avewright/proofwiki-math huggingface dataset:
Theorem: Sequence of Implications of Compactness Properties in Hausdorff Space

Let $P_1$ and $P_2$ be compactness properties.
Let:
:$P_1 \implies P_2$
mean:
:If a $T_2$ (Hausdorff) space $T$ satisfies property $P_1$, then $T$ also satisfies property $P_2$.
Then the following sequence of implications holds:
{|
|-
| align="center" | Fully $T_4$ ||
| align="center" | $\iff$ ||
| align="center" | Paracompact ||
|-
| align="center" | ||
| align="center" | ||
| align="center" | $\Big \Downarrow$ ||
|-
| align="center" | Weakly $\sigma$-Locally Compact ||
| align="center" | $\implies$ ||
| align="center" | $T_4$ ||
|-
| align="center" | $\Big \Downarrow$ ||
| align="center" | ||
| align="center" | $\Big \Downarrow$ ||
|-
| align="center" | Weakly Locally Compact ||
| align="center" | $\implies$ ||
| align="center" | $T_{3 \frac 1 2}$ ||
|-
| align="center" | ||
| align="center" | ||
| align="center" | $\Big \Downarrow$ ||
|-
| align="center" | First-Countable and Countably Compact||
| align="center" | $\implies$ ||
| align="center" | $T_3$ ||
|}

Proof:
From $T_2$ Space is $T_1$ and $T_1$ Space is $T_0$, we note that in a $T_2$ (Hausdorff) space:
:A fully $T_4$ space is a fully normal space
:A $T_4$ space is a normal space
:A $T_{3 \frac 1 2}$ space is a completely regular space
:A $T_3$ space is a regular space
all by definition.
From $T_2$ Space is Fully $T_4$ iff Paracompact we have that:
:Fully $T_4$ $\iff$ Paracompact.
From that same result and Fully $T_4$ Space is $T_4$:
:Paracompact $\implies$ $T_4$
From Normal Space is Completely Regular it follows that:
:$T_4$ $\implies$ $T_{3 \frac 1 2}$
From First-Countable Space is Sequentially Compact iff Countably Compact and Sequentially Compact Hausdorff Space is Regular it follows that:
:First-Countable and Countably Compact $\implies$ $T_3$
{{TheoremWanted|Replace the above with the below:}}
:Countably Compact First-Countable Hausdorff Space is $T_3$
The further justifications are listed as follows:
:Weakly $\sigma$-Locally Compact $T_2$ Space is $T_4$
:Weakly Locally Compact $T_2$ Space is $T_{3 \frac 1 2}$
:Weakly $\sigma$-Locally Compact implies Weakly Locally Compact by definition.
:$T_{3 \frac 1 2}$ Space is $T_3$
{{ProofWanted|Some results still to be written.}}
{{qed}}
Theorem: Zero Locus of Set is Zero Locus of Generated Ideal

Let $k$ be a field.
Let $n \ge 1$ be a natural number.
Let $A = k \sqbrk {X_1, \ldots, X_n}$ be the ring of polynomial functions in $n$ variables over $k$.
Let $T \subseteq A$ be a subset, and $\map V T$ the zero locus of $T$.
Let $J = \ideal T$ be the ideal generated by $T$.
Then:
:$\map V T = \map V J$

Proof:
Let $x \in \map V T$, so $\map f x = 0$ for all $f \in T$.
By definition, $J$ is the set of linear combinations of elements of $T$ over $k$.
So any $g \in J$ is of the form:
:$g = k_1 t_1 + \cdots + k_r t_r$
with $k_i \in k$ and $t_i \in T$.
Therefore:
{{begin-eqn}}
{{eqn | l = \map g x
      | r = k_1 \map {t_1} x + \cdots + k_r \map {t_r} x
}}
{{eqn | r = 0
      | c = because $\map f x = 0$ for all $f \in T$
}}
{{end-eqn}}
Therefore:
:$x \in \map V J$
Conversely, if $x \in \map V J$, then $\map f x = 0$ for all $f \in J$.
But $T \subseteq J$, so in particular $\map f x = 0$ for all $f \in T$.
So:
:$x \in \map V t$
{{qed}}
Category:Algebraic Geometry
1f1osixrmvym9qbri63fnzm3emco22v
Theorem: Sequence of Implications of Compactness Properties in T3 Space

Let $P_1$ and $P_2$ be compactness properties.
Let:
:$P_1 \implies P_2$
mean:
:If a $T_3$ space $T$ satisfies property $P_1$, then $T$ also satisfies property $P_2$.
Then the following sequence of implications holds:
{|
|-
| align="center" | Second-Countable ||
| align="center" | $\implies$ ||
| align="center" | Lindelöf ||
|-
| align="center" | $\Big \Downarrow$ ||
| align="center" | ||
| align="center" | $\Big \Downarrow$ ||
|-
| align="center" | $\Downarrow$||
| align="center" | ||
| align="center" | Fully $T_4$ $\iff$ Paracompact ||
|-
| align="center" | $\Big \Downarrow$ ||
| align="center" | ||
| align="center" | $\Big \Downarrow$ ||
|-
| align="center" | $T_5$ ||
| align="center" | $\implies$ ||
| align="center" | $T_4$ ||
|-
|}

Proof:
The justifications are listed as follows:
:Second-Countable $T_3$ Space is $T_5$
:Second-Countable Space is Lindelöf
:$T_3$ Lindelöf Space is Paracompact
:$T_3$ Space is Fully $T_4$ iff Paracompact
:Fully $T_4$ Space is $T_4$
:$T_5$ Space is $T_4$
{{qed}}
Theorem: Finite Product of Sigma-Compact Spaces is Sigma-Compact

Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\set {\struct {S_i, \tau_i}: 1 \le i \le n}$ be a finite set of topological spaces.
Let $\ds \struct {S, \tau} = \prod_{i \mathop = 1}^n \struct {S_i, \tau_i}$ be the product space of $\set {\struct {S_i, \tau_i}: 1 \le i \le n}$.
Let each of $\struct {S_i, \tau_i}$ be $\sigma$-compact.
Then $\struct {S, \tau}$ is also $\sigma$-compact.

Proof:
Let $\struct {S_1, \tau_1}$ and $\struct {S_2, \tau_2}$ be $\sigma$-compact.
Then $\ds S_1 = \bigcup_{n \mathop = 1}^\infty C_n$ and $\ds S_2 = \bigcup_{n \mathop = 1}^\infty D_n$ where all the $C_n$ and $D_n$ are compact sets.
From Tychonoff's Theorem, the set $\KK = \set {C_n \times D_m: n, m \in \N}$ is a countable set of compact sets of $S_1 \times S_2$ with the product topology.
Consider $\tuple {a, b} \in S_1 \times S_2$.
Then $a \in S_1$ and $b \in S_2$.
Using the $\sigma$-compactness of $S_1$ and $S_2$, $\exists n, m$ such that $a \in C_n$ and $b \in D_m$.
So $\tuple {a, b} \in C_n \times D_m \in \KK$.
Thus, $\bigcup \KK = S_1 \times S_2$ and then $S_1 \times S_2$ is $\sigma$-compact.
Using induction on the number of factors of the product we get the result.
{{finish|Complete the inductive proof.}}
Theorem: Countable Product of Sequentially Compact Spaces is Sequentially Compact

Let $I$ be an indexing set with countable 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 $\struct {S, \tau}$ is also sequentially compact.

Proof:
{{Tidy|By the very nature of the proof, the indices are a pain to look at.}}
Let $\sequence { {\mathbf x}_n}$ be a sequence in $S$.
That is, for each $n \in \N$, ${\mathbf x}_n$ is a tuple $\family {x_n^i}_{i \mathop \in I}$ indexed by $I$.
We claim there is a strictly increasing sequence $\sequence {n_r}$ in $\N$ such that the subsequence $\sequence { {\mathbf x}_{n_r} }$ converges in $\struct {S, \tau}$.
By Sequence on Product Space Converges to Point iff Projections Converge to Projections of Point, it suffices to construct a strictly increasing sequence $\sequence {n_r}$ in $\N$ such that the subsequence $\sequence { \map {\pr_i} { {\mathbf x}_{n_r} } }$ converges in $\struct {S_i, \tau_i}$ for every $i \in I$.
Let $\set {i_0, i_1, \ldots}$ be an enumeration of $I$.
By induction, for each $m \in \N$ we construct a strictly increasing sequence $\sequence {n^m_r}$ in $\N$ and an element $x^{i_m} \in S_{i_m}$ as follows.
By sequential compactness of $\struct {S_{i_0}, \tau_{i_0} }$, there is a strictly increasing sequence $\sequence {n^0_r}$ in $\N$ such that the subsequence $\sequence { x^{i_0}_{n^0_r} }$ converges in $\struct {S_{i_0}, \tau_{i_0} }$ to some point $x^{i_0} \in S_{i_0}$.
Suppose inductively that the strictly increasing sequence $\sequence {n^m_r}$ has been defined for some $m \in \N$.
Applying sequential compactness of $\struct {S_{i_{m+1} }, \tau_{i_{m+1} } }$ to the sequence $\sequence { x^{i_{m+1} }_{n^m_r} }$, there is a subsequence $\sequence {n^{m+1}_r}$ of $\sequence {n^m_r}$ such that the subsequence $\sequence { x^{i_{m+1} }_{n^{m+1}_r} }$ converges in $\struct {S_{i_{m+1} }, \tau_{i_{m+1} } }$ to some point $x^{i_{m+1} } \in S_{i_{m+1} }$.
Moreover, by each $\sequence { x^{i_m}_{n^{m'}_r} }$ is a subsequence of $\sequence { x^{i_m}_{n^m_r} }$ whenever $m,m' \in \N$ such that $m' \ge m$.
Since $I$ is countable, consider separately the cases where $I$ is finite and $I$ is countably infinite.
Theorem: Finite Product of Weakly Locally Compact Spaces is Weakly Locally Compact

Let $n \in \Z_{\ge 0}$ be a (strictly) positive integer.
Let $\set {\struct {S_i, \tau_i}: 1 \le i \le n}$ be a finite set of topological spaces.
Let $\ds \struct {S, \tau} = \prod_{i \mathop = 1}^n \struct {S_i, \tau_i}$ be the product space of $\set {\struct {S_i, \tau_i}: 1 \le i \le n}$.
Let each of $\struct {S_i, \tau_i}$ be weakly locally compact.
Then $\struct {S, \tau}$ is also weakly locally compact.

Proof:
Pick any $\tuple {x_1, \dots, x_n} \in S$.
Since each of $\struct {S_i, \tau_i}$ is weakly locally compact, $x_i$ has a compact neighborhood.
That is, there is a compact set $N_i$ such that:
:$\exists U_i \in \tau_i: x_i \in U_i \subseteq N_i$
From Cartesian Product of Family of Subsets:
:$\ds \tuple {x_1, \dots, x_n} \in \prod_{i \mathop = 1}^n U_i \subseteq \prod_{i \mathop = 1}^n N_i$
By definition of product space:
:$\ds \prod_{i \mathop = 1}^n U_i \in \tau$
From Finite Topological Product of Compact Spaces:
:$\ds \prod_{i \mathop = 1}^n N_i$ is compact
Hence:
:$\ds \prod_{i \mathop = 1}^n N_i$ is a compact neighborhood of $\tuple {x_1, \dots, x_n}$.
As $\tuple {x_1, \dots, x_n}$ is arbitrary, $\struct {S, \tau}$ is weakly locally compact.
{{qed}}
Theorem: Countable Product of First-Countable Spaces is First-Countable

Let $I$ be an indexing set with countable 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 first-countable.
Then $\struct {S, \tau}$ is also first-countable.

Proof:
Let $x = \family {x_\alpha} \in S$.
Let $\BB_\alpha = \set {N^\alpha_i : i \in \N}$ be a countable local basis for $x_\alpha$ in $\struct {S_\alpha, \tau_\alpha}$ for each $\alpha \in I$.
Let $\pr_\alpha$ denote the projection of $S$ onto $S_\alpha$.
Let $\LL_\alpha = \set {\map {\pr_\alpha^{-1} } {N^\alpha_i}: N^\alpha_i \in B_\alpha}$.
Let $\KK_J = \ds \set {\bigcap_{\alpha \mathop \in J} L_\alpha : \forall \alpha \in J: L_\alpha \in \LL_\alpha}$ for $J \subset I$, $\size J < \infty$.
From Product Space Local Basis Induced from Factor Spaces Local Bases:
:$\ds \BB_x = \bigcup_{J \mathop \subset I \mathop , \size J \mathop < \infty} \KK_J$ forms a local basis of $x$ in the product space $\struct {S_\alpha, \tau_\alpha}$
Now, each of the $\LL_\alpha$'s is countable. 
Since the $\KK_J$'s can be identified with a finite product of countable sets, they are each countable
From Countable Union of Countable Sets is Countable, $\BB_x$ forms a countable local basis of $x$ in the product space $\struct {S_\alpha, \tau_\alpha}$.
The result follows.
{{qed}}
Theorem: Countable Product of Second-Countable Spaces is Second-Countable

Let $I$ be an indexing set with countable 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 second-countable.
Then $\struct {S, \tau}$ is also second-countable.

Proof:
Let $\BB_\alpha = \set {N^\alpha_i : i \in \N}$ be a countable basis for $\tau_\alpha$ for each $\alpha \in I$.
Let $\pr_\alpha$ denote the projection of $S$ onto $S_\alpha$.
Let $\LL_\alpha = \set {\map {\pr_\alpha^{-1} } {N^\alpha_i}: N^\alpha_i \in B_\alpha}$.
Let $\ds \KK_J = \set {\bigcap_{\alpha \mathop \in J} L_\alpha : \forall \alpha \in J: L_\alpha \in \LL_\alpha}$ for $J \subset I$, $\size J < \infty$.
From Product Space Basis Induced from Factor Space Bases:
:$\ds \BB = \bigcup_{J \mathop \subset I \mathop , \size J \mathop < \infty} \KK_J$ forms a basis of the product space.
Now, each of the $\LL_\alpha$'s is countable. 
Since the $\KK_J$'s can be identified with a finite product of countable sets, they are each countable
From Countable Union of Countable Sets is Countable, $\BB$ forms a countable basis of $S$.
{{qed}}
Theorem: Countable Product of Separable Spaces is Separable

Let $I$ be an indexing set with countable 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 separable.
Then $\struct {S, \tau}$ is also separable.

Proof:
For each $\alpha \in I$, let $D_\alpha$ be a countable everywhere dense subset of $\struct {S_\alpha, \tau_\alpha}$.
Let $D = \ds \prod_{\alpha \mathop \in I} D_\alpha$.
From Countable Union of Countable Sets is Countable, $D$ is a countable set of $S$.
From Natural Basis of Product Topology, the set $\BB$ of cartesian products of the form $\ds \prod_{\alpha \mathop \in I} U_\alpha$ where:
:for all $\alpha \in I : U_\alpha \in \tau_\alpha$
:for all but finitely many indices $\alpha : U_\alpha = X_\alpha$
is a basis for $\tau$.
From Analytic Basis Characterization of Denseness, to show that $D$ is an everywhere dense subset in $\struct {S, \tau}$ it is sufficient to show that:
:$\forall U \in \BB : U \cap D \ne \O$
Let $U = \ds \prod_{\alpha \mathop \in I} U_\alpha \in \BB$.
Then:
:$\forall \alpha \in I : U_\alpha \cap D_\alpha \ne \O$
For all $\alpha \in I$, let $x_\alpha \in U_\alpha \cap D_\alpha$.
Then:
{{begin-eqn}}
{{eqn | l = \family{x_\alpha}_{\alpha \in I}
      | o = \in 
      | r = \prod_{\alpha \mathop \in I} \paren {U_\alpha \cap D_\alpha}
}}
{{eqn | r = \paren {\prod_{\alpha \mathop \in I} U_\alpha} \bigcap \paren {\prod_{\alpha \mathop \in I}  D_\alpha}
      | c = General Case of Cartesian Product of Intersections
}}
{{eqn | r = U \cap D
}}
{{end-eqn}}
Thus:
:$U \cap D \ne \O$
The result follows.
{{qed}}
Theorem: Compactness Properties Preserved under Continuous Surjection

Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a continuous surjection.
If $T_A$ has one of the following properties, then $T_B$ has the same property:
:Compact Space
:$\sigma$-Compact Space
:Countable Compact Space
:Sequential Compact Space
:Lindelöf Space

Proof:
=== Proof for Compactness ===
{{:Compactness is Preserved under Continuous Surjection}}
Theorem: Projection from Product Topology is Open

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $T = \struct {T_1 \times T_2, \tau}$ be the product space of $T_1$ and $T_2$, where $\tau$ is the product topology on $S$.
Let $\pr_1: T \to T_1$ and $\pr_2: T \to T_2$ be the first and second projections from $T$ onto its factors.
Then both $\pr_1$ and $\pr_2$ are open.

Proof:
Let $U \in \tau$.
It follows from the definition of product topology that $U$ can be expressed as:
:$\ds U = \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} \map {\pr_{i_{k, j} }^{-1} } {U_{k, j} }$
where:
:$J$ is an arbitrary index set
:$n_j \in \N$
:$i_{k, j} \in I$
:$U_{k, j} \in \tau_{i_{k, j} }$.
For all $i' \in I$, define $V_{i', k, j} \in \tau_{i'}$ by:
:$V_{i', k, j} = \begin {cases} U_{k, j} & : i' = i_{k, j} \\ S_{i'} & : i' \ne i_{k, j} \end {cases}$
For all $i \in I$, we have:
{{begin-eqn}}
{{eqn | l = \map {\pr_i} U
      | r = \bigcup_{j \mathop \in J} \map {\pr_i} {\bigcap_{k \mathop = 1}^{n_j} \map {\pr_{i_{k,j} }^{-1} } { U_{k,j} } }
      | c = Image of Union under Relation: Family of Sets
}}
{{eqn | r = \bigcup_{j \mathop \in J} \map {\pr_i} {\bigcap_{k \mathop = 1}^{n_j} \prod_{i' \mathop \in I} V_{i', k, j} }
      | c = {{Defof|Projection (Mapping Theory)|Projection}}
}}
{{eqn | r = \bigcup_{j \mathop \in J} \map {\pr_i} {\prod_{i' \mathop \in I} \bigcap_{k \mathop = 1}^{n_j} V_{i', k, j} }
      | c = Cartesian Product of Intersections: General Case
}}
{{eqn | r = \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} V_{i,k,j}
      | c = {{Defof|Projection (Mapping Theory)|Projection}}
}}
{{end-eqn}}
As:
:$\ds \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} V_{i, k, j} \in \tau_i$
it follows that $\pr_i$ is open.
Theorem: Projection from Product Topology is Open

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $T = \struct {T_1 \times T_2, \tau}$ be the product space of $T_1$ and $T_2$, where $\tau$ is the product topology on $S$.
Let $\pr_1: T \to T_1$ and $\pr_2: T \to T_2$ be the first and second projections from $T$ onto its factors.
Then both $\pr_1$ and $\pr_2$ are open.

Proof:
Let $U \in \tau$.
It follows from the definition of product topology that $U$ can be expressed as:
:$\ds U = \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} \map {\pr_{i_{k,j} }^{-1}} { U_{k,j} }$
where $J$ is an arbitrary index set, $n_j \in \N$, $i_{k, j} \in \set {1, 2}$, and $U_{k, j} \in \tau_{i_{k, j} }$.
For all $i \in \set {1, 2}$, define $V_{i, k, j} \in \tau_i$ by:
:$V_{i, k, j} = \begin {cases} U_{k, j} & : i = i_{k, j} \\ S_i & : i \ne i_{k, j} \end {cases}$
By definition of projection:
:$\map {\pr_{i_{k, j} }^{-1} } {U_{k, j} } = \ds \prod_{i \mathop = 1}^2 V_{i, k, j}$
{{WLOG}}, we consider $\pr_1$.
We have:
{{begin-eqn}}
{{eqn | l = \map {\pr_1} U
      | r = \bigcup_{j \mathop \in J} \map {\pr_1} {\bigcap_{k \mathop = 1}^{n_j} \map {\pr_{i_{k, j} }^{-1} } {U_{k, j} } }
      | c = Image of Union under Relation: Family of Sets
}}
{{eqn | r = \bigcup_{j \mathop \in J} \map {\pr_1} {\bigcap_{k \mathop = 1}^{n_j} \prod_{i \mathop = 1}^2 V_{i, k, j} }
}}
{{eqn | r = \bigcup_{j \mathop \in J} \map {\pr_1} {\prod_{i \mathop = 1}^2 \bigcap_{k \mathop = 1}^{n_j} V_{i, k, j} }
      | c = Cartesian Product of Intersections
}}
{{eqn | r = \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} V_{1, k, j}
      | c = {{Defof|Projection (Mapping Theory)|Projection}}
}}
{{end-eqn}}
As $\ds \bigcup_{j \mathop \in J} \bigcap_{k \mathop = 1}^{n_j} V_{1, k, j} \in \tau_1$, it follows that $\pr_1$ is open.
The proof for $\pr_2$ is the same, {{mutatis}}.
Theorem: Weak Local Compactness is Preserved under Open Continuous Surjection

Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a continuous mapping which is also an open mapping and a surjection.
If $T_A$ is weakly locally compact, then $T_B$ is also weakly locally compact.

Proof:
Let $\phi$ be a mapping which is surjective, continuous and open.
Let $T_A$ be weakly locally compact.
Take $b \in S_B$.
Let $V$ be a neighbourhood of $b$.
Since $\phi$ is surjective:
:$\forall y \in S_B: \exists x \in S_A: x \in \map {\phi^{-1} } y$
From the weak local compactness of $T_A$ and the continuity of $\phi$, there exists a compact neighbourhood $K$ of $x$ such that $\phi \sqbrk K \subseteq V$.
Since $K$ is a neighbourhood of $x$, then $x \in K^\circ$ and $y \in \phi \sqbrk {K^\circ} \subseteq \phi \sqbrk K$, where $K^\circ$ is the interior of $K$.
$\phi$ is an open mapping and $K^\circ$ is an open set, so $\phi \sqbrk {K^\circ}$ is also open.
Finally we get that $y \in \phi \sqbrk K \subseteq V$, where $\phi \sqbrk K$ is a compact neighbourhood.
Thus $T_B$ is weakly locally compact.
{{qed}}
Theorem: Compactness Properties Preserved under Projection Mapping

Let $I$ be an indexing set with countable 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 $\pr_\alpha: \struct {S, \tau} \to \struct {S_\alpha, \tau_\alpha}$ be the projection on the $\alpha$ coordinate.
Then $\pr_\alpha$ preserves the following compactness properties.
That is, if $\struct {S, \tau}$ has one of the following properties, then each of $\struct {S_\alpha, \tau_\alpha}$ has the same property:
:Compact Space
:$\sigma$-Compact Space
:Countably Compact Space
:Sequentially Compact Space
:Lindelöf Space
:Locally Compact Space
:Weakly Locally Compact Space
:Paracompact Space

Proof:
First note that Projection from Product Topology is Continuous.
Also note that Projection is Surjection.
It follows from Compactness Properties Preserved under Continuous Surjection that:
:Compact Space
:$\sigma$-Compact Space
:Countably Compact Space
:Sequentially Compact Space
:Lindelöf Space
are all preserved under projections.
Next note that Projection from Product Topology is Open.
It follows from Local Compactness is Preserved under Open Continuous Surjection that local compactness is preserved under projections.
It follows from Weak Local Compactness is Preserved under Open Continuous Surjection that weak local compactness is preserved under projections.
The final result is that Paracompactness is Preserved under Projections.
{{finish|Finish these results off.}}
Theorem: Countability Properties Preserved under Projection Mapping

Let $\sequence {\struct {S_\alpha, \tau_\alpha} }$ be a sequence of topological spaces.
Let $\ds \struct {S, \tau} = \prod \struct {S_\alpha, \tau_\alpha}$ be the product space of $\sequence {\struct {S_\alpha, \tau_\alpha} }$.
Let $\pr_\alpha: \struct {S, \tau} \to \struct {S_\alpha, \tau_\alpha}$ denote the projection on the $\alpha$ coordinate.
Then $\pr_\alpha$ preserves the following countability properties.
That is, if $\struct {S, \tau}$ has one of the following properties, then each of $\struct {S_\alpha, \tau_\alpha}$ has the same property.
:Separability
:First-Countability
:Second-Countability

Proof:
First note that Projection from Product Topology is Continuous.
It follows from Continuous Image of Separable Space is Separable that separability is preserved under projections.
Next note that Projection from Product Topology is Open.
It follows from Countability Axioms Preserved under Open Continuous Surjection that:
:First-Countability
:Second-Countability
are preserved under projections.
{{qed}}
Theorem: Countability Axioms Preserved under Open Continuous Surjection

Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a surjective open mapping which is also continuous.
If $T_A$ has one of the following properties, then $T_B$ has the same property:
:First-Countability
:Second-Countability

Proof:
=== Proof for First-Countability ===
{{:First-Countability is Preserved under Open Continuous Surjection}}
Theorem: Connectedness of Points is Equivalence Relation

Let $T = \struct {S, \tau}$ be a topological space.
Let $a \sim b $ denote the relation:
:$a \sim b \iff a$ is connected to $b$
where $a, b \in S$.
Then $\sim$ is an equivalence relation.

Proof:
Checking in turn each of the criteria for equivalence:
Theorem: Atlas Belongs to Unique Differentiable Structure

Let $M$ be a locally Euclidean space of dimension $d$.
Let $\AA$ be an atlas on $M$.
Then there exists a unique differentiable structure $\FF$ on $M$ with $\AA \in \FF$.

Proof:
Let $\FF$ be the equivalence class of $\AA$ under the relation of compatibility.
By Compatibility of Atlases is Equivalence Relation, this is indeed an equivalence relation.
By definition we have $\AA \in \FF$.
By Relation Partitions Set iff Equivalence, $\FF$ is an element of the partition of equivalence classes.
By definition, the elements of a partition are pairwise disjoint.
Therefore if $\GG \ne \FF$ is an element of the partition, we must have:
:$\AA \notin \GG$
Therefore $\AA$ belongs to exactly one differentiable structure.
{{qed}}
Category:Manifolds
Category:Topological Manifolds
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