hcju/proofwikiqa huggingface dataset:
Let $\struct {D, +, \times}$ be a Euclidean domain whose zero is $0$ and whose Euclidean valuation is $\nu$. We need to show that every ideal of $\struct {D, +, \times}$ is a principal ideal. Let $U$ be an ideal of $\struct {D, +, \times}$ such that $U \ne \set 0$. Let $d \in U$ such that $d \ne 0$ and $\map \nu d$ is as small as possible for elements of $U$. By definition, $\nu$ is defined as $\nu : D \setminus \set 0 \to \N$, so the codomain of $\nu$ is a subset of the natural numbers. By the Well-Ordering Principle, such an element $d$ exists as an element of the preimage of the least member of the image of $U$. <!--{{explain|Need to ensure that the Well-Ordering Principle can be applied here.}}--> Let $a \in U$. Let us write $a = d q + r$ where either $r = 0$ or $\map \nu r < \map \nu d$. Then $r = a - d q$ and so $r \in U$. Suppose $r \ne 0$. That would mean $\map \nu r < \map \nu d$ contradicting $d$ as the element of $U$ with the smallest $\nu$. So $r = 0$, which means $a = q d$. That is, every element of $U$ is a multiple of $d$. So $U$ is the principal ideal generated by $d$. This deduction holds for all ideals of $D$. Hence the result. {{qed}}
Let $y \in D$ be non-zero. Then, using the principal ideal property, for some $f \in D \sqbrk X$ we have: :$\gen {y, X} = \gen f \subseteq D \sqbrk X$ Therefore: :$\exists p, q \in D \sqbrk X: y = f p, X = f q$ By Properties of Degree we conclude that $f = a$ and $q = b + c X$ for some $a, b, c \in D$. Substituting into the equation $X = f q$ we obtain: :$X = a b + a c X$ which implies that: :$a c = 1$ That is: :$a \in D^\times$ where $D^\times$ denotes the group of units of $D$. Therefore: :$\gen f = \gen 1 = D \sqbrk X$ Therefore: :$\exists r, s \in D \sqbrk X: r y + s X = 1$ If $d$ is the constant term of $r$, then we have $y d = 1$. Therefore $y \in D^\times$. Our choice of $y$ was arbitrary, so this shows that $D^\times \supseteq D \setminus \set 0$, which says precisely that $D$ is a field. {{qed}}
Let $a, b \in D$. Let $U \subseteq D$ be the set of all elements $h a + k b$ of $D$ where $h, k \in D$. Then $U$ is an ideal of $D$. Note that $U = \ideal a + \ideal b$ where $\ideal a$ and $\ideal b$ are Principal Ideal. By Sum of Ideals is Ideal, $U$ is an ideal. By Euclidean Domain is Principal Ideal Domain, $U$ is a principal ideal, $\ideal d$ say. As $a, b \in U$ it follows that $d$ is a divisor of $a$ and $b$, that is: :$d \divides a \land d \divides b$ Since $d$ itself is in $U$, we have: :$\exists s, t \in D: s a + t b = d$ By Common Divisor in Integral Domain Divides Linear Combination: :$x \divides a \land x \divides b \implies x \divides d$ So $d$ is a greatest common divisor of $a$ and $b$. If $d$ and $d'$ are both greatest common divisors of $a$ and $b$, then $d \divides a \land d \divides b$ and so $d \divides d'$. Similarly $d' \divides d$. So $d$ and $d'$ are associates. {{qed}}
From the definition of associate, there exist $\map e X$ and $\map {e'} X$ \in $F \sqbrk X$ such that: :$\map d X = \map e X \cdot \map {d'} X$ :$\map {d'} X = \map {e'} X \cdot \map d X$ From Field is Integral Domain, $F$ is an integral domain. From Degree of Product of Polynomials over Integral Domain it follows that necessarily $\deg e = \deg e' = 0$, as $F$ has no proper zero divisors. Thus for some $c, c' \in F$, it must be that $\map e X = c$ and $\map {e'} X = c'$. From the two equations above it follows that $c \cdot c' = 1_F$, where $1_F$ is the unity of $F$. Hence, it follows that $c \ne 0_F$. The result follows. {{qed}}
By definition: :$\ideal {0_R} = \set {r \circ 0_R: r \in R}$ but for each $r \in R$ we have by Ring Product with Zero that $r \circ 0_R = 0_R$ for all $r \in R$. Therefore $\ideal {0_R}$ is the null ideal. {{qed}} Category:Ideal Theory g0bxzwsul4zh97e4us5incfu0yaqszb
We have that Integers form Integral Domain. Therefore the concept of irreducible is defined. Let $p$ be a prime number. By definition, the only divisors of $p$ are $1, -1, p, -p$. From Units of Ring of Integers, $1$ and $-1$ are (the only) units of $\Z$. From Associates are Unit Multiples, $p$ and $-p$ are (the only) associates of each other. Hence the result, from the definition of irreducible. {{qed}}
{{begin-eqn}} {{eqn | l = a       | o = \perp       | r = b       | c =  }} {{eqn | ll= \leadsto       | l = \gcd \set {a, b}       | r = 1       | c = {{Defof|Coprime Elements of Euclidean Domain}} }} {{eqn | ll= \leadsto       | lo= \exists x, y \in D:       | l = a \times x + b \times y       | r = 1       | c = Bézout's Lemma on Euclidean Domain }} {{eqn | ll= \leadsto       | l = c       | r = c \times \paren {a \times x + b \times y}       | c =  }} {{eqn | r = c \times a \times x + c \times b \times y       | c =  }} {{eqn | r = \paren {a \times c} \times x + \paren {b \times c} \times y       | c =  }} {{eqn | ll= \leadsto       | l = a       | o = \divides       | r = c \times \paren {a \times x + b \times y}       | c = as $a \divides a \times c$ and $a \divides b \times c$ }} {{eqn | r = c \times 1       | c = Bézout's Lemma on Euclidean Domain }} {{eqn | r = c       | c =  }} {{end-eqn}} {{qed}} {{Namedfor|Euclid}}
=== Proof of Existence === {{:Rational Polynomial is Content Times Primitive Polynomial/Existence}}{{qed|lemma}} === Proof of Uniqueness === {{:Rational Polynomial is Content Times Primitive Polynomial/Uniqueness}}
From Rational Polynomial is Content Times Primitive Polynomial: :$\cont h \cdot \map {h^*} X = \cont f \cont g \cdot \map {f^*} X \, \map {g^*} X$ and this expression is unique. By Gauss's Lemma on Primitive Rational Polynomials we have that $\map {f^*} X \, \map {g^*} X$ is primitive. From Content of Rational Polynomial is Multiplicative: :$\cont f \cont g = \cont f \cont g > 0$. The result follows. {{qed}}
Let $T = \struct {S, \tau}$ be a topological space and let $H \subseteq S$. === Definition 1 implies Definition 2 === Suppose that $H = H'$ where $H'$ is the derived set of $H$. By definition of derived set, $H'$ is the set of all limit points of $H$. So $H$ contains all its limit points and so by definition is closed. Now we also have that any point not in $H'$ is an isolated point. But there are no points in $H$ which are not in $H'$, so $H$ has no isolated points. Therefore $H = H'$ implies that $H$ is closed and has no isolated points. {{qed|lemma}} === Definition 2 implies Definition 1 === Suppose $H$ is closed and has no isolated points. By Closed Set iff Contains all its Limit Points we have that $H' \subseteq H$ where $H'$ is the derived set of $H$. As $S$ has no isolated points, all its elements are elements of its derived set. Therefore $H \subseteq H'$. So by definition of set equality it follows that $H = H'$. {{qed|lemma}} === Definition 2 iff Definition 3 === Let $H$ be closed with no isolated points. By definition, $H$ is closed {{iff}} it contains all its limit points. Also by definition, $H$ is dense-in-itself {{iff}} it has no isolated points. {{qed|lemma}} Hence the result. {{qed}}
Let $\tau$ be the topology on $T$. Let $\mathbb K = \set {K \in \tau: K \subseteq H}$. Then: {{begin-eqn}} {{eqn | l = T \setminus H^\circ       | r = T \setminus \bigcup_{K \mathop \in \mathbb K} K       | c = {{Defof|Interior (Topology)}} of $H$ }} {{eqn | r = \bigcap_{K \mathop \in \mathbb K} \paren {T \setminus K}       | c = De Morgan's Laws: Difference with Union }} {{end-eqn}} By the definition of closed set, $K$ is open in $T$ {{iff}} $T \setminus K$ is closed in $T$. Also, from Set Complement inverts Subsets we have that $T \setminus K \supseteq T \setminus H$. Now consider the set $\mathbb K'$ defined as: :$\mathbb K' := \set {K' \subseteq T: \paren {T \setminus H} \subseteq K', K' \text { closed in } T}$ From the above we see that: :$K \in \mathbb K \iff T \setminus K \in \mathbb K'$ Thus: {{begin-eqn}} {{eqn | l = T \setminus H^\circ        | r = \bigcap_{K' \mathop \in \mathbb K'} K' }} {{eqn | r = \paren {T \setminus H}^-       | c = {{Defof|Closure (Topology)|index = 2|Closure}} of $T \setminus H$ }} {{end-eqn}} {{qed|lemma}} Then we note that: {{begin-eqn}} {{eqn | l = H^\circ       | r = T \setminus \paren {T \setminus H^\circ}       | c = Relative Complement of Relative Complement }} {{eqn | r = T \setminus \paren {\paren {T \setminus H}^-}       | c = from above }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = \paren {T \setminus H}^\circ       | r = T \setminus \paren {\paren {T \setminus \paren {T \setminus H} }^-}       | c = from above }} {{eqn | r = T \setminus H^-       | c = Relative Complement of Relative Complement }} {{end-eqn}} {{qed}}
In the following, $H_i^-$ denotes the closure of the set $H_i$. {{begin-eqn}} {{eqn | l = \paren {\bigcap_{i \mathop = 1}^n H_i}^\circ       | r = T \setminus \paren {T \setminus \bigcap_{i \mathop = 1}^n H_i}^-       | c = Complement of Interior equals Closure of Complement }} {{eqn | r = T \setminus \paren {\paren {\bigcup_{i \mathop = 1}^n \paren {T \setminus H_i} }^-}       | c = De Morgan's Laws: Difference with Intersection }} {{eqn | r = T \setminus \paren {\bigcup_{i \mathop = 1}^n \paren {T \setminus H_i}^-}       | c = Closure of Finite Union equals Union of Closures }} {{eqn | r = T \setminus \paren {\bigcup_{i \mathop = 1}^n T \setminus H_i^\circ}       | c = Complement of Interior equals Closure of Complement }} {{eqn | r = T \setminus \paren {T \setminus \paren {\bigcap_{i \mathop = 1}^n H_i^\circ} }       | c = De Morgan's Laws: Difference with Intersection }} {{eqn | r = \bigcap_{i \mathop = 1}^n H_i^\circ       | c = Relative Complement of Relative Complement }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \paren {\bigcap_{i \mathop = 1}^n H_i}^{- \circ}       | r = \paren {T \setminus \paren {T \setminus \bigcap_{i \mathop = 1}^n H_i}^\circ}^\circ       | c = Complement of Interior equals Closure of Complement }} {{eqn | r = \paren {T \setminus \paren {\bigcup_{i \mathop = 1}^n \paren {T \setminus H_i} }^\circ}^\circ       | c = De Morgan's Laws: Difference with Intersection }} {{eqn | r = \paren {\paren {T \setminus \bigcup_{i \mathop = 1}^n \paren {T \setminus H_i} }^-}^\circ       | c = Complement of Interior equals Closure of Complement }} {{eqn | r = \paren {\bigcap_{i \mathop = 1}^n \paren {T \setminus \paren {T \setminus H_i} }^-}^\circ       | c = De Morgan's Laws: Difference with Intersection }} {{eqn | r = \paren {\bigcap_{i \mathop = 1}^n H_i^-}^\circ       | c = Relative Complement of Relative Complement }} {{eqn | r = \bigcap_{i \mathop = 1}^n H_i^{- \circ}       | c = Interior of Finite Intersection equals Intersection of Interiors }} {{eqn | r = \bigcap_{i \mathop = 1}^n H_i       | c = as all $H_i$ are regular open in $T$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \paren {\bigcup_{i \mathop = 1}^n H_i}^{\circ -}       | r = \paren {T \setminus \paren {T \setminus \bigcup_{i \mathop = 1}^n H_i}^-}^-       | c = Complement of Interior equals Closure of Complement }} {{eqn | r = \paren {T \setminus \paren {\bigcap_{i \mathop = 1}^n \paren {T \setminus H_i} }^-}^-       | c = De Morgan's Laws: Difference with Union }} {{eqn | r = \paren {\paren {T \setminus \bigcap_{i \mathop = 1}^n \paren {T \setminus H_i} }^\circ}^-       | c = Complement of Interior equals Closure of Complement }} {{eqn | r = \paren {\bigcup_{i \mathop = 1}^n \paren {T \setminus \paren {T \setminus H_i} }^\circ}^-       | c = De Morgan's Laws: Difference with Intersection }} {{eqn | r = \paren {\bigcup_{i \mathop = 1}^n H_i^\circ}^-       | c = Relative Complement of Relative Complement }} {{eqn | r = \bigcup_{i \mathop = 1}^n H_i^{\circ -}       | c = Closure of Finite Union equals Union of Closures }} {{eqn | r = \bigcup_{i \mathop = 1}^n H_i       | c = as all $H_i$ are regular closed in $T$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \partial X       | r = X^- \setminus X^\circ       | c = {{Defof|Boundary (Topology)|index = 1}}: $X^\circ$ is the interior of $X$ }} {{eqn | r = X^- \cap \overline {\paren {X^\circ} }       | c = Set Difference as Intersection with Relative Complement }} {{eqn | r = X^- \cap \paren {\overline X}^-       | c = Complement of Interior equals Closure of Complement }} {{end-eqn}} {{qed}}
From Boundary is Intersection of Closure with Closure of Complement: :$\partial H = H^- \cap \paren {T \setminus H}^-$ where $H^-$ is the closure of $H$. Hence from Intersection is Subset we have that: :$\partial H \subseteq H^-$ Then from Closed Set Equals its Closure, $H$ is closed in $T$ {{iff}} $H = H^-$. Hence the result. {{qed}}
From Boundary is Intersection of Closure with Closure of Complement: :$\partial H = H^- \cap \paren {T \setminus H}^-$ where $H^-$ is the closure of $H$. Hence from Intersection is Subset we have that: :$\partial H \subseteq \left({T \setminus H}\right)^-$ But from Closed Set Equals its Closure, $\paren {T \setminus H}^- = T \setminus H$ {{iff}} $T \setminus H$ is closed in $T$. That is, {{iff}} $H$ is open in $T$. So $\partial H \subseteq T \setminus H$ {{iff}} $H$ is open in $T$. From Intersection with Complement is Empty iff Subset it follows that $\partial H \cap H = \O$ {{iff}} $H$ is open in $T$. {{qed}}
From Set is Open iff Disjoint from Boundary we have that: :$H$ is open in $T$ {{iff}} $\partial H \cap H = \O$ From Set is Closed iff it Contains its Boundary we have that: :$H$ is closed in $T$ {{iff}} $\partial H \subseteq H$ From the above it follows that $\partial H = \O$ {{iff}} $H$ is closed and $H$ is open. {{qed}}
From Boundary is Intersection of Closure with Closure of Complement: :$\partial H = H^- \cap \paren {T \setminus H}^-$ where $H^-$ is the closure of $H$ From Topological Closure is Closed, both $H^-$ and $\paren {T \setminus H}^-$ are closed in $T$. From Topology Defined by Closed Sets, the intersection of arbitrarily many (in particular two) closed sets of $T$ is a closed set of $T$. As $\partial H$ is the intersection of $H^-$ and $\left({T \setminus H}\right)^-$ the result follows. {{qed}}
Let $B = \partial H$. From Boundary of Set is Closed we have that $B$ is closed in $T$. Let $B^-$ denote the closure of $B$. From Boundary is Intersection of Closure with Closure of Complement: :$\partial B = B^- \cap \paren {T \setminus B}^-$ and so from Intersection is Subset: :$\partial B \subseteq B^-$ But from Closed Set Equals its Closure: :$B = B^-$ and so: :$\partial B \subseteq B$ That is: :$\map \partial {\partial H} \subseteq \partial H$ {{qed}}
Let $H^e$ be defined as: :$H^e$ is the complement of the closure of $H$ in $T$. Then: {{begin-eqn}} {{eqn | l = H^e       | r = T \setminus H^-       | c =  }} {{eqn | r = \paren {T \setminus H}^\circ       | c = Complement of Interior equals Closure of Complement }} {{end-eqn}} Thus: :$H^e$ is the interior of the complement of $H$ in $T$. {{qed}}
{{begin-eqn}} {{eqn | l = \paren {H^e}^e       | r = \paren {T \setminus H^e}^\circ       | c = {{Defof|Exterior (Topology)|Exterior}} }} {{eqn | r = \paren {T \setminus \paren {T \setminus H^-} }^\circ       | c = Equivalence of Definitions of Exterior }} {{eqn | r = \paren {H^-}^\circ       | c = Relative Complement of Relative Complement }} {{eqn | o = \supseteq       | r = H^\circ       | c = Interior is Subset of Interior of Closure }} {{end-eqn}} {{qed}}
From Set is Subset of its Topological Closure, we have $H \subseteq H^-$. The result follows directly from Interior of Subset. {{qed}} Category:Set Closures Category:Set Interiors hpi3i0ga2cmk8ylx6ricx5jzzr61qz3
In the following, $H_i^\circ$ denotes the interior of the set $H_i$. {{begin-eqn}} {{eqn | l = \paren {\bigcup_{i \mathop = 1}^n H_i}^e       | r = \paren {T \setminus \bigcup_{i \mathop = 1}^n H_i}^\circ       | c = {{Defof|Exterior (Topology)|Exterior}} }} {{eqn | r = \paren {\bigcap_{i \mathop = 1}^n \paren {T \setminus H_i} }^\circ       | c = De Morgan's Laws: Difference with Union }} {{eqn | r = \bigcap_{i \mathop = 1}^n \paren {T \setminus H_i}^\circ       | c = Interior of Finite Intersection equals Intersection of Interiors }} {{eqn | r = \bigcap_{i \mathop = 1}^n H_i^e       | c = {{Defof|Exterior (Topology)|Exterior}} }} {{end-eqn}} {{qed}}
Let $\displaystyle K = \bigcup_{H \mathop \in \mathbb H} \map \cl H$ and $\displaystyle L = \bigcup_{H \mathop \in \mathbb H} H$. We have: :$\forall H \in \mathbb H: H \subseteq L$ so from Topological Closure of Subset is Subset of Topological Closure: :$\map \cl H \subseteq \map \cl L$ It follows from Union is Smallest Superset: General Result that: :$K \subseteq \map \cl L$ {{qed}}
From the definition of closure, we have: :$H^-$ is the union of $H$ and its limit points. From Subset of Union it follows directly that: :$H \subseteq H^-$ {{qed}} Category:Set Closures 43vh957rsenz5hj9tjjazua8jtrws71
What needs to be proved here is: :$\displaystyle H^- = \bigcap \mathbb K$ where: : $H^- = H \cup H'$ : $H'$ denotes the set of all limit points of $H$ : $\displaystyle \bigcap \mathbb K$ is the intersection of all closed sets of $T$ which contain $H$. Let $K \in \mathbb K$. From Topological Closure of Subset is Subset of Topological Closure, we have: : $H^- \subseteq K^-$ Thus from Closed Set Equals its Closure, we have: : $K^- = K$ Thus: :$\displaystyle \forall K \in \bigcap \mathbb K: H^- \subseteq K$ since the choice of $K$ is arbitrary.  So from Intersection is Largest Subset: : $\displaystyle H^- \subseteq \bigcap \mathbb K$ {{qed|lemma}} Conversely, from Topological Closure is Closed, $H^-$ is closed. From Set is Subset of its Topological Closure, $H \subseteq H^-$. So $H^-$ is, by definition, a closed set of $T$ which contains $H$. But we have by its definition that $\displaystyle \bigcap \mathbb K$ is the intersection of '''all''' closed sets in $T$ that contain $H$. So from Intersection is Subset it follows that: : $\displaystyle \bigcap \mathbb K \subseteq H^-$ {{qed|lemma}} Finally, we have that: : $\displaystyle H^- \subseteq \bigcap \mathbb K$ : $\displaystyle \bigcap \mathbb K \subseteq H^-$ So by definition of set equality: : $\displaystyle H^- = \bigcap \mathbb K$ which is what we needed to prove. {{qed}}
Define: :$\mathbb K := \leftset {K \supseteq H: K}$ is closed in $\rightset T$ That is, let $\mathbb K$ be the set of all supersets of $H$ that are closed in $T$. The claim is that $H^-$ is the smallest set of $\mathbb K$. From Set is Subset of its Topological Closure: :$H \subseteq H^-$ From Topological Closure is Closed, $H^-$ is closed in $T$. Thus $H^- \in \mathbb K$. Let $K \in \mathbb K$. From Set Closure as Intersection of Closed Sets: :$\displaystyle H^- = \bigcap \mathbb K$ Therefore, from Intersection is Subset: General Result: :$H^- \subseteq K$ Thus by definition $H^-$ is the smallest set of $\mathbb K$. {{qed}}
Let $\mathbb K$ be defined as: :$\mathbb K := \set {K \in \tau: K \subseteq H}$ That is, let $\mathbb K$ be the set of all open sets of $T$ contained in $H$. Then from definition 1 of the interior of $H$, we have: :$\displaystyle H^\circ = \bigcup_{K \mathop \in \mathbb K} K$ That is, $H^\circ$ is the union of all the open sets of $T$ contained in $H$. Let $K \subseteq T$ such that $K$ is open in $T$ and $K \subseteq H$. That is, let $K \in \mathbb K$. Then from Subset of Union it follows directly that $K \subseteq H^\circ$. So any open set in $T$ contained in $H$ is a subset of $H^\circ$, and so $H^\circ$ is the largest open set of $T$ contained in $H$. That is, $H^\circ$ is also the interior of $H$ by definition 2. Hence both definitions are equivalent. {{qed}} {{expand|to include definition 3}}
In the following, $H^-$ denotes the closure of the set $H$. {{begin-eqn}} {{eqn | l = \paren {\bigcap_{H \mathop \in \mathbb H} H}^\circ       | r = T \setminus \paren {T \setminus \bigcap_{H \mathop \in \mathbb H} H}^-       | c = Complement of Interior equals Closure of Complement }} {{eqn | r = T \setminus \paren {\paren {\bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H} }^-}       | c = De Morgan's Laws: Difference with Intersection }} {{end-eqn}} At this point we note that: :$(1): \quad \displaystyle \paren {\bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H} }^- \supseteq \bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H}^-$ from Closure of Union contains Union of Closures. Then we note that: :$\displaystyle T \setminus \paren {\paren {\bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H} }^-} \subseteq T \setminus \paren {\bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H}^-}$ from $(1)$ and Set Complement inverts Subsets. Then we continue: {{begin-eqn}} {{eqn | l = T \setminus \paren {\bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H}^-}       | r = T \setminus \paren {\bigcup_{H \mathop \in \mathbb H} T \setminus H^\circ}       | c = Complement of Interior equals Closure of Complement }} {{eqn | r = T \setminus \paren {T \setminus \paren {\bigcap_{H \mathop \in \mathbb H} H^\circ} }       | c = De Morgan's Laws: Difference with Intersection }} {{eqn | r = \bigcap_{H \mathop \in \mathbb H} H^\circ       | c = Relative Complement of Relative Complement }} {{end-eqn}} {{qed}}
In the following, $H^\circ$ denotes the interior of the set $H$. {{begin-eqn}} {{eqn | l = \paren {\bigcup_{H \mathop \in \mathbb H} H}^e       | r = \paren {T \setminus \bigcup_{H \mathop \in \mathbb H} H}^\circ       | c = {{Defof|Exterior (Topology)|Exterior}} }} {{eqn | r = \paren {\bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H} }^\circ       | c = De Morgan's Laws: Difference with Union }} {{eqn | o = \subseteq       | r = \bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H}^\circ       | c = Intersection of Interiors contains Interior of Intersection }} {{eqn | r = \bigcap_{H \mathop \in \mathbb H} H^e       | c = {{Defof|Exterior (Topology)|Exterior}} }} {{end-eqn}} {{qed}}
We have: {{begin-eqn}} {{eqn | l = \bigcup_{H \mathop \in \mathbb H} H^e       | r = \bigcup_{H \mathop \in \mathbb H} \paren {T \setminus H^-})       | c = {{Defof|Exterior (Topology)|Exterior}} }} {{eqn | r = T \setminus \bigcap_{H \mathop \in \mathbb H} H^-       | c = De Morgan's Laws: Difference with Intersection }} {{end-eqn}} From Closure of Intersection is Subset of Intersection of Closures: :$\displaystyle \paren {\bigcap_{H \mathop \in \mathbb H} H^-} \subseteq \bigcap_{H \mathop \in \mathbb H} H^-$ From Set Complement inverts Subsets: :$\displaystyle T \setminus \paren {\bigcap_{H \mathop \in \mathbb H} H^-} \supseteq T \setminus \bigcap_{H \mathop \in \mathbb H} H^-$ From the definition of exterior: :$\displaystyle T \setminus \paren {\bigcup_{H \mathop \in \mathbb H} H}^- = \paren {\bigcup_{H \mathop \in \mathbb H} H}^e$ Putting this together: :$\displaystyle \bigcup_{H \mathop \in \mathbb H} H^e \subseteq \paren {\bigcap_{H \mathop \in \mathbb H} H}^e$ {{qed}}
Let $A$ and $B$ be separated in $T$. Then: {{begin-eqn}} {{eqn | l = A^- \cap B       | r = \O       | c = {{Defof|Separated Sets}}: $A^-$ is the closure of $A$ }} {{eqn | ll= \leadsto       | l = \paren {A \cup A'} \cap B       | r = \O       | c = {{Defof|Closure (Topology)|Set Closure}}: $A'$ is the derived set of $A$ }} {{eqn | ll= \leadsto       | l = \paren {A \cap B} \cup \paren {A' \cap B}       | r = \O       | c = Intersection Distributes over Union }} {{eqn | ll= \leadsto       | l = A \cap B       | r = \O       | c = Union is Empty iff Sets are Empty }} {{end-eqn}} {{qed}}
By definition $T$ is second-countable {{iff}} its topology has a countable basis. Consider the entire set $S$ as an open set. From Set is Open iff Neighborhood of all its Points, $S$ has that property. As $T$ has a countable basis, then (trivially) every point in $T$ has a countable local basis. So a second-countable space is trivially first-countable. {{qed}}
From the definition of second-countable, $\struct {S, \tau}$ has a countable basis. That is, $\exists \BB \subseteq \tau$ such that: :for all $U \in \tau$, $U$ is a union of sets from $\BB$ :$\BB$ is countable. As $H \subseteq S$ it follows that a $H$ itself is a union of sets from $\BB$. The result follows from Basis for Topological Subspace. {{qed}}
From the definition of first-countable, every point in $S$ has a countable local basis in $T$. The intersection of $H$ with the countable local basis of $S$ provides a countable local basis for $H$. As every point in $H$ is also a point in $S$, the result follows from Basis for Topological Subspace. {{qed}}