nyuuzyou/wikis huggingface dataset:
Ultraproduct is Well-Defined

== Theorem ==
Ultraproduct is well-defined.
== Proof ==
Specifically, following the definitions on ultraproduct, it is to be proved that:
:$(1) \quad f^\MM$ is well-defined
:$(2) \quad R^\MM$ is well-defined
First of all, we need to prove:
=== Lemma ===
Following the definitions on ultraproduct:
:$\eqclass {m_{k, i} } \UU = \eqclass {m'_{k, i} } \UU$, $k = 1, \dotsc, n$
:
:$\set {i \in I: \tuple {m_{1, i}, \dots, m_{n, i} } = \tuple {m'_{1, i}, \dots, m'_{n, i} } } \in \UU$
=== Proof ===
Let:
Suppose:
:$\eqclass {m_{k, i} } \UU = \eqclass {m'_{k, i} } \UU$ for $k = 1, \dotsc, n$
We have:
:$I_k \in \UU$ for $k = 1, \dotsc, n$
Since $\UU$ is closed under intersection:
:$I^* \in \UU$
On the other hand, suppose:
:$I^* \in \UU$
Since $\UU$ is upward-closed:
{{explain|Check whether the above link should be Upper Section or Upper Closure}
:$I_k \in \UU$ for $k = 1, \dotsc, n$
Therefore:
:$\eqclass {m_{k, i} } \UU = \eqclass {m'_{k, i} } \UU$
=== Proposition 1 ===
The definition of $f^\MM$ is consistent. 
That is, for $\eqclass {m_{k, i} } \UU = \eqclass {m'_{k, i} } \UU$, $k = 1, \dotsc, n$:
:$\eqclass {\map {f^{\MM_i} } {m_{1, i}, \dotsc, m_{n, i} } } \UU = \eqclass {\map {f^{\MM_i} } {m'_{1, i}, \dotsc, m'_{n, i} } } \UU$
=== Proof ===
Firstly note that: 
:$\set {i \in I: \map {f^{\MM_i} } {m_{1, i}, \dotsc, m_{n, i} } = \map {f^{\MM_i} } {m'_{1, i}, \dots, m'_{n, i} } } \supseteq \set {i \in I: \tuple {m_{1, i}, \dotsc, m_{n, i} } = \tuple {m'_{1, i}, \dotsc, m'_{n, i} } }$
and by $\UU$ is an ultrafilter on $I$:
:$\set {i \in I: \tuple {m_{1, i}, \dotsc, m_{n, i} } = \tuple {m'_{1, i}, \dotsc, m'_{n, i} } } \in \UU$
implies:
:$\set {i \in I: \map {f^{\MM_i} } {m_{1, i}, \dotsc, m_{n, i} } = \map {f^{\MM_i} } {m'_{1, i}, \dotsc, m'_{n, i} } } \in \UU$
Therefore:
:$\eqclass {m_{k, i} } \UU = \eqclass {m'_{k, i} } \UU$, $k = 1, \dotsc, n$
by the lemma, which is equvalent to:
:$\set {i \in I: \tuple {m_{1, i}, \dotsc, m_{n, i} } = \tuple {m'_{1, i}, \dotsc, m'_{n, i} } } \in \UU$
implies:
:$\eqclass {\map {f^{\MM_i} } {m_{1, i}, \dotsc, m_{n, i} } } \UU = \eqclass {\map {f^{\MM_i} } {m'_{1, i}, \dotsc, m'_{n, i} } } \UU$
=== Proposition 2 ===
The definition of $R^\MM$ is consistent. 
That is, for $\eqclass {m_{k, i} } \UU = \eqclass {m'_{k, i} } \UU$, $k = 1, \dotsc, n$:
:$\set {i \in I: \tuple {m_{1, i}, \dotsc, m_{n, i} } \in R^{\MM_i} } \in \UU$
:
:$\set {i \in I: \tuple {m'_{1, i}, \dotsc, m'_{n, i} } \in R^{\MM_i} } \in \UU$
=== Proof ===
Let:
From the lemma:
:$I^* \in \UU$
therefore:
:$S \in \UU \implies T \in \UU$
Note that:
:$\tuple {m_{1, i}, \dots, m_{n, i} } = \tuple {m'_{1, i}, \dotsc, m'_{n, i} }$ for $i \in I^*$
we have:
:$T = T'$
Hence:
:$T' \in \UU$
and:
:$S' \in \UU$ since $S' \supseteq T'$
So far we have proved:
:$S \in \UU \implies S' \in \UU$
By symmetry:
:$S' \in \UU \implies S \in \UU$
Category:Model Theory for Predicate Logic
Unary Operation/Examples/All Mappings are Unary

== Example of Unary Operation ==
To a set theorist, all mappings are unary operations.
== Proof ==
Let $\ds \prod_{i \mathop = 1}^n S_i$ be the cartesian product of sets $S_1$ to $S_n$.
Let $f: \ds \prod_{i \mathop = 1}^n S_i \to T$ be a mapping from $\ds \prod_{i \mathop = 1}^n S_i$ to $T$.
Thus:
:$f \subseteq \paren {S_1 \times S_2 \times \dotsb \times S_n} \times T$
Hence to consider $f$ as a unary operation, one would consider $\ds \prod_{i \mathop = 1}^n S_i$ as:
:a set whose elements are ordered $n$-tuples
as opposed to:
:an $n$-dimensional cartesian product of sets.
A set theorist, to a certain level of approximation, considers a set to be an aggregation of any objects.
Hence a set of ordered $n$-tuples is a convenient way to regard the domain of $f$.
== Sources ==
* : $\S 1$: Naive Set Theory: $\S 1.6$: Functions: Exercise $1.6.1$
Category:Examples of Operations
Unary Representation of Natural Number

== Theorem ==
Let $x \in \N$ be a natural number.
Then:
:$x = \map s {\map s {\dots \map s 0} }$
where there are $x$ applications of the successor mapping to the constant $0$.
== Proof ==
We shall proceed by induction.
=== Base Case ===
When $x = 0$, we have:
:$0 = 0$
which follows from Equality is Reflexive.
But that is zero applications of the successor mapping to the constant $0$.
=== Induction Step ===
Let $x \in \N$ satisfy the theorem.
Then:
:$x = \map s {\map s {\dots \map s 0} }$
where there are $x$ applications of the successor mapping.
Denote the term as $\phi$.
But by Substitution Property of Equality:
:$\map s x = \map s \phi$
But $\map s \phi$ consists of $x + 1 = \map s x$ applications of the successor mapping to the constant $0$.
Thus, by Principle of Mathematical Induction, the theorem holds for every $x \in \N$.
Category:Natural Numbers
Category:Proofs by Induction
Unary Truth Functions

== Theorem ==
There are $4$ distinct unary truth functions:
:$(1): \quad$ The constant function $\map f p = \F$
:$(2): \quad$ The constant function $\map f p = \T$
:$(3): \quad$ The identity function $\map f p = p$
:$(4): \quad$ The logical not function $\map f p = \neg p$
== Proof ==
From Count of Truth Functions there are $2^{\paren {2^1} } = 4$ distinct truth functions on $1$ variable.
These can be depicted in a truth table as follows:
:$\begin{array}{|c|cccc|} \hline
p & \circ_1 & \circ_2 & \circ_3 & \circ_4 \\
\hline
\T & \T & \T & \F & \F \\
\F & \T & \F & \T & \F \\
\hline
\end{array}$
$\circ_1$: Whether $p = \T$ or $p = \F$, $\map {\circ_1} p = \T$.
Thus $\circ_1$ is the constant function $\map {\circ_1} p = \T$.
$\circ_2$: We have:
:$(1): \quad p = \T \implies \map {\circ_2} p = \T$
:$(2): \quad p = \F \implies \map {\circ_2} p = \F$
Thus $\circ_2$ is the identity function $\map {\circ_2} p = p$.
$\circ_3$: We have:
:$(1): \quad p = \T \implies \map {\circ_3} p = \F$
:$(2): \quad p = \F \implies \map {\circ_3} p = \T$
Thus $\circ_3$ is the logical not function $\map {\circ_3} p = \neg p$.
$\circ_4$: Whether $p = \T$ or $p = \F$, $\map {\circ_4} p = \F$.
Thus $\circ_4$ is the constant function $\map {\circ_4} p = \F$.
All four have been examined, and there are no other unary truth functions.
== Linguistic Note ==
== Sources ==
* : Chapter $2$: Propositional Calculus: $\S 2.1$: Boolean operators
* : $\S 2.4.1$
Category:Truth Functions
Unbiased Estimator/Examples/Sample Variance

== Example of Unbiased Estimator ==
For a random sample of $n$ observations $x_i$ for $1 = 1, 2, \ldots, n$, an unbiased estimator for the population variance $\sigma^2$ is given by:
:$\ds \dfrac 1 {n - 1} \sum_i \paren {x_i - \bar x}^2$
or presented as:
:$\ds \dfrac n {n - 1} {s_x}^2$
where ${s_x}^2 is the sample variance.
Compare with the plug-in estimator of the same thing:
:$\ds \sum_i \dfrac {\paren {x_i - \bar x}^2} n$
This is a biased estimator of $\sigma^2$.
== Sources ==
* 
* 
* 
* 
Category:Examples of Unbiased Estimators
Unbounded Closed Real Interval/Examples/Example 1

== Example of Unbounded Closed Real Interval ==
Let $I$ be the unbounded closed real interval defined as:
:$I := \hointl \gets 3$
Then $2 \in I$.
== Proof ==
By definition of open real interval:
:$I = \set {x \in \R: x \le 3}$
As $2 \le 3$ it follows that $2 \in I$.
== Sources ==
* : $\S 2$: Continuum Property: Exercise: $\S 2.10 \ (1) \ \text{(ii)}$
Category:Examples of Intervals
Unbounded Monotone Sequence Diverges to Infinity

== Theorem ==
Let $\sequence {x_n}$ be a sequence in $\R$.
Let $\sequence {x_n}$ be monotone, that is either increasing or decreasing.
=== Increasing ===
=== Decreasing ===
== Sources ==
* : $\S 4$: Convergent Sequences: Exercise $\S 4.29 \ (5)$
Category:Limits of Sequences
Category:Unbounded Monotone Sequence Diverges to Infinity
Unbounded Monotone Sequence Diverges to Infinity/Decreasing

== Theorem ==
Let $\sequence {x_n}$ be a sequence in $\R$.
Let $\sequence {x_n}$ be decreasing and unbounded below.
Then $x_n \to -\infty$ as $n \to \infty$.
== Proof ==
Let $H > 0$.
As $\sequence {x_n}$ is unbounded above:
:$\exists N: x_N > H$
As $\sequence {x_n}$ is decreasing:
:$\forall n \ge N: x_n \le x_N < H$
It follows from the definition of divergence to $-\infty$ that $x_n \to -\infty$ as $n \to \infty$.
== Sources ==
* : $\S 4$: Convergent Sequences: Exercise $\S 4.29 \ (5)$
Category:Unbounded Monotone Sequence Diverges to Infinity
Unbounded Monotone Sequence Diverges to Infinity/Increasing

== Theorem ==
Let $\sequence {x_n}$ be a sequence in $\R$.
Let $\sequence {x_n}$ be increasing and unbounded above.
Then $x_n \to +\infty$ as $n \to \infty$.
== Proof ==
Let $H > 0$.
As $\sequence {x_n}$ is unbounded above:
:$\exists N: x_N > H$
As $\sequence {x_n}$ is increasing:
:$\forall n \ge N: x_n \ge x_N > H$
It follows from the definition of divergence to $+\infty$ that $x_n \to +\infty$ as $n \to \infty$.
== Sources ==
* : $\S 4$: Convergent Sequences: Exercise $\S 4.29 \ (5)$
Category:Unbounded Monotone Sequence Diverges to Infinity
Unbounded Real-Valued Function/Examples/-1^n by n

== Example of Unbounded Real-Valued Function ==
The function $f$ defined on the integers $\Z$:
:$\forall n \in \Z: \map f n := \paren {-1}^n n$
is unbounded.
== Sources ==
* : Chapter $\text I$: Functions and Limits: $\S 3$: Bounds of a Function
Category:Examples of Unbounded Real-Valued Functions
Unbounded Real-Valued Function/Examples/1 over x

== Example of Unbounded Real-Valued Function ==
The function $f$ defined on the positive real numbers $\openint 0 \to$:
:$\forall x \in \openint 0 \to: \map f x := \dfrac 1 x$
is bounded below (by $0$) but unbounded above.
Hence $f$ is unbounded.
== Sources ==
* 
* 
Category:Examples of Unbounded Real-Valued Functions
Unbounded Real-Valued Function/Examples/x sin x

== Example of Unbounded Real-Valued Function ==
The function $f$ defined on the positive real numbers $\openint 0 \to$:
:$\forall x \in \openint 0 \to: \map f x := x \sin x$
is both unbounded below and unbounded above.
Hence $f$ is unbounded.
== Sources ==
* 
* 
Category:Examples of Unbounded Real-Valued Functions
Unbounded Set of Real Numbers is not Compact

== Theorem ==
Let $\R$ be the set of real numbers considered as a Euclidean space.
Let $S \subseteq \R$ be unbounded in $\R$.
Then $S$ is not a compact subspace of $\R$.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Compact Spaces
Category:Analysis
Category:Unbounded Set of Real Numbers is not Compact
Unbounded Set of Real Numbers is not Compact/Normed Vector Space

== Theorem ==
== Proof ==
We have that $\struct {\R, \size {\, \cdot \,} }$ is a normed vector space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence such that $x_n = n$.
 $\sequence {x_n}_{n \mathop \in \N}$ poseses a convergent subsequence $\sequence {x_{n_k}}_{k \mathop \in \N}$.
By Convergent Sequence is Cauchy Sequence, $\sequence {x_{n_k}}_{k \mathop \in \N}$ is Cauchy.
However:
This contradicts the definition of Cauchy sequence.
Hence, there is no convergent subsequence of $\sequence {x_n}_{n \mathop \in \N}$.
By definition, $\struct {\R, \size {\, \cdot \,} }$ is not compact.
== Sources ==
* : Chapter $\S 1.5$: Normed and Banach spaces. Compact sets
Category:Unbounded Set of Real Numbers is not Compact
Category:Compact Spaces
Unbounded Set of Real Numbers is not Compact/Proof 1

== Theorem ==
== Proof ==
By the rule of transposition, it suffices to show that if $S$ is a compact subspace of $\R$, then $S$ is bounded.
Let $\CC$ be the set of all open $\epsilon$-balls of $0$ in $\R$:
:$\CC = \set {\map {B_\epsilon} 0: \epsilon \in \R_{>0}}$
We have that:
:$\ds \bigcup \CC = \R \supseteq S$
From Open Ball of Metric Space is Open Set, it follows that $\CC$ is an open cover for $S$.
Let $\FF$ be a finite subcover of $\CC$ for $S$.
Then $\ds \bigcup \FF$ is the largest open $\epsilon$-ball in $\FF$.
Thus:
:$S \subseteq \ds \bigcup \FF \in \CC$
Hence, $S$ is bounded.
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: Compactness
Category:Unbounded Set of Real Numbers is not Compact
Unbounded Space Minus Bounded Space is Unbounded

== Theorem ==
Let $M$ be a metric space.
Let $A \subseteq M$ be unbounded in $M$.
Let $B \subseteq M$ be bounded in $M$.
Then $A \setminus B$ is unbounded in $M$.
== Proof 1 ==
== Proof 2 ==
Category:Bounded Metric Spaces
Unbounded Space Minus Bounded Space is Unbounded/Proof 1

== Theorem ==
== Proof ==
 $A \setminus B$ is bounded.
Then by definition of bounded:
:$\exists K \in \R: \forall x, y \in A \setminus B: \map d {x, y} \le K$
By definition of bounded, choose some $a_b \in A$ and $d_b \in \R$ such that:
:$\forall x \in B: \map d {x, a_b} \le d_b$
By definition of unbounded, $A$ is not bounded is $M$.
Thus, by the negation of the definition of bounded:
:$\exists a_x \in A: \map d {a_x, a_b} > d_b$
Let $d_x = \map d {a_x, a_b}$.
Again, by the negation of the definition of bounded:
:$\exists a_y \in A: \map d {a_y, a_b} > d_x + K$
Let $d_y = \map d {a_y, a_b}$.
As $\map d {a_x, a_b} > d_b$, it follows that $a_x \notin B$.
Likewise, as $\map d {a_y, a_b} > d_x > d_b$, $a_y \notin B$.
Thus, both $a_x, a_y \in A \setminus B$.
But:
Then, $a_y, a_x \in A \setminus B$ are two points such that $\map d {a_y, a_x} > K$, contradicting our assumption.
Thus, by Proof by Contradiction, it follows that $A \setminus B$ is unbounded.
Unbounded Space Minus Bounded Space is Unbounded/Proof 2

== Theorem ==
== Proof ==
 $A \setminus B$ is bounded.
By Finite Union of Bounded Subsets:
:$\paren {A \setminus B} \cup B$
is bounded.
On the other hand, by :
:$A \subseteq \paren {A \setminus B} \cup B$
Thus by Subset of Bounded Subset of Metric Space is Bounded, $A$ must be bounded, too.
This is a contradiction.
Uncertainty/Examples/Example 1

== Example of Uncertainty ==
Let $R_1$ and $R_2$ be horseraces.
Let $R_1$ have $7$ runners:
:$3$ of which each have probability $\dfrac 1 6$ of winning
:$4$ of which each have probability $\dfrac 1 8$ of winning.
Let $R_2$ have $8$ runners:
:$2$ of which each have probability $\dfrac 1 4$ of winning
:$6$ of which each have probability $\dfrac 1 {12}$ of winning.
Then $R_1$ and $R_2$ have equal uncertainty.
== Proof ==
Let $H_1$ and $H_2$ denote the uncertainty of $R_1$ and $R_2$ respectively.
Recall the definition of uncertainty:
:$\map H X = \ds -\sum_k p_k \lg p_k$
Thus:
and:
Hence the result.
== Sources ==
* : $\S 1$: Entropy = uncertainty = information: $1.1$ Uncertainty: Exercises $1.1$: $1.$
Category:Examples of Uncertainty
Uncertainty Function satisfies Axioms of Uncertainty

== Theorem ==
Let $X$ be a random variable.
Let $X$ take a finite number of values with probabilities $p_1, p_2, \dotsc, p_n$.
Let $\map H X$ be the uncertainty function of $X$:
:$\map H X = \ds -\sum_k p_k \lg p_k$
where:
:$\lg$ denotes logarithm base $2$
:the summation is over those $k$ where $p_k > 0$.
Then the uncertainty function satisfies the Axioms of Uncertainty.
== Proof ==
== Sources ==
* : $\S 1$: Entropy = uncertainty = information: $1.1$ Uncertainty: Exercises $1.1$: $2.$
Category:Uncertainty
Unconditional Inequality/Examples/Arbitrary Example 1

== Example of Unconditional Inequality ==
The inequality:
:$2 x^2 + 1 > x - 1$
is an unconditional inequality.
== Proof ==
It remains to be shown that $(1)$ is true for all $x \in \R$.
Differentiating $y = 2 x^2 - x + 2$ $x$ gives us:
:$\dfrac {\d y} {\d x} = 4 x - 1$
which is increasing throughout and is zero at $x = \dfrac 1 4$.
At $x = \dfrac 1 4$ we have that:
:$y = 2 \times \paren {\dfrac 1 4}^2 - \dfrac 1 4 + 2 = 1 \frac 7 8 > 0$
As the coefficient of $x$ is positive, this means $y$ has a minimum at $x = \dfrac 1 4$.
As that is positive, that means $2 x^2 - x + 2$ never goes negative.
Hence we have that:
:$\forall x \in \R: 2 x^2 + 1 > x - 1$
as we were required to show.
== Sources ==
* 
* 
Category:Examples of Unconditional Inequalities
Uncorrelated Data with Bivariate Normal Distribution is Independent

== Theorem ==
Let $X$ and $Y$ be random variables with bivariate normal distribution.
Let $X$ and $Y$ be uncorrelated.
Then $X$ and $Y$ are independent.
== Proof ==
== Sources == 
* 
* 
Category:Uncorrelated Data
Category:Bivariate Distributions
Category:Normal Distribution
Uncountable Closed Ordinal Space is Countably Compact

== Theorem ==
Let $\Omega$ denote the first uncountable ordinal.
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
Then $\closedint 0 \Omega$ is a countably compact space.
== Proof ==
We have:
:Closed Ordinal Space is Compact
:Compact Space is Countably Compact
== Sources ==
* : Part $\text {II}$: Counterexamples: $43$. Closed Ordinal Space $[0, \Omega]$: $8$
Category:Ordinal Spaces
Category:Countably Compact Spaces
Uncountable Closed Ordinal Space is Lindelöf

== Theorem ==
Let $\Omega$ denote the first uncountable ordinal.
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
Then $\closedint 0 \Omega$ is a Lindelöf space.
== Proof ==
We have:
:Closed Ordinal Space is Compact
:Compact Space is Lindelöf
== Sources ==
* : Part $\text {II}$: Counterexamples: $43$. Closed Ordinal Space $[0, \Omega]$: $10$
Category:Ordinal Spaces
Category:Lindelöf Spaces
Uncountable Closed Ordinal Space is Sigma-Compact

== Theorem ==
Let $\Omega$ denote the first uncountable ordinal.
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
Then $\closedint 0 \Omega$ is a $\sigma$-compact space.
== Proof ==
We have:
:Closed Ordinal Space is Compact
:Compact Space is $\sigma$-Compact
== Sources ==
* : Part $\text {II}$: Counterexamples: $43$. Closed Ordinal Space $[0, \Omega]$: $10$
Category:Ordinal Spaces
Category:Sigma-Compact Spaces
Uncountable Closed Ordinal Space is not First-Countable

== Theorem ==
Let $\Omega$ denote the first uncountable ordinal.
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
Then $\closedint 0 \Omega$ is not a first-countable space.
== Proof ==
From Omega is Closed in Uncountable Closed Ordinal Space but not G-Delta Set, $\set \Omega$ cannot be expressed as a countable intersection of open sets of $\closedint 0 \Omega$.
Thus, by definition, $\Omega$ does not have a countable local basis.
Hence the result by definition of first-countable space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $43$. Closed Ordinal Space $[0, \Omega]$: $2$
Category:Ordinal Spaces
Category:First-Countable Spaces
Uncountable Closed Ordinal Space is not Perfectly Normal

== Theorem ==
Let $\Omega$ denote the first uncountable ordinal.
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
Then $\closedint 0 \Omega$ is not a perfectly normal space.
== Proof ==
From Omega is Closed in Uncountable Closed Ordinal Space but not $G_\delta$ Set, $\set \Omega$ is not a $G_\delta$ set.
From Ordinal Space is Completely Normal, $\closedint 0 \Omega$ is a $T_1$ (Fréchet) space.
Thus by definition $\set \Omega$ is closed in $\closedint 0 \Omega$.
Thus we have that $\set \Omega$ is a closed set of $\closedint 0 \Omega$ which is not a $G_\delta$ set.
The result follows by definition of perfectly normal space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $43$. Closed Ordinal Space $[0, \Omega]$: $4$
Category:Ordinal Spaces
Category:Perfectly Normal Spaces
Uncountable Closed Ordinal Space is not Second-Countable

== Theorem ==
Let $\Omega$ denote the first uncountable ordinal.
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
Then $\closedint 0 \Omega$ is not a second-countable space.
== Proof ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $43$. Closed Ordinal Space $[0, \Omega]$: $5$
Category:Ordinal Spaces
Category:Second-Countable Spaces
Uncountable Closed Ordinal Space is not Separable

== Theorem ==
Let $\Omega$ denote the first uncountable ordinal.
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
Then $\closedint 0 \Omega$ is not a separable space.
== Proof ==
Let $H \subseteq \closedint 0 \Omega$ be a countable subset of $\closedint 0 \Omega$.
From Uncountable Open Ordinal Space is not Separable, there exists an open interval $\openint \sigma \Omega$ in the complement of $H^-$ in $\hointr 0 \Omega$, and so also in $\closedint 0 \Omega$.
Thus the closure of $H$ in $\closedint 0 \Omega$ does not equal $\closedint 0 \Omega$.
Thus $H$ is not everywhere dense in $\closedint 0 \Omega$.
Hence, by definition, $\closedint 0 \Omega$ is not separable.
== Sources ==
* : Part $\text {II}$: Counterexamples: $43$. Closed Ordinal Space $[0, \Omega]$: $3$
Category:Ordinal Spaces
Category:Separable Spaces
Uncountable Discrete Space is not Lindelöf

== Theorem ==
Let $T = \struct {S, \tau}$ be an uncountable discrete topological space.
Then $T$ is not a Lindelöf space.
== Proof ==
Consider the set $\CC$ of all singleton subsets of $S$:
:$\CC := \set {\set x: x \in S}$
From Discrete Space has Open Locally Finite Cover, $\CC$ is an open cover of $S$ which is finer than any other open cover of $S$.
That is, $\CC$ is an open cover of $S$ which is uncountable and has no countable subcover.
(Note that a subcover is a refinement of a cover.)
So by definition $T$ can not be a Lindelöf space.
== Also see ==
* Countable Discrete Space is Lindelöf
== Sources ==
* : Part $\text {II}$: Counterexamples: $3$. Uncountable Discrete Topology: $8$
Category:Lindelöf Spaces
Category:Discrete Topology
Category:Uncountable Sets
Uncountable Discrete Space is not Second-Countable

== Theorem ==
Let $T = \struct {S, \tau}$ be an uncountable discrete topological space.
Then $T$ is not second-countable.
== Proof ==
We have that an Uncountable Discrete Space is not Separable.
From Second-Countable Space is Separable, it follows that $T$ can not be second-countable.
== Also see ==
* Countable Discrete Space is Second-Countable
== Sources ==
* : Part $\text {II}$: Counterexamples: $3$. Uncountable Discrete Topology: $8$
Category:Second-Countable Spaces
Category:Discrete Topology
Category:Uncountable Sets
Uncountable Discrete Space is not Separable

== Theorem ==
Let $T = \struct {S, \tau}$ be an uncountable discrete topological space.
Then $T$ is not separable.
== Proof ==
Let $H \subseteq S$ be everywhere dense in $T$.
Then by definition of everywhere dense:
:$H^- = S$
where $H^-$ denotes the closure of $H$.
We have that $T$ is a discrete space.
Hence from Interior Equals Closure of Subset of Discrete Space:
:$H^- = H$
So:
:$H^- = S \implies H = S$
But $S$ is uncountable.
So there exists no $H \subseteq S$ such that $H$ is both countable and everywhere dense.
Hence by definition of separable space, if $T$ is an uncountable discrete space it can not be separable.
== Also see ==
* Countable Discrete Space is Separable
== Sources ==
* : Part $\text {II}$: Counterexamples: $3$. Uncountable Discrete Topology: $8$
Category:Separable Spaces
Category:Discrete Topology
Category:Uncountable Sets
Uncountable Discrete Space is not Sigma-Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be an uncountable discrete topological space.
Then $T$ is not $\sigma$-compact.
== Proof ==
We have that an Uncountable Discrete Space is not Lindelöf.
But a $\sigma$-compact space is Lindelöf.
So an uncountable discrete space can not be $\sigma$-compact.
== Also see ==
* Countable Discrete Space is Sigma-Compact
== Sources ==
* : Part $\text {II}$: Counterexamples: $3$. Uncountable Discrete Topology: $8$
Category:Sigma-Compact Spaces
Category:Discrete Topology
Category:Uncountable Sets
Uncountable Excluded Point Space is not Second-Countable

== Theorem ==
Let $T = \struct {S, \tau_{\bar p} }$ be an uncountable excluded point space.
Then $T$ is not second-countable.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $15$. Uncountable Excluded Point Topology: $6$
Category:Second-Countable Spaces
Category:Excluded Point Topology
Category:Uncountable Excluded Point Space is not Second-Countable
Uncountable Excluded Point Space is not Second-Countable/Proof 1

== Theorem ==
== Proof ==
Let $H = S \setminus \left\{{p}\right\}$ where $\setminus$ denotes set difference.
By definition, $H$ is an uncountable discrete space.
The result follows from Uncountable Discrete Space is not Second-Countable.
Category:Uncountable Excluded Point Space is not Second-Countable