint {y_{j - 1} } {y_j} }$
for each $1 \le j \le l$. 
Consider a pair of elements $\tuple {x_{i - 1}, x_i}$ in $P$. 
Since $P \subseteq Q$, there exists $a_i, b_i$ such that: 
:$\tuple {x_{i - 1}, x_i} = \tuple {y_{a_i}, y_{b_i} }$
We can see that: 
:$a_1 = 0$
and:
:$b_k = l$ 
We also clearly have: 
:$b_{i - 1} = a_i$ for each $1 \le i \le k$. 
Note that: 
:$\closedint {y_{j - 1} } {y_j} \subseteq \closedint {x_{i - 1} } {x_i}$
for all $a_i + 1 \le j \le b_i$. 
So:
:$\set {\map f x : x \in \closedint {y_{j - 1} } {y_j} } \subseteq \set {\map f x : x \in \closedint {x_{i - 1} } {x_i} }$
for all $a_i + 1 \le j \le b_i$. 
So, from Supremum of Subset, we have:
:$\sup \set {\map f x : x \in \closedint {y_{j - 1} } {y_j} } \le \sup \set {\map f x : x \in \closedint {x_{i - 1} } {x_i} }$
for all $a_i + 1 \le j \le b_i$.
That is: 
:${\tilde M}_j \le M_i$
for each $\tuple {i, j}$ with $a_i + 1 \le j \le b_i$
We can then write: 
:$\ds x_i - x_{i - 1} = \sum_{j \mathop = a_i + 1}^{b_i} \paren {y_j - y_{j - 1} }$
for each $1 \le i \le k$, giving:
{{begin-eqn}}
{{eqn	| l = \map U {f, P}
		| r = \sum_{i \mathop = 1}^k M_i \paren {x_i - x_{i - 1} }
		| c = {{Defof|Upper Darboux Sum}}
}}
{{eqn	| r = \sum_{i \mathop = 1}^k M_i \paren {\sum_{j \mathop = a_i + 1}^{b_i} \paren {y_j - y_{j - 1} } }
}}
{{eqn	| r = \sum_{i \mathop = 1}^k \paren {\sum_{j \mathop = a_i + 1}^{b_i} M_i \paren {y_j - y_{j - 1} } }
}}
{{eqn	| o = \ge
		| r = \sum_{i \mathop = 1}^k \paren {\sum_{j \mathop = a_i + 1}^{b_i} {\tilde M}_j \paren {y_j - y_{j - 1} } }
		| c = since $M_i \ge {\tilde M}_j$ for each $\tuple {i, j}$ with $a_i + 1 \le j \le b_i$
}}
{{eqn	| r = \sum_{j \mathop = 1}^l {\tilde M}_j \paren {y_j - y_{j - 1} }
}}
{{eqn	| r = \map U {f, Q}
}}
{{end-eqn}}
{{qed}}
Theorem: Dirac Delta Distribution is Tempered Distribution

Let $\delta$ be the Dirac delta distribution.
Let $\map {\SS'} \R$ be the tempered distribution space.
Then:
:$\delta \in \map {\SS'} \R$

Proof:
Let $\phi \in \map \SS \R$ be a Schwartz test function.
Consider the mapping $\phi \stackrel \delta \longrightarrow \map \phi 0 : \map \SS \R \to \C$.
We have that the Schwartz space is a vector space.
Let $\phi$ be a linear superposition of Schwartz test functions.
Then $\map \phi 0$ is also a linear supperposition evaluated at $x = 0$.
Hence, the mapping $\phi \stackrel \delta \longrightarrow \map \phi 0 : \map \SS \R \to \C$ is linear.
Suppose, $\sequence {\phi_n}_{n \mathop \in \N}$ converges in $\map \SS \R$:
:$\phi_n \stackrel \SS \longrightarrow \mathbf 0$
We have that:
:$\ds \forall x \in \R : \forall l, m \in \N : 0 \le \size {x^l \map {\phi_n^{\paren m} } x} \le \sup_{x \mathop \in \R} \size {x^l \map {\phi_n^{\paren m} } x}$
Specifically, for $x = 0$, $l = 0$, and $m = 0$ it holds that:
:$\ds 0 \le \size {\map {\phi_n} 0} \le \sup_{x \mathop \in \R} \size {\map {\phi_n} x}$
Taking the limit $n \to \infty$ and applying the squeeze theorem yields:
:$\ds \lim_{n \mathop \to \infty} \map {\phi_n} 0 = 0$.
{{qed}}
Theorem: Condition for Valid Time Indication

Consider an analog clock $C$ with an hour hand $H$ and a minute hand $M$.
Let $\theta \degrees$ be the angle made by the minute hand with respect to twelve o'clock.
Let $\phi \degrees$ be the angle made by the hour hand with respect to twelve o'clock.
Then $C$ displays a '''valid time indication''' {{iff}}:
:$12 \phi \mod 360 = \theta$

Proof:
Let $T$ be a time of day specified in hours $h$ and minutes $m$, where:
:$1 \le h \le 12$ is an integer
:$0 \le m < 60$ is a real number
whether a.m. or p.m. is immaterial.
From Speed of Minute Hand, $M$ travels $6 \degrees$ per minute.
So at time $m$ minutes after the hour, $\theta = 6 m$.
From Speed of Hour Hand, $H$ travels $\dfrac 1 2 \degrees$ per minute. 
The hour marks are at $30 \degrees$ intervals.
So at time $m$ minutes after the hour, $\phi = 30 h + \dfrac m 2 \degrees$ past hour $h$.
That is:
:$\phi = 30 h + \dfrac 1 2 \dfrac \theta 6$
or:
:$12 \phi = 360 h + \theta$
where $h$ is an integer.
Thus:
:$12 \phi \mod 360 = \theta$
{{qed}}
Category:Clocks
4p90luiuhbwgh6w6g8elv6ybqyfvb9g
Theorem: Speed of Minute Hand

Consider an analog clock $C$.
The minute hand of $C$ rotates at $6$ degrees per minute.

Proof:
It takes one hour, that is $60$ minutes, for the minute hand to go round the dial one time.
That is, in $60$ minutes the minute hand travels $360 \degrees$.
So in $1$ minute, the minute hand travels $\dfrac {360} {60} \degrees$, that is, $6 \degrees$.
{{qed}}
Theorem: Speed of Hour Hand

Consider an analog clock $C$.
The hour hand of $C$ rotates at $\dfrac 1 2$ of a degree per minute.

Proof:
It takes $12$ hours, for the hour hand to go round the dial one time.
That is, in $12$ minutes the hour hand travels $360 \degrees$.
So in $1$ hour, the hour hand travels $\dfrac {360} {12} \degrees$, that is, $30 \degrees$.
So in $1$ minute, the hour hand travels $\dfrac 1 {60} \times 30 \degrees$, that is, $\dfrac 1 2 \degrees$.
{{qed}}
Theorem: Condition for Valid Time Indication/Corollary

Let $\theta \degrees$ be the angle made by the minute hand with respect to twelve o'clock.
Let $\rho \degrees$ be the angle made by the hour hand with respect to the hour just past.
Then $C$ displays a '''valid time indication''' {{iff}}:
:$\rho = \dfrac \theta {12}$

Proof:
Follows directly.
{{qed}}
Category:Clocks
3n89lyit4y4z63d6v3s43q859ytvh70
Theorem: Expectation of Random Variable as Integral with respect to Probability Distribution

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be an integrable real-valued random variable on $\struct {\Omega, \Sigma, \Pr}$. 
Let $P_X$ be the probability distribution of $X$.
Then: 
:$\ds \expect X = \int_\R x \map {\rd P_X} x$
where $\expect X$ is the expected value of $X$.

Proof:
From the definition of expectation: 
:$\ds \expect X = \int_\Omega X \rd \Pr$
We can write: 
:$\ds \int_\Omega X \rd \Pr = \int_\Omega I_\R \circ X \rd \Pr$
where $I_\R$ is the identity map for $\R$. 
From the definition of probability distribution, we have: 
:$P_X = X_* \Pr$
where $X_* \Pr$ is the pushforward of $\Pr$ on $\tuple {\R, \map \BB \R}$, where $\map \BB \R$ denotes the Borel $\sigma$-algebra on $\R$.
So, from Integral with respect to Pushforward Measure: Corollary, we have: 
:$I_\R$ is $P_X$-integrable
and:
:$\ds \int_\Omega I_\R \circ X \rd \Pr = \int_\R I_\R \rd P_X$
That is: 
:$\ds \int_\Omega X \rd \Pr = \int_\R x \map {\rd P_X} x$
{{qed}}
Theorem: Expectation of Real-Valued Discrete Random Variable

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space. 
Let $X$ be a discrete real-valued random variable. 
Then $X$ is integrable {{iff}}:
:$\ds \sum_{x \in \Img X} \size x \map \Pr {X = x} < \infty$
in which case: 
:$\ds \expect X = \sum_{x \in \Img X} x \map \Pr {X = x}$

Proof:
From Characterization of Integrable Functions, we have: 
:$X$ is $\Pr$-integrable {{iff}} $\size X$ is $\Pr$-integrable.
That is, $X$ is integrable {{iff}}:
:$\ds \int \size X \rd \Pr < \infty$
Theorem: Linear Transformation of Real-Valued Random Variable is Real-Valued Random Variable

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be a real-valued random variable on $\struct {\Omega, \Sigma, \Pr}$. 
Let $a$ and $b$ be real numbers.
Then: 
:$a X + b$ is a real-valued random variable.

Proof:
From the definition of a real-valued random variable, we have: 
:$X$ is $\Sigma$-measurable.
We want to verify that $a X + b : \Omega \to \R$ is a $\Sigma$-measurable function.
From Pointwise Scalar Multiple of Measurable Function is Measurable, we have: 
:$a X$ is $\Sigma$-measurable.
From Constant Function is Measurable, we have: 
:$x \mapsto b$ is $\Sigma$-measurable.
From Pointwise Sum of Measurable Functions is Measurable, we have: 
:$a X + b$ is $\Sigma$-measurable.
So:
:$a X + b$ is a real-valued random variable.
{{qed}}
Category:Random Variables
tfamsem1glmkvn4qudszvc2wtxbdp7s
Theorem: Singular Random Variable is not Absolutely Continuous

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space. 
Let $X$ be a singular random variable on $\struct {\Omega, \Sigma, \Pr}$.
Then $X$ is not absolutely continuous.

Proof:
Let $P_X$ be the probability distribution of $X$. 
Let $\map \BB \R$ be the Borel $\sigma$-algebra on $\R$.
Let $\lambda$ be the Lebesgue measure for $\struct {\R, \map \BB \R}$. 
From the definition of an absolutely continuous random variable, we have that $X$ is absolutely continuous {{iff}}:
:$P_X$ is absolutely continuous with respect to $\lambda$. 
That is: 
:for all $A \in \map \BB \R$ with $\map \lambda A = 0$ we have $\map {P_X} A = 0$.
where $\map \BB \R$ is the Borel $\sigma$-algebra on $\R$.
Since $X$ is singular:
:there exists $B \in \map \BB \R$ with $\map \lambda B = 0$ and $\map \Pr {X \in B} = 1$.
That is, from the definition of probability distribution: 
:there exists $B \in \map \BB \R$ with $\map \lambda B = 0$ and $\map {P_X} B = 1$.
So:
:for all $A \in \map \BB \R$ with $\map \lambda A = 0$ we have $\map {P_X} A = 0$.
does not hold and so $X$ is not absolutely continuous.
{{qed}}
Category:Singular Random Variables
9iw5ug386nix8q7b589pwc0lgann9hs
Theorem: Cumulative Distribution Function is Right-Continuous

:$F_X$ is right-continuous.

Proof:
Let $x \in \R$. 
We show that $F_X$ is right-continuous at $x$. 
We use Monotonic Sequential Right-Continuity is Equivalent to Right-Continuity in the Reals, and will show that: 
:for all monotone sequences $\sequence {x_n}_{n \mathop \in \N}$, with $x_n > x$ for each $n$, that converge to $x$ we have:
::$\map {F_X} {x_n} \to \map {F_X} x$
Let $\sequence {x_n}_{n \mathop \in \N}$ be a monotone sequences with $x_n > x$ for each $n$ that converges to $x$.
Then $\sequence {x_n}_{n \mathop \in \N}$ is a decreasing sequence. 
So: 
:$\sequence {\hointl {-\infty} {x_n} }_{n \mathop \in \N}$ is a decreasing sequence of sets.
From Limit of Decreasing Sequence of Unbounded Below Closed Intervals: 
:$\ds \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} = \hointl {-\infty} x$
Let $P_X$ be the probability distribution of $X$. 
Note that $P_X$ is a finite measure. 
So, from Measure of Limit of Decreasing Sequence of Measurable Sets, we therefore have: 
:$\ds \map {P_X} {\hointl {-\infty} x} = \lim_{n \mathop \to \infty} \map {P_X} {\hointl {-\infty} {x_n} }$
So we obtain: 
{{begin-eqn}}
{{eqn	| l = \map {F_X} x
		| r = \map \Pr {X \le x}
		| c = {{Defof|Cumulative Distribution Function}}
}}
{{eqn	| r = \map {P_X} {\hointl {-\infty} x}
		| c = {{Defof|Probability Distribution}}
}}
{{eqn	| r = \lim_{n \mathop \to \infty} \map {P_X} {\hointl {-\infty} {x_n} }
}}
{{eqn	| r = \lim_{n \mathop \to \infty} \map \Pr {X \le x_n}
		| c = {{Defof|Probability Distribution}}
}}
{{eqn	| r = \lim_{n \mathop \to \infty} \map {F_X} {x_n}
		| c = {{Defof|Cumulative Distribution Function}}
}}
{{end-eqn}}
So, since $\sequence {x_n}_{n \mathop \in \N}$ was arbitrary:
:for all monotone sequences $\sequence {x_n}_{n \mathop \in \N}$, with $x_n > x$ for each $n$, that converge to $x$ we have:
::$\map {F_X} {x_n} \to \map {F_X} x$
So, from Monotonic Sequential Right-Continuity is Equivalent to Right-Continuity in the Reals:
:$F_X$ is right-continuous at $x$.
{{qed}}
Category:Right-Continuous Functions
Category:Cumulative Distribution Functions
Category:Cumulative Distribution Function is Right-Continuous
ru2hm86iaetx32xu26sctgfcym4es4a
Theorem: Linear Transformation of Continuous Random Variable is Continuous Random Variable

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space. 
Let $a$ be a non-zero real number. 
Let $b$ be a real number. 
Let $X$ be a continuous real variable.
Let $F_X$ be the cumulative distribution function of $X$. 
Then $a X + b$ is a continuous real variable.
Further, if $a > 0$, the cumulative distribution function of $a X + b$, $F_{a X + b}$. is given by: 
:$\ds \map {F_{a X + b} } x = \map {F_X} {\frac {x - b} a}$
for each $x \in \R$.
If $a < 0$, $F_{a X + b}$ is given by: 
:$\ds \map {F_{a X + b} } x = 1 - \map {F_X} {\frac {x - b} a}$
for each $x \in \R$.

Proof:
From Linear Transformation of Real-Valued Random Variable is Real-Valued Random Variable, $a X + b$ is a real-valued random variable.
Since $X$ is a continuous real variable, we have that: 
:$F_X$ is continuous.
We use this fact to show that $F_{a X + b}$ is continuous, showing that $a X + b$ is a continuous real variable.
We split up into cases.
Suppose that $a > 0$. 
Then, for each $x \in \R$, we have: 
{{begin-eqn}}
{{eqn	| l = \map {F_{a X + b} } x
		| r = \map \Pr {a X + b \le x}
}}
{{eqn	| r = \map \Pr {a X \le x - b}
}}
{{eqn	| r = \map \Pr {X \le \frac {x - b} a}
}}
{{eqn	| r = \map {F_X} {\frac {x - b} a}
		| c = {{Defof|Cumulative Distribution Function}}
}}
{{end-eqn}}
From Composite of Continuous Mappings is Continuous and Linear Function is Continuous, we therefore have: 
:$F_{a X + b}$ is continuous
in the case $a > 0$.
Now suppose that $a < 0$. 
Then, for each $x \in \R$, we have: 
{{begin-eqn}}
{{eqn	| l = \map {F_{a X + b} } x
		| r = \map \Pr {a X + b \le x}
}}
{{eqn	| r = \map \Pr {a X \le x - b}
}}
{{eqn	| r = \map \Pr {X \ge \frac {x - b} a}
}}
{{eqn	| r = \map \Pr {\Omega \setminus \set {X < \frac {x - b} a} }
}}
{{eqn	| r = 1 - \map \Pr {X < \frac {x - b} a}
		| c = Probability of Event not Occurring
}}
{{eqn	| r = 1 - \map \Pr {X \le \frac {x - b} a}
		| c = Probability of Continuous Random Variable Lying in Singleton Set is Zero
}}
{{eqn	| r = 1 - \map {F_X} {\frac {x - b} a}
}}
{{end-eqn}}
From Composite of Continuous Mappings is Continuous and Linear Function is Continuous, we therefore have: 
:$F_{a X + b}$ is continuous
in the case $a < 0$.
{{qed}}
Category:Continuous Random Variables
Category:Cumulative Distribution Functions
m1394q3o54anql5yx8a332vi3nsty8b
Theorem: Probability of Continuous Random Variable Lying in Singleton Set is Zero

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space. 
Let $X$ be a continuous real variable on $\struct {\Omega, \Sigma, \Pr}$.
Then, for each $x \in \R$, we have: 
:$\map \Pr {X \le x} = \map \Pr {X < x}$
In particular: 
:$\map \Pr {X = x} = 0$

Proof:
Let $F_X$ be the cumulative distribution function of $X$ so that: 
:$\map {F_X} x = \map \Pr {X \le x}$
for each $x \in \R$.
Let $P_X$ be the probability distribution of $X$. 
Since $X$ is a continuous real variable, we have: 
:$F_X$ is continuous. 
From Sequential Continuity is Equivalent to Continuity in the Reals, we have:
:for each real sequence $\sequence {x_n}_{n \mathop \in \N}$ converging to $x$, we have $\map {F_X} {x_n}\to \map {F_X} x$.
For each $n \in \N$, let: 
:$\ds x_n = x - \frac 1 n$
We have that $x_n \to x$, so: 
:$\ds \map {F_X} x = \lim_{n \mathop \to \infty} \map {F_X} {x - \frac 1 n}$ 
We also have that $\sequence {x_n}_{n \mathop \in \N}$ is a increasing sequence. 
So, we have: 
:$\ds \hointl {-\infty} {x - \frac 1 n} \subseteq \hointl {-\infty} {x - \frac 1 {n + 1} }$
for each $n \in \N$. 
So:
:$\ds \sequence {\hointl {-\infty} {x - \frac 1 n} }_{n \mathop \in \N}$ is a increasing sequence of sets.
We can see that: 
:$\ds \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x - \frac 1 n} = \openint {-\infty} x$
Theorem: Cumulative Distribution Function as Integral of Probability Density Function

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be an absolutely continuous random variable. 
Let $f_X$ be a probability density function for $X$. 
Let $F_X$ be the cumulative distribution function for $X$. 
Let $\map \BB \R$ be the Borel $\sigma$-algebra on $\R$. 
Let $\lambda$ be the Lebesgue measure on $\struct {\R, \map \BB \R}$. 
Then: 
:$\ds \map {F_X} x = \int_{-\infty}^x f_X \rd \lambda$
for each $x \in \R$, where $\ds \int_{-\infty}^x f_X \rd \lambda$ denotes the Lebesgue integral of $f$ over $\hointl {-\infty} x$.

Proof:
Let $P_X$ be the probability distribution of $X$. 
Since $f_X$ is a probability density function for $X$, $f_X$ is a Radon-Nikodym derivative of $P_X$ with respect to $\lambda$. 
Then, we have: 
{{begin-eqn}}
{{eqn | l = \map {F_X} x
      | r = \map \Pr {X \le x}
      | c = {{Defof|Cumulative Distribution Function}}
}}
{{eqn | r = \map \Pr {X \in \hointl {-\infty} x}
}}
{{eqn | r = \map {P_X} {\hointl {-\infty} x}
      | c = {{Defof|Probability Distribution}}
}}
{{eqn | r = \int_{\hointl {-\infty} x} \map {f_X} t \rd \map \lambda t
      | c = {{Defof|Radon-Nikodym Derivative}}
}}
{{eqn | r = \int_{-\infty}^x f_X \rd \lambda
}}
{{end-eqn}}
{{qed}}
Theorem: Probability of Continuous Random Variable Lying in Singleton Set is Zero/Corollary

Let $C$ be a countable subset of $\R$. 
Then: 
:$\map \Pr {X \in C} = 0$

Proof:
Since $C$ is countable, there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ of distinct real numbers such that: 
:$C = \set {x_n : n \mathop \in \N}$
That is: 
:$\ds C = \bigcup_{n \mathop = 1}^\infty \set {x_n}$
where $\set {\set {x_1}, \set {x_2}, \ldots}$ is pairwise disjoint.
We then have: 
{{begin-eqn}}
{{eqn	| l = \map \Pr {X \in C}
		| r = \map {P_X} C
}}
{{eqn	| r = \map {P_X} {\bigcup_{n \mathop = 1}^\infty \set {x_n} }
}}
{{eqn	| r = \sum_{n \mathop = 1}^\infty \map {P_X} {\set {x_n} }
		| c = using the countable additivity of $P_X$
}}
{{eqn	| r = \sum_{n \mathop = 1}^\infty \map \Pr {X \in \set {x_n} }
		| c = {{Defof|Probability Distribution}}
}}
{{eqn	| r = \sum_{n \mathop = 1}^\infty \map \Pr {X = x_n}
}}
{{eqn	| r = 0
		| c = Probability of Continuous Random Variable Lying in Singleton Set is Zero
}}
{{end-eqn}}
{{qed}}
Category:Probability of Continuous Random Variable Lying in Singleton Set is Zero
o7z95g96q0ujv9bqsg98nbfwt7fu2pn
Theorem: Real Sequence with all Subsequences having Convergent Subsequence to Limit Converges to Same Limit

Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence. 
Let $x \in \R$. 
Suppose that:
:every subsequence $\sequence {x_{n_j} }_{j \in \N}$ of $\sequence {x_n}_{n \in \mathop \N}$ has a subsequence $\sequence {x_{n_{j_k} } }_{k \in \N}$ such that: 
::$x_{n_{j_k} } \to x$ 
Then: 
:$x_n \to x$

Proof:
{{AimForCont}}, suppose that: 
:$x_n$ does not converge to $x$.
Then, there exists some $\epsilon > 0$ such that for every $k \in \N$ there exists $n_k \ge k$ such that: 
:$\size {x_{n_k} - x} \ge \epsilon$
Let $\sequence {x_{n_{j_k} } }_{k \in \N}$ be a subsequence of $\sequence {x_{n_j} }_{j \in \N}$.
Then: 
:$\size {x_{n_{j_k} } - x} \ge \epsilon$ 
for each $k \in \N$.
In particular, there exists no $N \in \N$ such that: 
:$\size {x_{n_{j_k} } - x} < \epsilon$
for each $k \ge N$, so: 
:$\sequence {x_{n_{j_k} } }_{k \mathop \in \N}$ does not converge to $x$.
So:
:$\sequence {x_{n_j} }_{j \mathop \in \N}$ is a subsequence of $\sequence {x_n}_{n \mathop \in \N}$, but has a subsequence that does not converge to $x$.
This contradicts our hypothesis on the subsequences of $\sequence {x_n}_{n \mathop \in \N}$. 
So $x_n$ converges to $x$.
{{qed}}
Category:Limits of Sequences
03dun1kyn8ftqoqlxazbr0b05n30279
Theorem: Sequential Continuity is Equivalent to Continuity in the Reals/Corollary

Let $I$ be a real interval. 
Let $x \in I$. 
Let $f : I \to \R$ be a real function.
Then $f$ is continuous at $x$ {{iff}}:
:for all monotone sequences $\sequence {x_n}_{n \mathop \in \N}$ converging to $x$ we have: 
::$\map f {x_n} \to \map f x$

Proof:
=== Necessary Condition ===
Suppose $f$ is continuous at $x$, then: 
:for all real sequences $\sequence {x_n}_{n \mathop \in \N}$ converging to $x$ we have:
::$\map f {x_n} \to \map f x$
from Sequential Continuity is Equivalent to Continuity in the Reals.
So in particular: 
:for all monotone sequences $\sequence {x_n}_{n \mathop \in \N}$ converging to $x$ we have: 
::$\map f {x_n} \to \map f x$
{{qed|lemma}}
Theorem: Monotonic Sequential Right-Continuity is Equivalent to Right-Continuity in the Reals

Let $\hointr a b$ be a real interval. 
Let $x \in \hointr a b$. 
Let $f : \hointr a b \to \R$ be a real function.
Then $f$ is right-continuous at $x$ {{iff}}:
:for all monotone sequences $\sequence {x_n}_{n \mathop \in \N}$, with $x_n > x$ for each $n$, that converge to $x$ we have:
::$\map f {x_n} \to \map f x$

Proof:
=== Necessary Condition ===
Suppose $f$ is right-continuous at $x$, then: 
:for each real sequence $\sequence {x_n}_{n \mathop \in \N}$, with $x_n > x$ for each $n$, converging to $x$ we have:
::$\map f {x_n} \to \map f x$
from Limit of Function by Convergent Sequences: Corollary.
So in particular: 
:for all monotone sequences $\sequence {x_n}_{n \mathop \in \N}$, with $x_n > x$ for each $n$, that converge to $x$ we have:
::$\map f {x_n} \to \map f x$
{{qed|lemma}}
Theorem: Limit of Decreasing Sequence of Unbounded Below Closed Intervals

Let $x \in \R$. 
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence converging to $x$ such that $x_n \ge x$ for each $n \in \N$. 
Then:
:$\ds \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} = \hointl {-\infty} x$

Proof:
We first show that: 
:$\ds \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} \subseteq \hointl {-\infty} x$
Let:
:$\ds t \in \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n}$
Then:
:$t \le x_n$
By the Squeeze Theorem, we then have:
:$t \le x$ 
taking $n \to \infty$. 
So $t \in \hointl {-\infty} x$. 
{{qed|lemma}}
We now show that:
:$\ds \hointl {-\infty} x \subseteq \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n}$
Now suppose that: 
:$t \in \hointl {-\infty} x$ 
Then $t \le x$. 
Since $x \le x_n$ for each $n \in \N$, we have $t \le x_n$ for each $n \in \N$.
That is, $t \in \hointl {-\infty} {x_n}$ for each $n \in \N$.
So:
:$\ds t \in \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n}$
{{qed|lemma}}
So, from the definition of set equality, we have: 
:$\ds \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} = \hointl {-\infty} x$
{{qed}}
Category:Decreasing Sequences of Sets
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Theorem: Sequential Characterization of Limit at Positive Infinity of Real Function

Let $f : \R \to \R$ be a real function. 
Let $L$ be a real number. 
Then: 
:$\ds \lim_{x \to \infty} \map f x = L$ 
{{iff}}: 
:for all real sequences $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to \infty$ we have $\map f {x_n} \to L$
where:
:$\ds \lim_{x \mathop \to \infty} \map f x$ denotes the limit at $+\infty$ of $f$.

Proof:
=== Necessary Condition ===
Suppose that: 
:$\ds \lim_{x \to \infty} \map f x = L$
Let $\sequence {x_n}_{n \mathop \in \N}$ be a real sequence with $x_n \to \infty$. 
Let $\epsilon > 0$. 
From the definition of limit at infinity, we have:
:there exists $M > 0$ such that for all $x > M$ we have $\size {\map f x - L} < \epsilon$.
Since $\sequence {x_n}_{n \mathop \in \N}$ diverges to $+\infty$, we have: 
:there exists $N \in \N$ such that $x_n > M$ for all $n \ge N$. 
That is, for all $n \ge N$, we have: 
:$\size {\map f {x_n} - L} < \epsilon$ 
Since $\epsilon$ was arbitrary, we have: 
:$\map f {x_n} \to L$ 
Since $\sequence {x_n}_{n \mathop \in \N}$ was arbitrary: 
:for all real sequences $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to \infty$ we have $\map f {x_n} \to L$. 
So, if: 
:$\ds \lim_{x \to \infty} \map f x = L$
then:
:for all real sequences $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to \infty$ we have $\map f {x_n} \to L$. 
{{qed|lemma}}