hcju/proofwikiqa huggingface dataset:
Consider the open cover of $T$: :$\CC = \set {\set {x, p}: x \in S, x \ne p}$ As $S$ is countable, then so is $\CC$, as we can set up a bijection from $\phi: S \setminus \set p \leftrightarrow \CC$: :$\forall x \in S \setminus \set p: \map \phi x = \set {x, p}$ Hence $\CC$ is its own countable subcover. The result follows by definition of Lindelöf apace. {{qed}}
By definition, $T$ is separable {{iff}} there exists a countable subset of $S$ which is everywhere dense in $T$. Consider $U := \set p \subseteq S$. By definition, $U$ is open in $T$. From Closure of Open Set of Particular Point Space we have that $U^- = S$, where $U^-$ is the closure of $U$. By definition, $U$ is everywhere dense in $T$. $U$ is (trivially) countable. Hence the result, by definition of a separable space. {{qed}}
{{explain|Most of this needs to be explained in more detail. Suffers from coherence issues.}} Let $\set {X_s: s \in S}$ be a family of indeterminates indexed by $S$. Let $\phi$ be the Evaluation Homomorphism such that $\phi \sqbrk {X_s} = s$. {{explain|Can the above Evaluation Homomorphism link be replaced by Definition:Polynomial Evaluation Homomorphism?}} By Ring Homomorphism Preserves Subrings, $\Img \phi$ is a subring of $F$ This subring contains $\phi \sqbrk {X_s} = s$ for each $s \in S$ and $\map \phi k = k$ for all $k \in K$. Moreover since it is obtained by evaluating polynomials on $S$ it is the set of all finite linear combinations of powers elements of $S$ with coefficients in $K$ as claimed. All these linear combinations must belong to any subring of $F$ that contains $K$ and $S$ (otherwise it is not closed), so $\Img \phi$ is the smallest such subring. By Universal Property for Field of Quotients, the inclusion $K \sqbrk S \to F$ extends uniquely to a homomorphism $\psi : Q \to F$, given by $\map \psi {a / b} = a b^{-1}$. We have: :$\Img \psi = \set {a b^{-1} \in F: a, b \in K \sqbrk S, \ b \ne 0} \simeq Q$ The isomorphism comes from the fact that a field homomorphism is injective and the First Isomorphism Theorem. Thus $\Img \phi$ is a subfield of $F$ containing $K \sqbrk S$ and $K \cup S$. Any subfield of $F$ containing $K$ and $S$ must contain $K \cup S$, $K \sqbrk S$ and all $a b^{-1}$ with $a, b \in K \sqbrk S$ (otherwise it would not be closed). Therefore $Q \simeq \Img \phi$ is the smallest such subfield. {{qed}} Category:Field Extensions 6huc5ywotbfw61onr0njwdp9aclt5dp
By definition, $H$ is separable {{iff}} there exists a countable subset of $S$ which is everywhere dense in $T$. Let $V \subseteq H$ where $V$ is countable. $V$ is not open in $T$ as it does not contain $p$. From Subset of Particular Point Space is either Open or Closed it follows that $V$ is closed. From Closed Set Equals its Closure, $V^- = V$. But $V^- \ne H$ as $V$ is countable and $H$ is uncountable. So whatever $V$ is, if it is countable it is not everywhere dense. The result follows from definition of separable. {{qed}}
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. {{qed}}
Let $x \in S: x \ne p$. Consider the set $U_x = \set {x, p} \subseteq S$. Now let $V \in \tau_p$ be an open set in $S$ such that $x \in V$. So $x \in V$, by definition of $V$, and $p \in V$ as $V$ is open. It follows directly that $U_x \subseteq V$. So $\set {U_x}$ is a local basis at $x$ which is (trivially) countable. If $x = p$ then $U_x = \set p$ and the same argument still applies. So every element of $T$ has a countable local basis. Hence $T$ is first-countable by definition. {{qed}}
Let $K \sqbrk X$ be the ring of polynomial forms over $K$. By Polynomial Forms over Field form Unique Factorization Domain, $K \sqbrk X$ is a unique factorisation domain. Therefore, we can write $f = u g_1 \cdots g_r$, where $u$ a unit of $K \sqbrk X$ and $g_i$ is irreducible for $i = 1, \ldots, r$. Clearly it is sufficient to find an extension of $K$ in which the irreducible factor $g_1$ of $f$ has a root. Let $L = K \sqbrk X / \gen {g_1}$, where $\gen {g_1}$ is the ideal generated by $g_1$. By Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal, $\gen {g_1}$ is maximal. Therefore by Maximal Ideal iff Quotient Ring is Field, $L$ is a field. Moreover, writing $\overline {\map p x} = \map p X + \gen {g_1}$ for $\map p x \in K \sqbrk X$: {{begin-eqn}} {{eqn | l = \map {g_1} {\overline X}       | r = \overline {\map {g_1} X}       | c =  }} {{eqn | r = \overline {0_{K \sqbrk X} }       | c =  }} {{eqn | r = 0_L       | c =  }} {{end-eqn}} So $\overline X$ is a root of $g_1$ in $L$. It remains to show that $\index L K \le n$. By the Euclidean Algorithm every polynomial $\map p X \in K \sqbrk X$ can be written as: :$\map p X = \map q X \map {g_1} X + \map r X$ with $\map \deg {\map r X} < \map \deg {\map {g_1} X} \le \map \deg {\map f X} = n$. Now we have by the definition of the quotient ring: {{begin-eqn}} {{eqn | l = \overline {\map p X}       | r = \overline {\map q X \map {g_1} X + \map r X}       | c =  }} {{eqn | r = \overline {\map r X}       | c =  }} {{end-eqn}} So if $\map r X = r_0 + r_1 X + \cdots + r_{n - 1} X^{n - 1}$, we have: :$\overline {\map p X} = \overline {r_0 + r_1 X + \cdots + r_{n - 1} X^{n - 1} }$ Our choice of $p$ was arbitrary, so every polynomial can be written in this form. In particular the set $\set {\overline 1, \overline X, \ldots, \overline {X^{n - 1} } }$ spans $L$ over $K$. Thus by Spanning Set Contains Basis, a basis of $L$ has at most $n$ elements. That is, $\index L K \le n$. {{qed}} {{Namedfor|Leopold Kronecker|cat = Kronecker}} Category:Field Extensions p81n7mn6mchiyahli6djva9jcnq4ctj
We can set up the mapping $\phi: S \to S$: :$\forall x \in S: \map \phi x = \begin{cases} q & : x = p \\ p & : x = q \\ x & : \text {otherwise} \end{cases}$ It is straightforward to show that $\phi$ is a homeomorphism. However, we have, for example, that $\set q \notin \tau_p$ and $\set p \notin \tau_q$. So neither $\tau_p$ nor $\tau_q$ are a subset of the other. Hence by definition $\tau_p$ and $\tau_q$ are not comparable.  {{qed}}
Let $\psi: K \left({\alpha}\right) \to L$ be a homomorphism. Let $\phi = \psi \big|_K$. Let $\overline \phi: K \left[{X}\right] \to L \left[{X}\right]$ be the Induced Homomorphism of Polynomial Forms. For any $f = a_0 + \dotsb + a_n X^n \in K \left[{X}\right]$ we have: {{begin-eqn}} {{eqn | l = \psi \left({f \left({\alpha}\right)}\right)       | r = \psi \left({a_0 + \cdots + a_n \alpha^n}\right)       | c =  }} {{eqn | r = \phi \left({a_0}\right) + \cdots + \phi \left({a_n}\right) \psi \left({\alpha}\right)^n       | c = by the homomorphism property, and the fact that $\phi = \psi \big\vert_K$ }} {{eqn | r = \overline \phi \left({f}\right) (\psi\left({\alpha}\right))       | c =  }} {{end-eqn}} Since $\mu_\alpha \left({\alpha}\right) = 0$ and a Ring Homomorphism Preserves Zero, the above yields: :$\overline \phi \left({f}\right) \left({\psi \left({\alpha}\right)}\right) = 0$ as required. {{qed|lemma}} Conversely let $L$ be a field. Let $\phi: K \to L$ be a homomorphism. Let $\beta$ be a root of $\overline \phi \left({\mu_\alpha}\right)$ in $L$. By Structure of Simple Algebraic Field Extension, there exists a unique isomorphism: :$\Delta: K \left[{\alpha}\right] \to K \left[{X}\right] / \langle \mu_\alpha \rangle$ such that: :$\Delta \left({\alpha}\right) = X + \left\langle{\mu_\alpha}\right\rangle$ and: :$\Delta \big|_K$ is the identity mapping. By Universal Property of Polynomial Ring there exists a unique homomorphism: :$\chi: K \left[{X}\right] \to L$ such that: :$\chi \left({X}\right) = \beta$ and: :$\chi \big|_K = \phi$ By Universal Property of Quotient Ring there exists a unique homomorphism: :$\psi: K \left[{X}\right] / \left\langle{\mu_\alpha}\right\rangle \to L$ such that: :$\chi = \psi \circ \pi$ So we have the following diagram: ::$\begin{xy}\xymatrix@L+2mu@+1em {  K \left[{X}\right] \ar[r]^*{\pi}                     \ar@{-->}[rd]_*{\exists ! \chi} &  \dfrac {K \left[{X}\right]} {\left\langle{\mu_\alpha}\right\rangle} \ar@{-->}[d]^*{\exists ! \psi}                                                                      \ar[r]^*{\Delta} &  K \left[{\alpha}\right] \\  &  L & }\end{xy}$ Now we have: :$\chi \left({X}\right) = \beta$ and: :$\pi \left({X}\right) = X + \left\langle{\mu_\alpha}\right\rangle$ Therefore, because: :$\chi = \psi \circ \pi$ we have: :$\psi \left({X + \left\langle{\mu_\alpha}\right\rangle}\right) = \beta$ Also: :$\Delta^{-1} \left({\alpha}\right) = X + \left\langle{\mu_\alpha}\right\rangle$ so by the above: : $\psi \circ \Delta^{-1}\left({ \alpha }\right) = \beta$ : $\psi \circ \Delta^{-1} \big|_K =  \psi\big|_K = \phi$ : the choice of $\Delta, \psi$ is unique. {{qed}} Category:Field Extensions Category:Universal Properties drpnzgmm7nkwt879f4v4gzg4wolq1pw
Let $x \in L$ be arbitrary. Let $n = \left[{L : K}\right]$ be the degree of $L$ over $K$. Since any set of $n+1$ vectors in $L$ is linearly dependent, there is a $K$-linear combination of $\left\{{1, \ldots, x^n}\right\}$ equal to $0$. Say $a_n x^n + \cdots + a_1 x + a_0 = 0$, $a_i \in K$, $i = 0, \ldots, n$. Therefore $x$ satisfies a polynomial with coefficients in $K$. {{explain|Clarify whether the above link is what it is supposed to be}} That is, $x$ is algebraic. Since $x \in L$ was chosen arbitrarily, $L / K$ is algebraic. {{qed}} {{MissingLinks|I believe that "any set of $n+1$ vectors in $L$ is linearly dependent" may already exist. If not, this link needs to be renamed as there's $\LaTeX$ in the title, which renders it invalid.}} Category:Field Extensions sm8umfdxlne2wj3wixstqy0vuq2zwsz
Let $S = \left\{{\alpha_1, \ldots, \alpha_n}\right\}$. We show by induction on $n$ that $K \left({S}\right) / K$ is finite. Clearly $K$ is finite over itself, so the result holds when $n = 0$. Now suppose that for all sets $T \subseteq L$ with $\left|{T}\right| \leq n - 1$ and each element of $T$ algebraic over $K$, $K \left({T}\right) / K$ is finite. We have: :$K \left({S}\right) = K \left({\alpha_1, \ldots, \alpha_{n-1}}\right) \left({\alpha_n}\right)$ By the induction hypothesis $K \left({\alpha_1, \ldots, \alpha_{n-1}}\right) / K$ is finite. By the Structure of Simple Algebraic Field Extension $K \left({S}\right) / K \left({\alpha_1, \ldots, \alpha_{n-1}}\right)$ is finite. Therefore, by the Tower Law, $K \left({S}\right) / K$ is finite. {{qed}} Category:Field Extensions dmf7bqm08ocbjnyhee81auwcc05dz5u
Let $x \in E$. There are $a_0, \ldots, a_n \in F$ such that $a_0 + \cdots + a_n x^n = 0$. Let $L = K \left({a_0, \ldots, a_n}\right)$. We have that $L / K$ is finitely generated and algebraic. Therefore by Finitely Generated Algebraic Extension is Finite this extension is finite. We have that $L \left({x}\right) / L$ is simple and algebraic. So by Structure of Simple Algebraic Field Extension, this extension is also finite. Therefore, by the Tower Law: :$L \left({x}\right) / K$ is finite. That is, $x$ is contained in a finite extension of $K$. Therefore because a Finite Field Extension is Algebraic, it follows that $x$ is algebraic over $K$, as was to be proved. {{qed}}
Let $\alpha,\beta \in K^a$. By Field Adjoined Algebraic Elements is Algebraic, $K(\alpha,\beta)/K$ is algebraic. By definition $K(\alpha,\beta)$ is a field. Therefore $\alpha\beta$, $\alpha^{-1}$ and $\alpha - \beta$ all lie in $K(\alpha,\beta)$, hence are algebraic over $K$. Also $K \subseteq K^a$ so $K^a \neq \emptyset$, and by the Subfield Test, $K^a \leq L$ is a subfield. {{qed}} Category:Field Extensions lavzxgixxbb8jiuzry8vjewgbuieyv9
=== Definition $(1)$ implies Definition $(2)$ === Let $K$ be algebraically closed by definition 1. Let $f$ be an irreducible polynomial over $K$. By Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal, the ideal $\left\langle{f}\right\rangle$ generated by $f$ is maximal. So by Maximal Ideal iff Quotient Ring is Field: : $L = K \left[{X}\right] / \left\langle{f}\right\rangle$ is a field where $L$ is a field extension over $K \left[{X}\right]$. Now: :$L = \left\{{g + \left\langle{f}\right\rangle: g \in K \left[{X}\right]}\right\}$ From Division Theorem for Polynomial Forms over Field: :$\forall g \in K \left[{X}\right]: \exists q, r \in K \left[{X}\right]: g = q f + r, \operatorname{deg} r < \operatorname{deg} f =: n$ Therefore: :$L = \left\{{r + \left\langle{f}\right\rangle: r \in K \left[{X}\right],\ \operatorname{deg} r < n}\right\}$ By Basis for Quotient of Polynomial Ring, this has basis: :$1 + \left\langle{f}\right\rangle, \ldots, X^{n-1} + \left\langle{f}\right\rangle$ span $L$. Thus $L$ is finite. By Finite Field Extension is Algebraic, $L$ is algebraic. Also $K \subseteq L$. So by hypothesis $K = L$. This implies: : $\left[{L : K}\right] = 1$ where $\left[{L : K}\right]$ is the degree of $L$ over $K$. Hence $n = \operatorname{deg}f = 1$. Thus $K$ is algebraically closed by definition 2. {{qed|lemma}} === Definition $(2)$ implies Definition $(3)$ === Let $K$ be algebraically closed by definition 2. Let $f$ be a polynomial in $K \left[{X}\right]$ of strictly positive degree. From Polynomial Forms over Field form Principal Ideal Domain, $K \left[{X}\right]$ is a principal ideal domain. From Principal Ideal Domain is Unique Factorization Domain, $K \left[{X}\right]$ is a unique factorization domain. So $f$ can be factorized $f = u g_1 \cdots g_r$ such that: : $u$ is a unit and: : $g_i$ are irreducible for $i = 1, \ldots, r$. By hypothesis, $g_1$ has degree $1$. Therefore by the Polynomial Factor Theorem $g_1$, and hence $f$, has a root in $K$. Thus $K$ is algebraically closed by definition 3. {{qed|lemma}} === Definition $(3)$ implies Definition $(1)$ === Let $K$ be algebraically closed by definition 3. Let $L / K$ be an algebraic field extension of $K$. Let $\alpha \in L$. By hypothesis, the minimal polynomial $\mu_\alpha$ of $\alpha$ over $K$ has a root $\beta$ in $K$. Therefore by the Polynomial Factor Theorem: : $\mu_\alpha = \left({X - \beta}\right) g$ for some $g \in K \left[{X}\right]$. Since $\mu_\alpha$ is irreducible and monic (see Minimal Polynomial) it follows that: :$\mu_\alpha = X - \beta$ {{explain|Review the link to Minimal Polynomial as the material inside may be moving elsewhere.}} Also: : $\mu_\alpha \left({\alpha}\right) = \alpha - \beta = 0$ so $\alpha = \beta$. Therefore $\alpha \in K$. Therefore $L = K$ as required. Thus $K$ is algebraically closed by definition 1. {{qed}} Category:Field Extensions ndwyuryxd6gwp2e6nmhtoqqe6ue081p
Let $H$ be closed in $T$. Then by definition $p \notin H$. Let $x \in H$. By definition, $x$ is a limit point of $H$ if every open set $U \in \tau$ such that $x \in U$ contains some point of $H$ other than $x$. Consider the set $U_x := \set {x, p} \subseteq S$. As $p \in U_x$ we have that $U_x$ is open in $T$. But there is no $y \in U_x: y \in H, y \ne x$ and so $x$ is not a limit point of $H$. So $H$ has no limit points. {{qed}}
Let $H \subseteq S$ such that $p \notin H$. From Subset of Particular Point Space is either Open or Closed, $H$ is closed in $T$. We have that Closed Set in Particular Point Space has no Limit Points. So if $p \notin H$ then $H$ has no limit points. So by definition, every element of such a set $H$ is an isolated point. So if $p \notin H$ then $H$ is by definition not dense-in-itself. Now let $H \subseteq S$ such that $p \in H$. We have that $\set p \in \tau_p$. That is, $p$ is an open point of $T$. From Point in Topological Space is Open iff Isolated, $p$ is an isolated point. So if $p \in H$ then $H$ is by definition not dense-in-itself. Hence the result, by definition of a scattered space. {{qed}}
By definition, $T = \struct {S, \tau_p}$ is irreducible {{iff}} every two non-empty open sets of $T$ have non-empty intersection. Let $U_1$ and $U_2$ be non-empty open sets of $T$. By definition of particular point space, $p \in U_1$ and $p \in U_2$. Thus: :$p \in U_1 \cap U_2$ and so: :$U_1 \cap U_2 \ne \O$ Hence the result. {{qed}}
Let $x, y \in S: x \ne p, y \ne p, x \ne y$. Consider $\set x$ and $\set y$. Neither are open as neither contain $p$. So from Subset of Particular Point Space is either Open or Closed they are both closed. We have that $\set x \cap \set y = \O$. The result follows by definition of ultraconnected. {{qed}}
Let $H = S \setminus \set p$. Let $T_H = \struct {H, \tau_H}$ be the topological subspace induced on $H$ by $\tau_p$. From Particular Point Space less Particular Point is Discrete, the space $T_H$ is discrete. We have Discrete Space is Locally Connected. Thus from Totally Disconnected and Locally Connected Space is Discrete we have that $S \setminus \set p$ is totally disconnected. Hence the result, from definition of dispersion point. {{qed}}
By definition, $T$ is weakly countably compact {{iff}} every infinite subset of $S$ has a limit point in $S$. Let $H \subseteq S$ be an infinite subset of $S$ where $p \notin H$. $H$ is not open in $T$ by definition. So from Subset of Particular Point Space is either Open or Closed, $H$ is closed in $T$. Then we have that a Closed Set in Particular Point Space has no Limit Points. The result follows from definition of weakly countably compact. {{qed}} If $T = \struct {S, \tau_p}$ is a finite particular point space, then Finite Space Satisfies All Compactness Properties applies.
Let $p \in R := K \sqbrk {x_1, \dotsc, x_n}$ be a symmetric polynomial. Let $\sigma_1, \dotsc, \sigma_n$ denote the $n$ elementary symmetric polynomials. Let $<$ be the degree reverse lexicographic monomial order. Let us denote by $\map {LM} f$ the leading monomial of any $f \in R$, that is to say the unique $<$-largest monomial appearing in $f$. Then it is straightforward to verify that $\map {LM} {\sigma_k} = x_1 \cdots x_k$ for $1 \le k \le n$: For $k < n$, any other monomial appearing in $\sigma_k$ must contain a factor $x_j^{\alpha_1}$ with $j > k$ and $\alpha_1 > 0$. This makes it smaller than $x_1 \cdots x_k$ with respect to $<$. The case $k = n$ is trivial because $\sigma_n$ consists of only one monomial. It follows the proof that $p$ may be written as a polynomial in $\sigma_1, \dotsc, \sigma_k$. Firstly, since $p$ is symmetric, it must have the property: :$\map {LM} p = x_1^{a_1} \dotsm x_n^{a_n}$ with $a_1 \ge a_2 \ge \ldots \ge a_n$. To see this, assume that $a_i > a_j$ for some $1 \le i < j \le n$. The monomial: :$x_1^{a_1} \dotsm x_{i - 1}^{a_{i - 1} } \cdot x_i^{a_j} x_j^{a_i} \cdot x_{j + 1}^{a_{j + 1} } \cdots x_n^{a_n}$ must also appear in $p$, and if $a_i > a_j$ then it would be larger than $\map {LM} p$, a contradiction. Now consider: :$\sigma :=  c \sigma_1^{a_1 - a_2} \cdot \sigma_2^{a_2 - a_3} \cdot \dotsm \cdot \sigma_{n - 1}^{a_{n - 1} - a_n} \cdot \sigma_n^{a_n}$ where $c$ is the coefficient of $\map {LM} p$ in $p$. Define $\bar p := p - \sigma$. Because: {{begin-eqn}} {{eqn | l = \map {LM} \sigma       | r = \prod_{i \mathop = 1}^n \map {LM} {\sigma_i^{a_i - a_{i + 1} } }       | c =  }} {{eqn | r = x_1^{a_1 - a_2} \cdot \paren {x_1 x_2}^{a_2 - a_3} \cdot \dotsm \cdot \paren {x_1 \ldots x_{n - 1} }^{a_{n - 1} - a_n} \cdot \paren {x_1 \ldots x_n}^{a_n}       | c =  }} {{eqn | r = x_1^{a_1} \dotsm x_n^{a_n}       | c =  }} {{eqn | r = \map {LM} p       | c =  }} {{end-eqn}} the leading monomials in the difference $\bar p$ cancel. Therefore: :$\map {LM} {\bar p} < \map {LM} p$ Iterate this method, and after finitely many steps one has decomposed $p$ into a polynomial $F \in K \sqbrk {y_1, \dotsc, y_n}$ with $\map F {\sigma_1, \dotsc, \sigma_n} = p$. The termination of this algorithm follows immediately from Dickson's Lemma. {{qed}} Category:Symmetric Polynomials Category:Fundamental Theorems dbvjw2s6ppn77j0avsq5cmja8ca59hz
We have that: :a Particular Point Space is Irreducible :an Irreducible Space is Pseudocompact. {{qed}}
Let $q \in S$. Let $\mathbb I$ be the closed unit interval in $\R$. Let $f: \mathbb I \to S$ be the mapping defined as: :$\forall x \in \mathbb I: \map f x = \begin{cases} p & : x \in \hointr 0 1 \\ q & : x = 1 \end{cases}$ Let $U \in \tau_p$. Then by definition of particular point space, $p \in U$ Either $q \in U$ or $q \notin U$. If $q \in U$ then $f^{-1} \sqbrk U = \closedint 0 1$ which is open in $\mathbb I$ because $\closedint 0 1 = \mathbb I$. If $q \notin U$ then $f^{-1} \sqbrk U = \hointr 0 1$ which is half open in $\R$ but open in $\mathbb I$. So $f: \mathbb I \to S$ is a continuous mapping and so a path from $p$ to $q$. As $q$ is any point in $S$, it follows from Path-Connected iff Path-Connected to Point that $T$ is path-connected. {{qed}}
Let $q \in S$ be such that $q \ne p$. Let $f: \closedint 0 1 \to T$ be an injection such that: :$\map f 0 = q$ :$\map f 1 = p$ Because $f$ is an injection, it must be that: :$\map {f^{-1} } {\set p} = \set 1$ where $f^{-1}$ denotes the preimage under $f$. Now $\set p$ is open in $\tau_p$ by definition of particular point topology. By Closed Real Interval is not Open Set, $\set 1 = \closedint 1 1$ is not open. Thus we have exhibited an open set whose preimage under $f$ is not open. Hence $f$ is not continuous. Since $f$ was an arbitrary injection, no arc between $p$ and $q$ can exist. Hence $T$ is not arc-connected. {{qed}}
Let $H \in \tau_p$ be open in $T$. Then: : $\forall y \in H: \exists \left\{{y, p}\right\} \in \mathcal B$ which also holds when $y = p$ as $\left\{{y, p}\right\} = \left\{{p}\right\} \in \mathcal B$. Thus: :$\displaystyle H = \bigcup_{y \mathop \in H} \left\{{y, p}\right\}$ So $\mathcal B$ is an analytic basis for $T$. {{qed}} It could equally well be shown that $\mathcal B$ is also a synthetic basis for $T$. Category:Particular Point Topology Category:Topological Bases 5l71n059ph20ei9lqy9eqgkkcte2dxf
Consider the set $\BB$ defined as: :$\BB = \set {\set {x, p}: x \in S}$ Then $\BB$ is a basis for $T$. Now consider the open set $\set {p, q} \in \BB$. Let $\mathbb I$ be the closed unit interval in $\R$. Let $f: \mathbb I \to S$ be the mapping defined as: :$\forall x \in \mathbb I: \map f x = \begin{cases} p & : x \in \hointr 0 1 \\ q & : x = 1 \end{cases}$ Let $U \in \tau_p$. Then either $q \in U$ or $q \notin U$. If $q \in U$ then $f^{-1} \sqbrk U = \closedint 0 1$ which is open in $\mathbb I$ because $\closedint 0 1 = \mathbb I$. If $q \notin U$ then $f^{-1} \sqbrk U = \hointr 0 1$ which is half open in $\R$ but open in $\mathbb I$. So $f: \mathbb I \to S$ is a continuous mapping and so a path from $p$ to $q$. As $\set {p, q}$ is any element of $\BB$ it follows that $T$ is locally path-connected. {{qed}}
Suppose $T$ were meager. Then it would be a countable union of subsets which are nowhere dense in $T$. Let $H \subseteq S$. From Closure of Open Set of Particular Point Space, the closure of $H$ is $S$. From the definition of interior, the interior of $S$ is $S$. So the interior of the closure of $H$ is not empty. So $T$ can not be the union of a countable set of subsets which are nowhere dense in $T$. Hence $T$ is not meager and so by definition must be non-meager. {{qed}}
By definition of particular point space, any subset of $S$ which contains $p$ is open in $T$. So $\left\{{p}\right\}$ itself is open in $T$. That is, $p$ is an open point. The result follows from Space with Open Point is Non-Meager. {{qed}}
By definition of particular point space, any subset of $S$ which contains $p$ is open in $T$. {{AimForCont}} $T$ is meager. By definition, $T$ is meager {{iff}} it is a countable union of subsets of $S$ which are nowhere dense in $T$. At least one such nowhere dense subset $U$ of $S$ must contain $p$. By definition, $U$ is nowhere dense in $T$ {{iff}}: :$U^-$ contains no open set of $T$ which is non-empty where $U^-$ denotes the closure of $U$. By definition of particular point space, $U$ is open in $T$. By Closure of Open Set of Particular Point Space, $U^- = S$. But $S$ is itself open in $T$ and non-empty, and so $U$ is not nowhere dense. From this contradiction it follows that $T$ is non-meager {{qed}}