nyuuzyou/wikis huggingface dataset:
Uncountable Excluded Point Space is not Second-Countable/Proof 2

== Theorem ==
== Proof ==
We have:
: Uncountable Discrete Space is not Second-Countable
: Excluded Point Topology is Open Extension Topology of Discrete Topology
The result follows from Condition for Open Extension Space to be Second-Countable
Category:Uncountable Excluded Point Space is not Second-Countable
Uncountable Excluded Point Space is not Separable

== Theorem ==
Let $T = \struct {S, \tau_{\bar p} }$ be an uncountable excluded point space.
Then $T$ is not separable.
== Proof ==
Let $H \subseteq S$ such that $H$ is countable.
Then $H \ne S$ as $S$ is uncountable by hypothesis.
From Limit Points in Excluded Point Space, the only limit point of $H$ is $p$.
So, by definition, the closure of $H$ is $H \cup \set p$.
From Countable Union of Countable Sets is Countable we have that $H \cup \set p$ is countable.
So $H \cup \set p \ne S$.
So $H$ is not everywhere dense in $T$.
Thus $T$ can have no countable subset of $S$ which is everywhere dense in $T$.
Hence the result by definition of separable space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $15$. Uncountable Excluded Point Topology: $6$
Category:Separable Spaces
Category:Excluded Point Topology
Uncountable Finite Complement Space is not First-Countable

== Theorem ==
Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$.
Then $T$ is not first-countable.
== Proof ==
 some $x \in S$ has a countable local basis.
That means:
:there exists a countable set $\BB_x \subseteq \tau$
such that:
:$\forall B \in \BB_x: x \in B$
and such that:
:every open neighborhood of $x$ contains some $B \in \BB_x$.
Let $y \in S$ with $y \ne x$.
Because its complement relative to $S$ is finite, we have that $S \setminus \set y$ is an open set containing $x$.
Because $\BB_x$ is a neighborhood basis at $x$, there exists some $B \in \BB_x$ such that:
:$B \subseteq S \setminus \set y$
whence $y \notin B$.
So:
:$y \notin \bigcap \BB_x$
So:
By definition, each of $S \setminus B$ is finite.
From Countable Union of Countable Sets is Countable it follows that $\ds \bigcup_{B \mathop \in \BB_x} \paren {S \setminus B}$ is countable.
So $S \setminus \set x$ and therefore $S$ is also countable.
From this contradiction (as we have specified that $S$ is uncountable) it follows that our assumption that $x \in S$ has a countable local basis must be false.
Hence by definition $T$ can not be first-countable.
== Sources ==
* : Part $\text {II}$: Counterexamples: $19$. Finite Complement Topology on an Uncountable Space: $4$
Category:First-Countable Spaces
Category:Finite Complement Topology
Uncountable Finite Complement Topology is not Perfectly T4

== Theorem ==
Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$.
Then $T$ is not a perfectly $T_4$ space.
== Proof ==
Recall the definition of a perfectly $T_4$ space
:Every closed set in $T$ can be written as a countable intersection of open sets of $T$.
Let $V$ be a closed set in $T$.
From Closed Set of Uncountable Finite Complement Topology is not $G_\delta$:
:$V$ is not a $G_\delta$ set.
The result follows by definition of perfectly $T_4$ space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $19$. Finite Complement Topology on an Uncountable Space: $3$
Category:Perfectly T4 Spaces
Category:Finite Complement Topology
Uncountable Fort Space is not First-Countable

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a Fort space on an uncountable set $S$.
Then $T$ is not a first-countable space.
== Proof ==
Let $\UU$ be a countable set of open neighborhoods of $p$.
Let $U \in \UU$.
Then as $p \in U$, $p \notin \relcomp S U$ and so for $U$ to be open it must follow that $\relcomp S U$ is finite.
From De Morgan's Laws: Complement of Intersection we have:
:$\ds H := \bigcup_{U \mathop \in \UU} \relcomp S U = \relcomp S {\bigcap \UU}$
From Countable Union of Countable Sets is Countable it follows that $H$ can be no more than countable.
So $\ds \bigcap \UU = \relcomp S H$ must be uncountable.
So (trivially):
:$H \ne \set p$
So:
:$\exists q \ne p: q \in \ds \bigcap \UU$
So $\relcomp S {\set q}$ is an open neighborhood of $P$ which does not contain any of the elements of $\UU$.
So $\UU$ is not a countable local basis of $p$.
Hence by definition $T$ is not a first-countable space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $24$. Uncountable Fort Space: $5$
Category:First-Countable Spaces
Category:Fort Spaces
Uncountable Fort Space is not Perfectly Normal

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a Fort space on an uncountable set $S$.
Then $T$ is not a perfectly normal space.
== Proof ==
From Clopen Points in Fort Space, $\set p$ is closed in $T$.
Consider a countable intersection of open sets of $T$ which contain $p$.
By definition, all these are cofinite in $S$ and so uncountable.
So this intersection must itself contain all but a countable number of points of $S$.
So $\set p$ is not a $G_\delta$ set.
Hence $T$ is not a perfectly normal space as not each one of its closed sets is a $G_\delta$ set.
== Sources ==
* : Part $\text {II}$: Counterexamples: $24$. Uncountable Fort Space: $3$
Category:Perfectly Normal Spaces
Category:Fort Spaces
Uncountable Fort Space is not Separable

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a Fort space on an uncountable set $S$.
Then $T$ is not a separable space.
== Proof ==
Let $C \subseteq S$ be a countable set.
Since $S$ is uncountable, by Uncountable Set less Countable Set is Uncountable, so is $\relcomp S C$.
Thus there exists some point $x \in \relcomp S C$ and $x \ne p$.
By Clopen Points in Fort Space, $\set x \in \tau_p$. 
By Empty Intersection iff Subset of Complement, we have $C \cap \set x = \O$.
Therefore $C$ is not everywhere dense.
Since $C$ is arbitrary, $T$ is not a separable space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $24$. Uncountable Fort Space: $4$
Category:Separable Spaces
Category:Fort Spaces
Uncountable Open Ordinal Space is Countably Compact

== Theorem ==
Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is a countably compact space.
== Proof ==
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
From Uncountable Closed Ordinal Space is Countably Compact, $\closedint 0 \Omega$ is a countably compact space.
So every sequence in $\hointr 0 \Omega$ has an accumulation point in $\closedint 0 \Omega$.
But $\Omega$ cannot be an accumulation point of any sequence in $\closedint 0 \Omega$.
So every sequence in $\hointr 0 \Omega$ has an accumulation point in $\hointr 0 \Omega$.
This means that $\hointr 0 \Omega$ is countably compact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $42$. Open Ordinal Space $[0, \Omega)$: $8$
Category:Ordinal Spaces
Category:Countably Compact Spaces
Uncountable Open Ordinal Space is First-Countable

== Theorem ==
Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is a first-countable space.
== Proof ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $42$. Open Ordinal Space $[0, \Omega)$: $3$
Category:Ordinal Spaces
Category:First-Countable Spaces
Uncountable Open Ordinal Space is Sequentially Compact

== Theorem ==
Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is a sequentially compact space.
== Proof ==
We have that:
:Uncountable Open Ordinal Space is First-Countable
:Uncountable Open Ordinal Space is Countably Compact
The result follows from First-Countable Space is Sequentially Compact iff Countably Compact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $42$. Open Ordinal Space $[0, \Omega)$: $11$
Category:Ordinal Spaces
Category:Sequentially Compact Spaces
Uncountable Open Ordinal Space is not Lindelöf

== Theorem ==
Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is not a Lindelöf space.
== Proof ==
 $\hointr 0 \Omega$ is a Lindelöf space.
From Ordinal Space is Completely Normal, $\hointr 0 \Omega$ is a completely normal.
From Sequence of Implications of Separation Axioms, $\hointr 0 \Omega$ is a $T_3$ space.
From Lindelöf $T_3$ Space is Paracompact, it follows that $\hointr 0 \Omega$ is a paracompact space.
But this contradicts the fact that from Uncountable Open Ordinal Space is not Paracompact, $\hointr 0 \Omega$ is not a paracompact space.
Hence the result by Proof by Contradiction.
== Sources ==
* : Part $\text {II}$: Counterexamples: $42$. Open Ordinal Space $[0, \Omega)$: $10$
Category:Ordinal Spaces
Category:Paracompact Spaces
Uncountable Open Ordinal Space is not Metacompact

== Theorem ==
Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is not a metacompact space.
== Proof ==
 $\hointr 0 \Omega$ is a metacompact space.
From Open Ordinal Space is not Compact in Closed Ordinal Space we have that $\hointr 0 \Omega$ is a countably compact space.
From Metacompact Countably Compact Space is Compact it follows that $\hointr 0 \Omega$ is a compact space.
But from Open Ordinal Space is not Compact in Closed Ordinal Space this contradicts the fact that $\hointr 0 \Omega$ is not a compact space.
Hence the result by Proof by Contradiction.
== Sources ==
* : Part $\text {II}$: Counterexamples: $42$. Open Ordinal Space $[0, \Omega)$: $9$
Category:Ordinal Spaces
Category:Metacompact Spaces
Uncountable Open Ordinal Space is not Paracompact

== Theorem ==
Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is not a paracompact space.
== Proof ==
 $\hointr 0 \Omega$ is a paracompact space.
From Paracompact Space is Metacompact, it follows that $\hointr 0 \Omega$ is a metacompact space.
But from Uncountable Open Ordinal Space is not Metacompact this contradicts the fact that $\hointr 0 \Omega$ is not a metacompact space.
Hence the result by Proof by Contradiction.
== Sources ==
* : Part $\text {II}$: Counterexamples: $42$. Open Ordinal Space $[0, \Omega)$: $9$
Category:Ordinal Spaces
Category:Paracompact Spaces
Uncountable Open Ordinal Space is not Second-Countable

== Theorem ==
Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is not a second-countable space.
== Proof ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $42$. Open Ordinal Space $[0, \Omega)$: $5$
Category:Ordinal Spaces
Category:Second-Countable Spaces
Uncountable Open Ordinal Space is not Separable

== Theorem ==
Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is not a separable space.
== Proof ==
Because $\Omega$ is the first uncountable ordinal, any ordinal which strictly precedes $\Omega$ is countable.
Let $H \subseteq \hointr 0 \Omega$ be a countable subset of $\hointr 0 \Omega$.
Let $\sigma$ be the supremum of $H$.
As $H$ by definition strictly precedes $\Omega$, $H$ itself is countable.
Thus $\sigma$ strictly precedes $\Omega$.
The closed interval $\closedint 0 \sigma$ is such that $H \subseteq \closedint 0 \sigma$.
By definition of the closure $H^-$ of $H$ as the smallest closed set of $\hointr 0 \Omega$ containing $H$, it follows that $H^- \subseteq \closedint 0 \sigma$.
Therefore, there exists an open interval $\openint \sigma \Omega$ in the complement of $H^-$ in $\hointr 0 \Omega$.
Thus the closure of $H$ does not equal $\hointr 0 \Omega$.
Thus $H$ is not everywhere dense in $\hointr 0 \Omega$.
Hence, by definition, $\hointr 0 \Omega$ is not separable.
== Sources ==
* : Part $\text {II}$: Counterexamples: $42$. Open Ordinal Space $[0, \Omega)$: $3$
Category:Ordinal Spaces
Category:Separable Spaces
Uncountable Open Ordinal Space is not Sigma-Compact

== Theorem ==
Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is not a $\sigma$-compact space.
== Proof ==
 $\hointr 0 \Omega$ is a $\sigma$-compact space.
From Sigma-Compact Space is Lindelöf, $\hointr 0 \Omega$ is a Lindelöf space.
But this contradicts the fact that from Uncountable Open Ordinal Space is not Lindelöf, $\hointr 0 \Omega$ is not a Lindelöf space.
Hence the result by Proof by Contradiction.
== Sources ==
* : Part $\text {II}$: Counterexamples: $42$. Open Ordinal Space $[0, \Omega)$: $10$
Category:Ordinal Spaces
Category:Sigma-Compact Spaces
Uncountable Particular Point Space is not Lindelöf

== Theorem ==
Let $T = \struct {S, \tau_p}$ be an uncountable particular point space.
Then $T$ is not a Lindelöf space.
== Proof ==
Consider the open cover of $T$:
:$\CC = \set {\set {x, p}: x \in S, x \ne p}$
As $S$ is uncountable, then so is $\CC$, as we can set up a bijection $\phi: S \setminus \set p \leftrightarrow \CC$:
:$\forall x \in S \setminus \set p: \map \phi x = \set {x, p}$
Hence $\CC$ has no countable subcover.
The result follows by definition of Lindelöf space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $10$. Uncountable Particular Point Topology: $5$
Category:Lindelöf Spaces
Category:Particular Point Topology
Uncountable Particular Point Space is not Second-Countable

== Theorem ==
Let $T = \struct {S, \tau_p}$ be an uncountable particular point space.
Then $T$ is not second-countable.
== Proof ==
Let $H = S \setminus \set p$ where $\setminus$ denotes set difference.
Every subset $V \subseteq H$ is a closed set from Subset of Particular Point Space is either Open or Closed.
Thus we can consider $H$ as an uncountable discrete space.
The result follows from Uncountable Discrete Space is not Second-Countable.
== Sources ==
* : Part $\text {II}$: Counterexamples: $10$. Uncountable Particular Point Topology: $7$
Category:Second-Countable Spaces
Category:Particular Point Topology
Uncountable Product of First-Countable Spaces is not always First-Countable

== Theorem ==
Let $I$ be an indexing set with uncountable cardinality.
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.
Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be a first-countable space.
Then it is not necessarily the case that $\struct {S, \tau}$ is also a first-countable space.
== Proof ==
Let $T = \struct {\Z_{\ge 0}, \tau}$ denote the topological space consisting of the set of positive integers $\Z_{\ge 0}$ under the discrete topology.
Let $I$ be an indexing set with uncountable cardinality.
Let $T' = \struct {\ds \prod_{\alpha \mathop \in \mathbb I} \struct {\Z_{\ge 0}, \tau}_\alpha, \tau'}$ be the uncountable Cartesian product of $\struct {\Z_{\ge 0}, \tau}$ indexed by $I$ with the product topology $\tau'$.
From Discrete Space is First-Countable, $T$ is a first-countable space.
But from Uncountable Cartesian Product of Discrete Topology on Positive Integers is not First-Countable, $T$ is not a first-countable space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties
Category:First-Countable Spaces
Category:Product Topology
Uncountable Product of Second-Countable Spaces is not always Second-Countable

== Theorem ==
Let $I$ be an indexing set with uncountable cardinality.
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.
Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be a second-countable space.
Then it is not necessarily the case that $\struct {S, \tau}$ is also a second-countable space.
== Proof ==
Let $T = \struct {\Z_{\ge 0}, \tau}$ denote the topological space consisting of the set of positive integers $\Z_{\ge 0}$ under the discrete topology.
Let $I$ be an indexing set with uncountable cardinality.
Let $T' = \struct {\ds \prod_{\alpha \mathop \in I} \struct {\Z_{\ge 0}, \tau}_\alpha, \tau'}$ be the uncountable Cartesian product of $\struct {\Z_{\ge 0}, \tau}$ indexed by $I$ with the product topology $\tau'$.
From Countable Discrete Space is Second-Countable, $T$ is a second-countable space.
But from Uncountable Cartesian Product of Discrete Topology on Positive Integers is not Second-Countable, $T$ is not a second-countable space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties
Category:Second-Countable Spaces
Category:Product Topology
Uncountable Product of Separable Spaces is not always Separable

== Theorem ==
Let $I$ be an indexing set with uncountable cardinality.
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.
Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be a separable space.
Then it is not necessarily the case that $\struct {S, \tau}$ is also a separable space.
== Proof ==
Let $T = \struct {\Z_{\ge 0}, \tau}$ denote the topological space consisting of the set of positive integers $\Z_{\ge 0}$ under the discrete topology.
Let $I$ be an indexing set with uncountable cardinality.
Let $\ds T' = \struct {\prod_{\alpha \mathop \in I} \struct {\Z_{\ge 0}, \tau}_\alpha, \tau'}$ be the uncountable Cartesian product of $\struct {\Z_{\ge 0}, \tau}$ indexed by $I$ with the product topology $\tau'$.
From Countable Discrete Space is Separable, $T$ is a separable space.
But from Uncountable Cartesian Product of Discrete Topology on Positive Integers is not Separable, $T$ is not a separable space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties
Category:Separable Spaces
Category:Product Topology
Uncountable Product of Sequentially Compact Spaces is not always Sequentially Compact

== Theorem ==
Let $I$ be an indexing set with uncountable cardinality.
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.
Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be sequentially compact.
Then it is not necessarily the case that $\struct {S, \tau}$ is also sequentially compact.
== Proof ==
Let $\mathbb I$ denote the closed unit interval.
Let $T = \struct {\mathbb I, \tau}$ be the topological space consisting of $\mathbb I$ under the usual (Euclidean) topology.
Let $T' = \mathbb I^{\mathbb I} = \struct {\ds \prod_{\alpha \mathop \in \mathbb I} \struct {\mathbb I, \tau}_\alpha, \tau'}$ be the uncountable Cartesian product of $\struct {\mathbb I, \tau}$ indexed by $\mathbb I$ with the product topology $\tau'$.
From Closed Real Interval is Sequentially Compact, $T$ is sequentially compact.
But from Uncountable Cartesian Product of Closed Unit Interval is not Sequentially Compact, $T$ is not a sequentially compact space.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties
Category:Sequentially Compact Spaces
Category:Product Topology
Uncountable Set less Countable Set is Uncountable

== Theorem ==
Let $S$ be an uncountable set.
Let $T \subseteq S$ be a countable subset of $S$.
Then:
:$S \setminus T$ is uncountable
where $\setminus$ denotes set difference.
== Proof ==
 $S \setminus T$ were countable.
By definition of relative complement:
:$S \setminus T = \relcomp S T$
Thus from Union with Relative Complement:
:$\paren {S \setminus T} \cup T = S$
But from Finite Union of Countable Sets is Countable it follows that $S$ is countable.
From this contradiction it follows that $S \setminus T$ is uncountable.
Category:Uncountable Sets
Category:Countable Sets
Uncountable Subset of Countable Complement Space Intersects Open Sets

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Let $H \subseteq S$ be an uncountable subset of $S$.
Then the intersection of $H$ with any non-empty open set of $T$ is uncountable.
== Proof ==
Let $U \in \tau$ be any non-empty open set of $T$.
Then $\relcomp S U$ is countable.
Suppose $H \cap U = \O$.
Then from Intersection with Complement is Empty iff Subset it follows that $H \subseteq \relcomp S U$ and so $H$ is countable.
So if $H$ is uncountable it is bound to have a non-empty intersection with every open set in $T$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $20$. Countable Complement Topology: $1$
Category:Countable Complement Topology
Uncountable Sum as Series

== Theorem ==
Let $X$ be an uncountable set.
Let $f: X \to \closedint 0 {+\infty}$ be an extended real-valued function.
The uncountable sum:
:$\ds \sum_{x \mathop \in X} \map f x = \sup \set {\sum_{x \mathop \in F} \map f x : F \subseteq X, F \text{ finite} }$
is:
:if $f$ has uncountably infinite support, then $+\infty$ 
:Otherwise, can be expressed as a (possibly divergent) series.
=== Corollary ===
== Proof ==
Define:
:$A_n = \set {x \in X: \map f x >\dfrac 1 n, n \in \N_{\ge 1} }$
Then $\sequence {A_n}_{n \mathop \in \N} \uparrow A$ is an exhausting sequence of sets, where:
:$A = \set {x \in X: \map f x > 0}$
Suppose $A$ is uncountable.
From Countable Union of Countable Sets is Countable, necessarily there is some $A_{n_0}$ which is uncountable.
Then:
Thus:
:$\ds \sum_{x \mathop \in X} \map f x = +\infty$
Otherwise, suppose then that $A$ is countably infinite.
Then there is a bijection $g: \N \to A$
Define $B_N = g \sqbrk {\set {1, 2, \ldots, N - 1, N} }$
Then every finite subset $F$ of $A$ is contained in some $B_N$.
This implies the inequality:
:$\ds \sum_{x \mathop \in F} \map f x \le \sum_{n \mathop = 1}^N \map f {\map g n} \le \sum_{x \mathop \in X} \map f x$
for each $F$ finite.
Taking the supremum of this inequality over $N$:
:$\ds \sum_{x \mathop \in F} \map f x \le \sum_{n \mathop = 1}^\infty \map f {\map g n} \le \sum_{x \mathop \in X} \map f x$
Taking the supremum of this inequality over $F$:
:$\ds \sum_{x \mathop \in X} \map f x \le \sum_{n \mathop = 1}^\infty \map f {\map g n} \le \sum_{x \mathop \in X} \map f x$
Thus:
:$\ds \sum_{x \mathop \in X} \map f x = \sum_{n \mathop = 1}^\infty \map f {\map g n}$ 
for some bijection $g: \N \to A$
== Sources ==
* : $\S P.5$
Category:Real Analysis
Category:Uncountable Sum as Series
Uncountable Sum as Series/Corollary

== Theorem ==
Let $X$ be an uncountable set.
Let $f: X \to \closedint 0 {+\infty}$ be an extended real-valued function.
Let $f: X \to \closedint 0 {+\infty}$ have uncountably infinite support.
Then:
:$\ds \sum_{x \mathop \in X} \map f x = +\infty$
== Proof ==
This is the first case of Uncountable Sum as Series.
== Sources ==
* : $\S P.5$
Category:Uncountable Sum as Series
Undecidability Theorem

== Theorem ==
Let $T$ be the set of theorems of some consistent theory in the language of arithmetic which contains minimal arithmetic $Q$.
Then $T$ is not recursive.
== Proof ==
Let $\Theta$ be the set of Gödel numbers of the theorems in $T$.
We have that $T$ is a consistent extension of $Q$.
By Set of Gödel Numbers of Arithmetic Theorems Not Definable in Arithmetic, $\Theta$ is not a definable set in $T$.
Since Recursive Sets are Definable in Arithmetic, this means that $\Theta$ is not recursive.
 $T$ were recursive.
Then by Gödel Numbering is Recursive, we could recursively go from $\Theta$ to $T$.
Then using the recursivity of $T$, we could then recursively go back to $\Theta$.
Thus $\Theta$ would be recursive.
This would be a contradiction.
Thus, by Proof by Contradiction, $T$ is not recursive.
== Sources ==
* 
Category:Mathematical Logic