topocorr

topo/topo2.html

@@ -206,11 +206,11 @@ <h1><a
 Let, if possible, $C$  be not connected. Let $C = D\cup E $ be a disconnection. 
 <p></p>
 Then $\forall \alpha\in \Lambda~~(C_ \alpha\subseteq D\mbox{ or } C_ \alpha\subseteq E).$
-<blockquote>
+<blockquote class="scrpt">
 <b>Because:</b>Otherwise, for some $\alpha\in \Lambda$  we have $C_ \alpha\cap D\neq\phi$  and
  $C_ \alpha\cap E\neq\phi.$  Then $(C_ \alpha\cap E)\cup(C_ \alpha\cap E)$  is a disconnection for $C_ \alpha.$</blockquote>
 Also, $\exists \alpha\in \Lambda~~C_ \alpha\subseteq D$  and $\exists \beta\in \Lambda~~C_ \beta\subseteq E.$
-<blockquote>
+<blockquote class="scrpt">
 <b>Because:</b>Otherwise, either $D=\phi$  or $E=\phi.$</blockquote>
 
 <p></p>
@@ -279,7 +279,7 @@ <h1><a
 Since $U\cap V=\phi,$  this is well-defined. 
 <p></p>
 This function is continuous.
-<blockquote>
+<blockquote class="scrpt">
 <b>Because:</b>
 Let $A\subseteq{\mathbb R}$  be open. 
 <ul>
@@ -353,7 +353,7 @@ <h1><a
 Let, if possible, $A\cup C$  be not connected. Let $(U,V)$  be a disconnection. Then either $C\subseteq U$  or
  $C\subseteq V,$  since $C$  is connected. Wlog, let $C\subseteq V.$  Then  $(U,V\cup B)$  is a disconnection.
  for $A\cup B\cup C.$  
-<blockquote>
+<blockquote class="scrpt">
 <b>Because:</b>
 $U$  is clopen in $A\cup C.$  Since $C\cap U=\phi,$  hence $U\subseteq A.$ 
  Since $A,B$  are separated, hence  $B$  is separated from $U.$  So $U$  is clopen in $A\cup B\cup C.$ 
@@ -433,7 +433,7 @@ <h1><a
 Ler $A$  be a connected component of a topological space $(X,\tau).$  
 <p></p>
 Then $A$  is the intersection of all its  clopen supersets.
-<blockquote>
+<blockquote class="scrpt">
 <b>Because:</b>
 Let $C$  be a clopen superset of $A.$  Let $p\in C^c.$   Then $C^c$  and
  $C$  provide a disconnection for $A\cup\{p\}.$ 
@@ -874,7 +874,7 @@ <h1><a
 Let $C = \cup_n B_n^c,$  which is a countable union of finite sets, and hence countable. Since $X$  is uncountable,
  $C^c\neq\phi.$  
 <p></p>
-We take any  $x\in C.$. Then  $\{x}^c$  is an open set containing $p.$  
+We take any  $x\in C.$. Then  $\{x\}^c$  is an open set containing $p.$  
 <p></p>
 So there must be some $n\in{\mathbb N}$  such that $B_n\subseteq\{x\}^c,$  ie, $c\in B_n^c(\contra\bc x\in C^c).$   
 <p></p>
@@ -997,7 +997,7 @@ <h1><a
 <p></p>
 <b>SOLUTION:</b>
 <b><font color="#0000cc">
-Let $G=\{(x,f(x))~:~x\in X\}.$  We take any $(a,b)\in (X\timesY)\seminus G.$
+Let $G=\{(x,f(x))~:~x\in X\}.$  We take any $(a,b)\in (X\times Y)\setminus G.$
 <p></p>
 So $b\neq f(a).$  Since $Y$  is $T_2,$  hence can find two disjoint open sets $U$  and $V$  in $Y$ 
  such that $f(a)\in U$  and $b\in V.$  
@@ -1105,7 +1105,7 @@ <h1><a
 Let $\epsilon = \alpha-\inf\{d(x,a)~:~a\in A\}&gt;0.$
 <p></p>
 Then $N\left(x,\frac \epsilon2\right)\subseteq G.$
-<blockquote>
+<blockquote class="scrpt">
 <b>Because:</b>
 Take any $y\in N\left(x,\frac \epsilon2\right)$  and any $a\in A.$
 <p></p>
@@ -1189,7 +1189,7 @@ <h1><a
 We assume that $f$  is onto.
 <p></p>
 Since $f$  is an isometry, it is one-one.
-<blockquote>
+<blockquote class="scrpt">
 <b>Because:</b>
 $x_1\neq x_2\Rightarrow d(x_1,x_2) &gt; 0 \Rightarrow d'(f(x_1),f(x_2))= d(x_1,x_2)&gt; 0\Rightarrow f(x_1)\neq f(x_2).$
 </blockquote>
@@ -1340,7 +1340,7 @@ <h1><a
 <p></p>
 Then the subcollection ${\cal D}=\{B_n~:~n\in{\mathbb N}\}$  is again a base.
 <p></p>
-Clearly, $B_n\in\{\cal B}\subseteq\tau.$  
+Clearly, $B_n\in{\cal B}\subseteq\tau.$  
 <p></p>
 Let $U\in\tau$  and $x\in U.$  Then $\exists n\in{\mathbb N} ~~x\in C_n\subseteq U.$  
 </font></b> ///
@@ -1406,7 +1406,7 @@ <h1><a
 It is not Hausdorff, because all non-empty open sets intersect. 
 <p></p>
 Each convergent sequence is eventually constant. 
-<blockquote>
+<blockquote class="scrpt">
 <b>Because:</b>
 Let $x_n\rightarrow L.$  Then $G = \{x_n\}^c\cup\{L\}$  is an open neighbourhood of $L.$  So
  $\exists K\in{\mathbb N}~~\forall n\geq K~~x_n\in G.$  

topo/topo2.rb

@@ -540,7 +540,7 @@
 Let <M>C = \cup_n B_n^c,</M>  which is a countable union of finite sets, and hence countable. Since <M>X</M>  is uncountable,
  <M>C^c\neq\phi.</M>  
 
-We take any  <M>x\in C.</M>. Then  <M>\{x}^c</M>  is an open set containing <M>p.</M>  
+We take any  <M>x\in C.</M>. Then  <M>\{x\}^c</M>  is an open set containing <M>p.</M>  
 
 So there must be some <M>n\in\nn</M>  such that <M>B_n\seq\{x\}^c,</M>  ie, <M>c\in B_n^c(\contra\bc x\in C^c).</M>   
 
@@ -630,7 +630,7 @@
 Y,</M> where <M>X\times Y</M> is endowed with the product topology.[2]
 </E>@}</EBODY><SOLN/>
 @{<WR>
-Let <M>G=\{(x,f(x))~:~x\in X\}.</M>  We take any <M>(a,b)\in (X\timesY)\seminus G.</M>
+Let <M>G=\{(x,f(x))~:~x\in X\}.</M>  We take any <M>(a,b)\in (X\times Y)\setminus G.</M>
 
 So <M>b\neq f(a).</M>  Since <M>Y</M>  is <M>T_2,</M>  hence can find two disjoint open sets <M>U</M>  and <M>V</M>  in <M>Y</M> 
  such that <M>f(a)\in U</M>  and <M>b\in V.</M>  
@@ -917,7 +917,7 @@
 
 Then the subcollection <M>{\cal D}=\{B_n~:~n\in\nn\}</M>  is again a base.
 
-Clearly, <M>B_n\in\{\cal B}\seq\tau.</M>  
+Clearly, <M>B_n\in{\cal B}\seq\tau.</M>  
 
 Let <M>U\in\tau</M>  and <M>x\in U.</M>  Then <M>\exists n\in\nn ~~x\in C_n\seq U.</M>  
 </WR>@}</EXM>

topo/topo2.rb.scrp

@@ -1,25 +0,0 @@
-<BECAUSE>
-
-</BECAUSE>
--------EOD-------
-<EXM ref="2023.1f" paper="2023.1f"><EBODY>@{<E>
-Let <M>(X,\tau)</M> be a co-countable space, where <M>X</M> is an
-uncountable set. Then which of the following is true?
-<VL>
-<LI><M>(X,\tau)</M> is a first countable space.</LI>
-<LI><M>(X,\tau)</M> is a Hausdorff space.</LI>
-<LI>There exists a convergent sequence in <M>X</M> whose limit
-is not unique.</LI>
-<LI>A sequence <M>\{x_n\}</M> in <M>X</M> is convergent if and
-only if there is some positive integer <M>m</M> such that for
-all <M>n\geq m</M> <M>x_n=</M> constant.</LI>
-</VL>
-</E>@}</EBODY><SOLN/>khAli sheSerTAi Thik.</EXM>
--------EOD-------
-<EXM ref="2021.3" paper="2021.3"><EBODY>@{<E>
-Prove that a topological invariant is a metric invariant. Is the
-converse ture? Justify.[3+2]
-</E>@}</EBODY><SOLN/></EXM>
--------EOD-------
-
--------EOD-------

topo/topo3.html

@@ -158,7 +158,7 @@ <h1><a
 
 <p>
 <b>EXAMPLE (2021.1h):</b>&nbsp;
-If $\tau = \{\phi, \{a\},\{\a,b\},X\}$ is a topology on $X =
+If $\tau = \{\phi, \{a\},\{a,b\},X\}$ is a topology on $X =
 \{a,b,c\},$ then $(X,\tau)$ is
 
 <li>compact and Hausdorff.</li>
@@ -317,8 +317,8 @@ <h1><a
 <p></p>
 <b>SOLUTION:</b>Continuous, কারণ এর গ্রাফ আঁকতে হলে পেন তুলতে হয় একমাত্র
  domain-এ ফাঁক আছে যেখানে, সেখানে৷ Open  নয়, কারণ
- $(2.1,2.9)$  এখানে $open,$  কিন্তু এর image  হল $\{2\},$ 
- যেক্ষা মোটেই ${\mathbb R}$-এর মধ্যে open  নয়৷ ///
+ $(2.1,2.9)$  এখানে open,  কিন্তু এর image  হল $\{2\},$ 
+ যেটা মোটেই ${\mathbb R}$-এর মধ্যে open  নয়৷ ///
 </p>
 
 <p></p>
@@ -385,7 +385,7 @@ <h1><a
  $a\in A$  and $b\in B.$  
 <p></p>
 Clearly, this relation is reflexive and symmetric. Also it is transitive.
-<blockquote>
+<blockquote class="scrpt">
 <b>Because:</b>
 Let $a\sim b$  and $b\sim c.$  Let, if possible, $a\not\sim c.$
 <p></p>
@@ -419,12 +419,18 @@ <h1><a
 <p></p>
 
 <u>Second part</u>:
-Since the elements of $\beta$  are all complements of closed sets (wrt $\tau$), hence $bb\subseteq \tau.$  
+Since the elements of $\beta$  are all complements of closed sets (wrt $\tau$), hence $\beta\subseteq \tau.$  
 <p></p>
 So $\tau'\subseteq \tau.$
 <p></p>
 
 <u>Third part</u>:
+<p></p>
+Let $\{U_ \alpha~:~ \alpha\in \Lambda\}\subseteq \tau'$  be a cover for $X.$  Pick some
+ $U$  in this cover (possible since this cover cannot be empty, as $X\neq\phi$). 
+Then $U^c$  is compact in $(X,\tau)$   and has $\{U_ \alpha~:~ \alpha\in \Lambda\}\setminus\{U\}$  as an open
+ cover (since $\tau'\subseteq\tau$). So we can extract a finite subcover $\{U_1,...,U_n\}$  for $U^c.$  Then $\{U_1,...,U_n\}\cup \{U\}$ 
+ is a finite subcover for $X$  in $(X,\tau'),$  completing the proof.
 </font></b> ///
 </p>
 
@@ -482,7 +488,7 @@ <h1><a
  So $\{F_ \alpha^c~:~ \alpha\in \Lambda\}$  is an open cover of $X.$
 <p></p>
 Since $X$  is compact, there is a finite subcover, $\{F_{\alpha_1}^c,...,F_{\alpha_n}^c\}$  ie, 
-$\cup _{i=1}^n F_ \alpha_i^c = X,$  ie, $\cap _{i=1}^n F_ \alpha_i = \phi (\contra$  FIP$).$
+$\cup _{i=1}^n F_ {\alpha_i}^c = X,$  ie, $\cap _{i=1}^n F_ {\alpha_i} = \phi (\contra$  FIP$).$
 <p></p>
 Conversely, if every family with FIP has non-empty intersection, then we shall show that $X$  must be compact.
 <p></p>
@@ -508,6 +514,8 @@ <h1><a
 <p></p>
 
 <div class="box">Target</div>$\forall x\in K^c~~\exists U\in \tau~~x\in U\subseteq K^c.$
+<p></p>
+
 <div class="box">$\forall x$</div>Take any $x\in K^c.$  
 <p></p>
 Then $\forall y\in K~~x\neq y.$
@@ -645,7 +653,7 @@ <h1><a
 Let $\tau = \big\{~\{-\infty\}\cup{\mathbb N},~\{\infty\}\cup{\mathbb N},~\{-\infty,\infty\}\cup{\mathbb N}~\big\}\cup {\mathscr P}({\mathbb N}).$
 <p></p>
 Then $\tau$  is a topology on $X.$  
-<blockquote>
+<blockquote class="scrpt">
 <b>Because:</b>
 Clearly, $\phi, X\in\tau.$
 <p></p>

topo/topo3.rb

@@ -7,7 +7,7 @@
 @{<HEAD1>Compact</HEAD1>@}
 
 <EXM ref="2021.1h" paper="2021.1h"><EBODY>@{<E>
-If <M>\tau = \{\phi, \{a\},\{\a,b\},X\}</M> is a topology on <M>X =
+If <M>\tau = \{\phi, \{a\},\{a,b\},X\}</M> is a topology on <M>X =
 \{a,b,c\},</M> then <M>(X,\tau)</M> is
 <VL>
 <LI>compact and Hausdorff.</LI>
@@ -110,8 +110,8 @@
 </VL>
 </E>@}</EBODY><SOLN/>@{<E>Continuous</E>@}, kAraN er grAf A,nkte hale pen tulte hay ekmAtra
  @{<E>domain</E>@}-e f,nAk Achhe yekhAne, sekhAne. @{<E>Open</E>@}  nay, kAraN
- @{<M>(2.1,2.9)</M>@}  ekhAne @{<M>open,</M>@}  kintu er @{<E>image</E>@}  hala @{<M>\{2\},</M>@} 
- yexA moTei @{<M>\rr</M>@}-er madhye @{<E>open</E>@}  nay.</EXM>
+ @{<M>(2.1,2.9)</M>@}  ekhAne @{<E>open,</E>@}  kintu er @{<E>image</E>@}  hala @{<M>\{2\},</M>@} 
+ yeTA moTei @{<M>\rr</M>@}-er madhye @{<E>open</E>@}  nay.</EXM>
 
 <EXM ref="2023.1i" paper="2023.1i"><EBODY>@{<E>
 Let <M>(X,\tau)</M> be an uncountable compact space
@@ -182,11 +182,17 @@
 Hence <M>B_1\cap B_2\in \beta.</M>  So <M>\beta</M>  is a basis.
 
 <U>Second part</U>:
-Since the elements of <M>\beta</M>  are all complements of closed sets (wrt <M>\tau</M>), hence <M>bb\seq \tau.</M>  
+Since the elements of <M>\beta</M>  are all complements of closed sets (wrt <M>\tau</M>), hence <M>\beta\seq \tau.</M>  
 
 So <M>\tau'\seq \tau.</M>
 
 <U>Third part</U>:
+
+Let <M>\{U_ \alpha~:~ \alpha\in \Lambda\}\seq \tau'</M>  be a cover for <M>X.</M>  Pick some
+ <M>U</M>  in this cover (possible since this cover cannot be empty, as <M>X\neq\phi</M>). 
+Then <M>U^c</M>  is compact in <M>(X,\tau)</M>   and has <M>\{U_ \alpha~:~ \alpha\in \Lambda\}\setminus\{U\}</M>  as an open
+ cover (since <M>\tau'\seq\tau</M>). So we can extract a finite subcover <M>\{U_1,...,U_n\}</M>  for <M>U^c.</M>  Then <M>\{U_1,...,U_n\}\cup \{U\}</M> 
+ is a finite subcover for <M>X</M>  in <M>(X,\tau'),</M>  completing the proof.
 </WR>@}</EXM>
 
 
@@ -232,7 +238,7 @@
  So <M>\{F_ \alpha^c~:~ \alpha\in \Lambda\}</M>  is an open cover of <M>X.</M>
 
 Since <M>X</M>  is compact, there is a finite subcover, <M>\{F_{\alpha_1}^c,...,F_{\alpha_n}^c\}</M>  ie, 
-<M>\cup _{i=1}^n F_ \alpha_i^c = X,</M>  ie, <M>\cap _{i=1}^n F_ \alpha_i = \phi (\contra</M>  FIP<M>).</M>
+<M>\cup _{i=1}^n F_ {\alpha_i}^c = X,</M>  ie, <M>\cap _{i=1}^n F_ {\alpha_i} = \phi (\contra</M>  FIP<M>).</M>
 
 Conversely, if every family with FIP has non-empty intersection, then we shall show that <M>X</M>  must be compact.
 
@@ -255,6 +261,7 @@
 Shall show that <M>K</M>  is closed, ie, <M>K^c</M>  is open, ie
 
 <TGT>\forall x\in K^c~~\exists U\in \tau~~x\in U\seq K^c.</TGT>
+
 <FLL>x</FLL>Take any <M>x\in K^c.</M>  
 
 Then <M>\forall y\in K~~x\neq y.</M>

topo/topo3.rb.bkp

@@ -7,7 +7,7 @@
 @{<HEAD1>Compact</HEAD1>@}
 
 <EXM ref="2021.1h" paper="2021.1h"><EBODY>@{<E>
-If <M>\tau = \{\phi, \{a\},\{\a,b\},X\}</M> is a topology on <M>X =
+If <M>\tau = \{\phi, \{a\},\{a,b\},X\}</M> is a topology on <M>X =
 \{a,b,c\},</M> then <M>(X,\tau)</M> is
 <VL>
 <LI>compact and Hausdorff.</LI>
@@ -110,8 +110,8 @@ Then which of the following is true?
 </VL>
 </E>@}</EBODY><SOLN/>@{<E>Continuous</E>@}, kAraN er grAf A,nkte hale pen tulte hay ekmAtra
  @{<E>domain</E>@}-e f,nAk Achhe yekhAne, sekhAne. @{<E>Open</E>@}  nay, kAraN
- @{<M>(2.1,2.9)</M>@}  ekhAne @{<M>open,</M>@}  kintu er @{<E>image</E>@}  hala @{<M>\{2\},</M>@} 
- yexA moTei @{<M>\rr</M>@}-er madhye @{<E>open</E>@}  nay.</EXM>
+ @{<M>(2.1,2.9)</M>@}  ekhAne @{<E>open,</E>@}  kintu er @{<E>image</E>@}  hala @{<M>\{2\},</M>@} 
+ yeTA moTei @{<M>\rr</M>@}-er madhye @{<E>open</E>@}  nay.</EXM>
 
 <EXM ref="2023.1i" paper="2023.1i"><EBODY>@{<E>
 Let <M>(X,\tau)</M> be an uncountable compact space
@@ -182,11 +182,17 @@ Hence <M>B_1\cap B_2 = (K_1\cup K_2)^c = K^c,</M>  where <M>K=K_1\cup K_2</M>  i
 Hence <M>B_1\cap B_2\in \beta.</M>  So <M>\beta</M>  is a basis.
 
 <U>Second part</U>:
-Since the elements of <M>\beta</M>  are all complements of closed sets (wrt <M>\tau</M>), hence <M>bb\seq \tau.</M>  
+Since the elements of <M>\beta</M>  are all complements of closed sets (wrt <M>\tau</M>), hence <M>\beta\seq \tau.</M>  
 
 So <M>\tau'\seq \tau.</M>
 
 <U>Third part</U>:
+
+Let <M>\{U_ \alpha~:~ \alpha\in \Lambda\}\seq \tau'</M>  be a cover for <M>X.</M>  Pick some
+ <M>U</M>  in this cover (possible since this cover cannot be empty, as <M>X\neq\phi</M>). 
+Then <M>U^c</M>  is compact in <M>(X,\tau)</M>   and has <M>\{U_ \alpha~:~ \alpha\in \Lambda\}\setminus\{U\}</M>  as an open
+ cover (since <M>\tau'\seq\tau</M>). So we can extract a finite subcover <M>\{U_1,...,U_n\}</M>  for <M>U^c.</M>  Then <M>\{U_1,...,U_n\}\cup \{U\}</M> 
+ is a finite subcover for <M>X</M>  in <M>(X,\tau'),</M>  completing the proof.
 </WR>@}</EXM>
 
 
@@ -232,7 +238,7 @@ Let, if possible, <M>\cap _{\alpha\in \Lambda} F_ \alpha = \phi.</M>  Then, de M
  So <M>\{F_ \alpha^c~:~ \alpha\in \Lambda\}</M>  is an open cover of <M>X.</M>
 
 Since <M>X</M>  is compact, there is a finite subcover, <M>\{F_{\alpha_1}^c,...,F_{\alpha_n}^c\}</M>  ie, 
-<M>\cup _{i=1}^n F_ \alpha_i^c = X,</M>  ie, <M>\cap _{i=1}^n F_ \alpha_i = \phi (\contra</M>  FIP<M>).</M>
+<M>\cup _{i=1}^n F_ {\alpha_i}^c = X,</M>  ie, <M>\cap _{i=1}^n F_ {\alpha_i} = \phi (\contra</M>  FIP<M>).</M>
 
 Conversely, if every family with FIP has non-empty intersection, then we shall show that <M>X</M>  must be compact.
 
@@ -336,13 +342,13 @@ samAdhAne khAli @{<M>f</M>@}-er @{<E>domain</E>@}-TA ye @{<M>T_2</M>@}, eTukui l
 <EXM ref="2023.15a" paper="2023.15a"><EBODY>@{<E>
 Prove that a real valued continuous function <M>f</M> on a
 compact set <M>(X,\tau)</M> attains its greatest value. [2]
-</E>@}</EBODY><SOLN/></EXM>
+</E>@}</EBODY><SOLN/>@{<E>Standard result.</E>@}</EXM>
 
 <EXM ref="2023.16a" paper="2023.16a"><EBODY>@{<E>
 Prove that every closed and bounded interval of the real
 line<M>\rr</M> (i.e., <M>\rr</M> with the usual topology) is
 compact. [3]
-</E>@}</EBODY><SOLN/></EXM>
+</E>@}</EBODY><SOLN/>@{<E>Standard result.</E>@}</EXM>
 
 
 <EXM ref="2021.13a" paper="2021.13a"><EBODY>@{<E>

topo/topo3.rb.mrk

@@ -1 +1 @@
-345
+265