hcju/proofwikiqa huggingface dataset:
By definition of the hyperoperation sequence: :$\forall n, x, y \in \Z_{\ge 0}: H_n \left({x, y}\right) = \begin{cases} y + 1 & : n = 0 \\ x & : n = 1, y = 0 \\ 0 & : n = 2, y = 0 \\ 1 & : n > 2, y = 0 \\ H_{n - 1} \left({x, H_n \left({x, y - 1}\right)}\right) & : n > 0, y > 0 \end{cases}$ Thus the $1$st hyperoperation is defined as: :$\forall x, y \in \Z_{\ge 0}: H_1 \left({x, y}\right) = \begin{cases} x & : y = 0 \\ H_0 \left({x, H_1 \left({x, y - 1}\right)}\right) & : y > 0 \end{cases}$ From Zeroth Hyperoperation is Successor Function: :$(1): \quad \forall x, y \in \Z_{\ge 0}: H_1 \left({x, y}\right) = \begin{cases} x & : y = 0 \\ H_1 \left({x, y - 1}\right) + 1 & : y > 0 \end{cases}$ The proof proceeds by induction. For all $y \in \Z_{\ge 0}$, let $P \left({y}\right)$ be the proposition: :$\forall x \in \Z_{\ge 0}: H_1 \left({x, y}\right) = x + y$ === Basis for the Induction === $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = H_1 \left({x, 0}\right)       | r = x       | c = from $(1)$ }} {{eqn | r = x + 0       | c =  }} {{end-eqn}} Thus $P \left({0}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$\forall x \in \Z_{\ge 0}: H_1 \left({x, k}\right) = x + k$ from which it is to be shown that: :$\forall x \in \Z_{\ge 0}: H_1 \left({x, k + 1}\right) = x + k + 1$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = H_1 \left({x, k + 1}\right)       | r = H_1 \left({x, \left({k + 1}\right) - 1}\right) + 1       | c = from $(1)$ }} {{eqn | r = H_1 \left({x, k}\right) + 1       | c =  }} {{eqn | r = \left({x + k}\right) + 1       | c = Induction Hypothesis }} {{eqn | r = x + k + 1       | c =  }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall x, y \in \Z_{\ge 0}: H_1 \left({x, y}\right) = x + y$ {{qed}} Category:Hyperoperation rr0qu34x9pl40qm570lji32pxeb72sr
By definition of the hyperoperation sequence: :$\forall n, x, y \in \Z_{\ge 0}: H_n \left({x, y}\right) = \begin{cases} y + 1 & : n = 0 \\ x & : n = 1, y = 0 \\ 0 & : n = 2, y = 0 \\ 1 & : n > 2, y = 0 \\ H_{n - 1} \left({x, H_n \left({x, y - 1}\right)}\right) & : n > 0, y > 0 \end{cases}$ Thus the $2$nd hyperoperation is defined as: :$\forall x, y \in \Z_{\ge 0}: H_2 \left({x, y}\right) = \begin{cases} 0 & : y = 0 \\ H_1 \left({x, H_2 \left({x, y - 1}\right)}\right) & : y > 0 \end{cases}$ From First Hyperoperation is Addition Operation: :$(1): \quad \forall x, y \in \Z_{\ge 0}: H_2 \left({x, y}\right) = \begin{cases} 0 & : y = 0 \\ x + H_2 \left({x, y - 1}\right) & : y > 0 \end{cases}$ The proof proceeds by induction. For all $y \in \Z_{\ge 0}$, let $P \left({y}\right)$ be the proposition: :$\forall x \in \Z_{\ge 0}: H_2 \left({x, y}\right) = x \times y$ === Basis for the Induction === $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = H_2 \left({x, 0}\right)       | r = 0       | c = from $(1)$ }} {{eqn | r = x \times 0       | c =  }} {{end-eqn}} Thus $P \left({0}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$\forall x \in \Z_{\ge 0}: H_2 \left({x, k}\right) = x \times k$ from which it is to be shown that: :$\forall x \in \Z_{\ge 0}: H_2 \left({x, k + 1}\right) = x \times \left({k + 1}\right)$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = H_2 \left({x, k + 1}\right)       | r = x + H_2 \left({x, \left({k + 1}\right) - 1}\right)       | c = from $(1)$ }} {{eqn | r = x + H_2 \left({x, k}\right)       | c =  }} {{eqn | r = x + \left({x \times k}\right)       | c = Induction Hypothesis }} {{eqn | r = x \times \left({k + 1}\right)       | c =  }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall x, y \in \Z_{\ge 0}: H_2 \left({x, y}\right) = x \times y$ {{qed}} Category:Hyperoperation q94fl08d4aqw0fa3kd18aojxjnbrwtx
By definition of the hyperoperation sequence: :$\forall n, x, y \in \Z_{\ge 0}: H_n \left({x, y}\right) = \begin{cases} y + 1 & : n = 0 \\ x & : n = 1, y = 0 \\ 0 & : n = 2, y = 0 \\ 1 & : n > 2, y = 0 \\ H_{n - 1} \left({x, H_n \left({x, y - 1}\right)}\right) & : n > 0, y > 0 \end{cases}$ Thus the $3$rd hyperoperation is defined as: :$\forall x, y \in \Z_{\ge 0}: H_3 \left({x, y}\right) = \begin{cases} 1 & : y = 0 \\ H_2 \left({x, H_3 \left({x, y - 1}\right)}\right) & : y > 0 \end{cases}$ From Second Hyperoperation is Multiplication Operation: :$(1): \quad \forall x, y \in \Z_{\ge 0}: H_3 \left({x, y}\right) = \begin{cases} 1 & : y = 0 \\ x \times H_3 \left({x, y - 1}\right) & : y > 0 \end{cases}$ The proof proceeds by induction. For all $y \in \Z_{\ge 0}$, let $P \left({y}\right)$ be the proposition: :$\forall x \in \Z_{\ge 0}: H_3 \left({x, y}\right) = x^y$ === Basis for the Induction === $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = H_3 \left({x, 0}\right)       | r = 1       | c = from $(1)$ }} {{eqn | r = x^0       | c = {{Defof|Integer Power}} }} {{end-eqn}} Thus $P \left({0}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$\forall x \in \Z_{\ge 0}: H_3 \left({x, k}\right) = x^k$ from which it is to be shown that: :$\forall x \in \Z_{\ge 0}: H_3 \left({x, k + 1}\right) = x^{k + 1}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = H_3 \left({x, k + 1}\right)       | r = x \times H_3 \left({x, \left({k + 1}\right) - 1}\right)       | c = from $(1)$ }} {{eqn | r = x \times H_3 \left({x, k}\right)       | c =  }} {{eqn | r = x \times \left({x^k}\right)       | c = Induction Hypothesis }} {{eqn | r = x^{k + 1}       | c = {{Defof|Integer Power}} }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall x, y \in \Z_{\ge 0}: H_3 \left({x, y}\right) = x^y$ {{qed}} Category:Hyperoperation ik9u4f67pheqyzzfhqs87wuipr3fv4e
By definition of the hyperoperation sequence: :$\forall n, x, y \in \Z_{\ge 0}: H_n \left({x, y}\right) = \begin{cases} y + 1 & : n = 0 \\ x & : n = 1, y = 0 \\ 0 & : n = 2, y = 0 \\ 1 & : n > 2, y = 0 \\ H_{n - 1} \left({x, H_n \left({x, y - 1}\right)}\right) & : n > 0, y > 0 \end{cases}$ Thus the $4$th hyperoperation is defined as: :$\forall x, y \in \Z_{\ge 0}: H_4 \left({x, y}\right) = \begin{cases} 1 & : y = 0 \\ H_3 \left({x, H_4 \left({x, y - 1}\right)}\right) & : y > 0 \end{cases}$ From Third Hyperoperation is Integer Power Operation: :$(1): \quad \forall x, y \in \Z_{\ge 0}: H_4 \left({x, y}\right) = \begin{cases} 1 & : y = 0 \\ x \uparrow H_3 \left({x, y - 1}\right) & : y > 0 \end{cases}$ The proof proceeds by induction. For all $y \in \Z_{\ge 0}$, let $P \left({y}\right)$ be the proposition: :$\forall x \in \Z_{\ge 0}: H_4 \left({x, y}\right) = x \uparrow \uparrow y$ === Basis for the Induction === $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = H_4 \left({x, 0}\right)       | r = 1       | c = from $(1)$ }} {{eqn | r = x \uparrow \uparrow 0       | c = {{Defof|Tetration}} }} {{end-eqn}} Thus $P \left({0}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$\forall x \in \Z_{\ge 0}: H_4 \left({x, k}\right) = x \uparrow \uparrow k$ from which it is to be shown that: :$\forall x \in \Z_{\ge 0}: H_4 \left({x, k + 1}\right) = x \uparrow \uparrow \left({k + 1}\right)$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = H_4 \left({x, k + 1}\right)       | r = x \uparrow H_4 \left({x, \left({k + 1}\right) - 1}\right)       | c = from $(1)$ }} {{eqn | r = x \uparrow H_4 \left({x, k}\right)       | c =  }} {{eqn | r = x \uparrow \left({x \uparrow \uparrow k}\right)       | c = Induction Hypothesis }} {{eqn | r = x \uparrow \uparrow \left({k + 1}\right)       | c = {{Defof|Tetration}} }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall x, y \in \Z_{\ge 0}: H_3 \left({x, y}\right) = x \uparrow \uparrow y$ {{qed}} Category:Hyperoperation 6isg3bn6b55qkkohlu8gqkztbyknyvd
Let $n$ be the number of digits in $9^{9^9}$ From Number of Digits in Number: :$n = 1 + \floor {\map {\log_{10} } {9^{9^9} } }$ where $\floor {\ldots}$ denotes the floor function. Then: {{begin-eqn}} {{eqn | l = \map {\log_{10} } {9^{9^9} }       | o = \approx       | r = 369 \, 693 \, 099 \cdotp 63157 \, 03685 \, 87876 \, 1       | c = Largest Integer Expressible by 3 Digits: Logarithm Base 10 }} {{eqn | ll= \leadsto       | l = n       | r = 1 + \floor {369 \, 693 \, 099 \cdotp 63157 \, 03685 \, 87876 \, 1}       | c =  }} {{eqn | r = 1 + 369 \, 693 \, 099       | c = {{Defof|Floor Function}} }} {{eqn | r = 369 \, 693 \, 100       | c =  }} {{end-eqn}} {{qed}}
Let $S$ be a set containing more than one element. Let $\tau = \set {S, \O}$ be the indiscrete topology on $S$. Let $x \in S$. Let $H = S \setminus \set x$. Then $H$ is a proper subset of $S$. Then from Subset of Indiscrete Space is Compact and Sequentially Compact, the subspace induced by $\tau$ on $H$ is a compact subspace of $\struct {S, \tau}$. But from Closed Sets in Indiscrete Topology, the only closed subsets of $\struct {S, \tau}$ are $S$ and $\O$. Thus $H$ is a compact subspace of $\struct {S, \tau}$ which is not closed in $S$. {{qed}}
It needs to be demonstrated that there exists a separable topological space which has a subspace which is closed but not separable. Consider an uncountable particular point space $T = \struct {S, \tau_p}$. From Particular Point Space is Separable, $T$ is separable. By definition, the particular point $p$ is an open point of $T$. Thus the subset $S \setminus \set p$ is by definition closed in $T$. But from Separability in Uncountable Particular Point Space, $S \setminus \set p$ is not separable. Thus by Proof by Counterexample, separability is not weakly hereditary. {{qed}}
=== Definition 1 implies Definition 3 === Let $\mathcal F$ be an ultrafilter on $S$ by definition 1. Thus $\mathcal F \subseteq \mathcal P \left({S}\right)$ is a filter on $S$ which fulfills the condition: :whenever $\mathcal G$ is a filter on $S$ and $\mathcal F \subseteq \mathcal G$ holds, then $\mathcal F = \mathcal G$. Let $A \subseteq S$. {{AimForCont}} $A \notin \mathcal F$ and $\complement_S \left({A}\right) \notin \mathcal F$. Consider the set $\mathcal B := \left\{{A \cap V: V \in \mathcal F}\right\}$. This is a basis of a filter $\mathcal G$ on $S$, for which $\mathcal F \subseteq \mathcal G$ holds. Let $U \in \mathcal F$. {{begin-eqn}} {{eqn | l = \complement_S \left({A}\right)       | o = \notin       | r = \mathcal F       | c = by hypothesis }} {{eqn | ll= \leadsto       | l = U       | o = \not \subseteq       | r = \complement_S \left({A}\right)       | c = {{Defof|Filter on Set/Definition 1|Filter}}: Axiom $(4)$: $U \subseteq \complement_S \left({A}\right) \subseteq S \implies \complement_S \left({A}\right) \in \mathcal F$ }} {{eqn | ll= \leadsto       | l = A \cap U       | o = \ne       | r = \varnothing       | c = Empty Intersection iff Subset of Complement }} {{eqn | ll= \leadsto       | l = A \cap U       | o = \in       | r = \mathcal G       | c = Definition of $\mathcal G$ by way of $\mathcal B$ }} {{eqn | ll= \leadsto       | l = A       | o = \in       | r = \mathcal G       | c = {{Defof|Filter on Set/Definition 1|Filter}}: Axiom $(4)$: $A \cap U \subseteq A \subseteq S \implies A \in \mathcal G$ }} {{end-eqn}} But we have: :$A \notin \mathcal F$ and so: :$\mathcal F \subsetneq \mathcal G$ Thus $\mathcal F$ is not an ultrafilter by Definition 1. This contradicts our initial hypothesis. Hence either: :$A \in \mathcal F$ or: :$\complement_S \left({A}\right) \in \mathcal F$ and so $\mathcal F$ fulfils the conditions to be an ultrafilter by Definition 3. {{qed|lemma}} === Definition 3 implies Definition 1 === Let $\mathcal F$ be an ultrafilter on $S$ by definition 3. That is, $\mathcal F \subseteq \mathcal P \left({S}\right)$ is a filter on $S$ which fulfills the condition: :for any $A \subseteq S$ either $A \in \mathcal F$ or $\complement_S \left({A}\right) \in \mathcal F$ holds. Let $\mathcal G$ be a filter on $X$ such that $\mathcal F \subseteq \mathcal G$. {{AimForCont}} $\mathcal F \subsetneq \mathcal G$. Then there exists $A \in \mathcal G \setminus \mathcal F$. By definition of filter, $\varnothing \notin \mathcal G$. But from Intersection with Relative Complement is Empty: :$A \cap \complement_S \left({A}\right)$ and so: :$\complement_S \left({A}\right) \notin \mathcal G$ By hypothesis: :$\mathcal F \subsetneq \mathcal G$ and so: :$\complement_S \left({A}\right) \notin \mathcal F$ Therefore neither $A \in \mathcal F$ nor $\complement_S \left({A}\right) \in \mathcal F$. This contradicts our assumption: :for any $A \subseteq S$ either $A \in \mathcal F$ or $\complement_S \left({A}\right) \in \mathcal F$ holds. Thus: : $\mathcal F = \mathcal G$ and so $\mathcal F$ fulfils the conditions to be an ultrafilter by Definition 1. {{qed}}
=== Necessary Condition === Suppose for every set of $\BB'$, there is a set of $\BB$ subset to it. Pick any $U \in \FF'$. Then from definition of a basis: :$\exists V' \in \BB': V' \subseteq U$ By our assumption: :$\exists V \in \BB: V \subseteq V' \subseteq U \subseteq S$ By definition of a filter: :$U \in \FF$. Hence we have $\FF' \subseteq \FF$. {{qed|lemma}} === Sufficient Condition === We prove the contrapositive. Suppose there is some set of $\BB'$ with no set of $\BB$ subset to it. Let $U$ be such a set. Then $U \in \BB' \subseteq \FF'$. From our assumption: :$\not \exists V \in \BB: V \subseteq U$ From the definition of a basis: :$U \notin \FF$ This gives $\FF' \not \subseteq \FF$. {{qed}}
Proof by Counterexample: Let $T$ be the topological space consisting of the double pointed topology on the countable complement topology on an uncountable set. From Double Pointed Countable Complement Topology fulfils no Separation Axioms, we have that $T$ satisfies none of the Tychonoff separation axioms. {{qed}}
Let $T$ be the overlapping interval space. From Overlapping Interval Space fulfils no Separation Axioms but $T_0$, we have that $T$ satisfies none of the Tychonoff separation axioms except for the $T_0$ (Kolmogorov) axiom. {{qed}}
Proof by Counterexample: Let $T$ be the finite complement topology on a countable space. From Finite Complement Space is $T_1$, $T$ is a $T_1$ (Fréchet) space. Hence from $T_1$ Space is $T_0$ Space, also a $T_0$ (Kolmogorov) space. The rest of the result follows from: :Finite Complement Space is not $T_2$ and: :Finite Complement Space is not $T_3$, $T_4$ or $T_5$ {{qed}}
Let $T$ be an irrational slope topological space. From Irrational Slope Space is $T_2$, $T$ is a $T_2$ (Hausdorff) space. Hence from $T_2$ Space is $T_1$ Space and $T_1$ Space is $T_0$ Space, also a $T_0$ (Kolmogorov) space and a $T_1$ (Fréchet) space. The rest of the result follows from: :Irrational Slope Space is not $T_3$, $T_4$ or $T_5$. {{qed}}
Proof by Counterexample: Let $T$ be the topological space consisting of the double pointed topology on the Tychonoff corkscrew topology. From Double Pointed Tychonoff Corkscrew fulfils no Separation Axioms but $T_3$, we have that $T$ satisfies none of the Tychonoff separation axioms except for the $T_3$ axiom. {{qed}}
Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$. Let $D = \struct {\set {0, 1}, \vartheta}$ be the indiscrete topology on two points. Let $T \times D$ be the double pointed topology on $T$. Let $\paren {T \times D}^*_{\bar p}$ be the open extension topology on $S \times \set {0, 1} \cup \set p$ where $p \notin S \times \set {0, 1}$. From Open Extension of Double Pointed Countable Complement Topology is $T_4$ Space, we have that $\paren {T \times D}^*_{\bar p}$ satisfies none of the Tychonoff separation axioms except for the $T_4$ axiom. {{qed}}
Let $T$ be the Tychonoff plank. From Tychonoff Plank is Compact, $T$ is a compact topological space. From Tychonoff Plank is Hausdorff, $T$ is a $T_2$ (Hausdorff) space. From Tychonoff Plank is not Completely Normal, $T$ is not a completely normal space. From $T_2$ Space is $T_1$ Space, $T$ is also a $T_1$ (Fréchet) space. As $T$ is a $T_1$ space, it follows by definition of completely normal space that $T$ is not a $T_5$ space. {{qed}}
Let $T$ be a Tychonoff corkscrew. From Tychonoff Corkscrew is Regular, $T$ is a regular space. From Tychonoff Corkscrew is not Completely Regular, $T$ is not a Tychonoff space. Hence the result. {{qed}}
Let $T$ be a Niemytzki's tangent disc space. From Niemytzki's Tangent Disc Space is Tychonoff, $T$ is a Tychonoff space. From Niemytzki's Tangent Disc Space is not Normal, $T$ is not a normal space. Hence the result. {{qed}}
Let $T$ be a Hjalmar Ekdal space. From Hjalmar Ekdal Space is $T_4$, $T$ is a $T_4$ space. From Hjalmar Ekdal Space is not $T_{3 \frac 1 2}$ Space, $T$ is not a $T_{3 \frac 1 2}$ space. Hence the result. {{qed}}
Let $T$ be a Tychonoff plank. From Tychonoff Plank is Normal, $T$ is a normal space. From Tychonoff Plank is not Completely Normal, $T$ is not a completely normal space. Hence the result. {{qed}}
Let $T$ be a half-disc space. From Half-Disc Space is Completely Hausdorff, $T$ is a completely Hausdorff space. From Half-Disc Space is not Regular, $T$ is not a regular space. Hence the result. {{qed}}
Let $T$ be an irrational slope topological space. From Irrational Slope Space is $T_2$, $T$ is a $T_2$ (Hausdorff) space. From Irrational Slope Space is not Completely Hausdorff, $T$ is not a completely Hausdorff space. {{qed}}
Let $\struct {\R, \tau_1}$ be the set of real numbers under the usual (Euclidean) topology. Let $\struct {\R, \tau_2}$ be the indiscrete rational extension of $\struct {\R, \tau_1}$. From Metric Space fulfils all Separation Axioms, $\struct {\R, \tau_1}$ is: :$T_3$ Space :Regular Space :$T_4$ Space :Completely Regular Space :$T_5$ Space :Normal Space :Completely Normal Space But we have: :Indiscrete Rational Extension of Real Number Line is not $T_3$ Space :Indiscrete Rational Extension of Real Number Line is not $T_4$ Space :Indiscrete Rational Extension of Real Number Line is not $T_5$ Space By definition, $\struct {\R, \tau_2}$ is an expansion of $\struct {\R, \tau_1}$. Hence the result. {{qed}}
Let $T$ be the Sorgenfrey line. Let $T' = T \times T$ be Sorgenfrey's half-open square topology. From Sorgenfrey Line satisfies all Separation Axioms, $T$ is a completely normal space. From Sorgenfrey's Half-Open Square Topology is Not Normal, $T'$ is not a normal space. Hence the result. {{qed}}
A regular space is a topological space which is both a $T_0$ (Kolmogorov) space and a $T_3$ space. Hence from $T_0$ Property is Hereditary and $T_3$ Property is Hereditary it follows that the property of being a regular space is also hereditary. A Tychonoff (completely regular) space is a topological space which is both a $T_0$ (Kolmogorov) space and a $T_3 \frac 1 2$ space. Hence from $T_0$ Property is Hereditary and $T_3 \frac 1 2$ Property is Hereditary it follows that the property of being a Tychonoff (completely regular) space is also hereditary. A completely normal space is a topological space which is both a $T_1$ (Fréchet) space and a $T_5$ space. Hence from $T_1$ Property is Hereditary and $T_5$ Property is Hereditary it follows that the property of being a completely normal space is also hereditary. {{qed}}
Let $T$ be an uncountable Fort space. From Fort Space is Completely Normal, $T$ is a completely normal space. From Uncountable Fort Space is not Perfectly Normal, $T$ is not a perfectly normal space. Hence the result. {{qed}}
Let $T$ be a simplified Arens square. From Simplified Arens Square is Semiregular, $T$ is a semiregular space. From Simplified Arens Square is not $T_3$, $T$ is not a $T_3$ space. Hence the result. {{qed}}
Let $T$ be a simplified Arens square. From Simplified Arens Square is Semiregular, $T$ is a semiregular space. From Simplified Arens Square is not Completely Hausdorff, $T$ is not a $T_3$ space. Hence the result. {{qed}}
Let $T$ be an Arens square. From Arens Square is Semiregular, $T$ is a semiregular space. From Arens Square is not Urysohn Space, $T$ is not a Urysohn space. Hence the result. {{qed}}
Let $T = \struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology. From Real Number Line is Sigma-Compact, $T$ is a $\sigma$-compact space. From Real Number Line is not Countably Compact, $T$ is not a countably compact space. The result follows from Compact Space is Countably Compact. {{qed}}
Let $T$ be the Sorgenfrey line. From Sorgenfrey Line is Lindelöf, $T$ is a Lindelöf space. From Sorgenfrey Line is not $\sigma$-Compact, $T$ is not a $\sigma$-compact space. Hence the result. {{qed}}
Let $T = \mathbb I^{\mathbb I}$ be the uncountable Cartesian product of the closed unit interval under the usual (Euclidean) topology. From Uncountable Cartesian Product of Closed Unit Interval is Compact, $T$ is a compact space. From Uncountable Cartesian Product of Closed Unit Interval is not Sequentially Compact, $T$ is not a sequentially compact space. Hence the result. {{qed}}
Let $T := \hointr 0 \Omega \times \mathbb I^{\mathbb I}$ be the Cartesian product of $\hointr 0 \Omega$ under the interval topology, with the uncountable Cartesian product of the closed unit interval under the usual (Euclidean) topology. From Product of Uncountable Ordinal Interval Space with Uncountable Cartesian Product of Closed Unit Interval is Countably Compact, $T$ is a countably compact space. From Product of Uncountable Ordinal Interval Space with Uncountable Cartesian Product of Closed Unit Interval is not Sequentially Compact, $T$ is not a sequentially compact space. Hence the result. {{qed}}
Let $T$ be the deleted integer topological space. From Deleted Integer Topology is Weakly Countably Compact, $T$ is a weakly countably compact space. From Deleted Integer Topology is not Countably Compact, $T$ is not a countably compact space. Hence the result. {{qed}}
Let $T$ be an infinite particular point space. From Particular Point Space is Pseudocompact, $T$ is a pseudocompact space. From Infinite Particular Point Space is not Weakly Countably Compact, $T$ is not a weakly countably compact space. From Countably Compact Space is Weakly Countably Compact, $T$ is not a countably compact space. Hence the result. {{qed}}
Let $T$ be the nested interval topological space. From Nested Interval Topology is Weakly Locally Compact, $T$ is a weakly locally compact space. From Nested Interval Topology is not Strongly Locally Compact, $T$ is not a strongly locally compact space. Hence the result. {{qed}}
Let $T$ be an uncountable discrete space. From Discrete Space is Strongly Locally Compact, $T$ is a strongly locally compact space. From Uncountable Discrete Space is not $\sigma$-Compact, $T$ is not a weakly $\sigma$-compact space. By definition, it follows that $T$ is not a weakly $\sigma$-locally compact space. Hence the result. {{qed}}
Let $T$ be the nested interval topological space. From Nested Interval Topology is Weakly $\sigma$-Locally Compact, $T$ is a weakly $\sigma$-locally compact space. From Nested Interval Topology is not Strongly Locally Compact, $T$ is not a strongly locally compact space. Hence the result. {{qed}}
Let $T$ be a Sorgenfrey line. From Sorgenfrey Line is First-Countable, $T$ is a first-countable space. From Sorgenfrey Line is not Second-Countable, $T$ is not a second-countable space. Hence the result. {{qed}}