Solutions To Problems on General Topology[MAT404 Test]

Solution To Problems On General Topology[MAT404 Continuous assessment]

Photo Credit: Wikimedia.org

This is a solution to problems on general topology(MAT404 C.A Test), the problems include construction of topologies from a given set, construction of topology from a given basis, proof of a compact topological space, and proof of a continuous topological function.
You can refer back to my previous posts on topology by clicking on any of the following links below:
(i) How to identify the open, closed and clopen sets in a topological space[Topology made simple]

(ii)  The Co-finite Topology on X is discrete if and only if X is finite[Theorem and Proof]

(iii)The Cofinite or Finite-Closed Topology[Definition and Examples]

(iv) General Topology[Discrete and Indiscrete Topology With Examples]



(1)  Lists all possible topologies on the set $X=\{a,b,c\}$.

(2)  Consider $S=\{\{a,b\},\{a,c\}\}$, where $X=\{a,b,c\}$. Obtain the topology generated by $S$ $(i.e \mathcal{T}(S))$. Is  $S$ a basis?

(3)  Let $A$ be a closed subset of a compact topological space $X$. Show that $A$ is compact.

(4)  Prove that $f$ is continuous if for each $x\in{X}$ and each neighbourhood $V$ of $f(x)$ there is a neighbourhood $U$ of $x$ such that $f(U)\subseteq{V}$.


Solutions

(1) There are $29$ possible topologies for the set $X$.
$X=\{a,b,c\}$
$\mathcal{T}_1=\{\phi,X\}$
$\mathcal{T}_2=\{\phi,X,\{a\}\}$
$\mathcal{T}_3=\{\phi,X,\{b\}\}$
$\mathcal{T}_4=\{\phi,X,\{c\}\}$
$\mathcal{T}_5=\{\phi,X,\{a\},\{b\},\{a,b\}\}$
$\mathcal{T}_6=\{\phi,X,\{a\},\{c\},\{a,c\}\}$
$\mathcal{T}_7=\{\phi,X,\{b\},\{c\},\{b,c\}\}$
$\mathcal{T}_8=\{\phi,X,\{a\},\{a,b\},\{a,c\}\}$
$\mathcal{T}_9=\{\phi,X,\{b\},\{a,b\},\{b,c\}\}$
$\mathcal{T}_{10}=\{\phi,X,\{c\},\{a,c\},\{b,c\}\}$
$\mathcal{T}_{11}=\{\phi,X,\{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\}\}$
$\mathcal{T}_{12}=\{\phi,X,\{a,b\}\}$
$\mathcal{T}_{13}=\{\phi,X,\{b,c\}\}$
$\mathcal{T}_{14}=\{\phi,X,\{a,c\}\}$
$\mathcal{T}_{15}=\{\phi,X,\{a\},\{b,c\}\}$
$\mathcal{T}_{16}=\{\phi,X,\{b\},\{a,c\}\}$
$\mathcal{T}_{17}=\{\phi,X,\{c\},\{a,b\}\}$
$\mathcal{T}_{18}=\{\phi,X,\{a\},\{a,b\}\}$
$\mathcal{T}_{19}=\{\phi,X,\{a\},\{a,c\}\}$
$\mathcal{T}_{20}=\{\phi,X,\{b\},\{b,c\}\}$
$\mathcal{T}_{21}=\{\phi,X,\{b\},\{a,b\}\}$
$\mathcal{T}_{22}=\{\phi,X,\{c\},\{a,c\}\}$
$\mathcal{T}_{23}=\{\phi,X,\{c\},\{b,c\}\}$
$\mathcal{T}_{24}=\{\phi,X,\{a\},\{b\},\{a,b\},\{a,c\}\}$
$\mathcal{T}_{25}=\{\phi,X,\{a\},\{c\},\{a,c\},\{a,b\}\}$
$\mathcal{T}_{26}=\{\phi,X,\{b\},\{c\},\{b,c\},\{a,b\}\}$
$\mathcal{T}_{27}=\{\phi,X,\{b\},\{c\},\{a,c\},\{b,c\}\}$
$\mathcal{T}_{28}=\{\phi,X,\{a\},\{c\},\{a,c\},\{b,c\}\}$
$\mathcal{T}_{29}=\{\phi,X,\{a\},\{b\},\{a,b\},\{b,c\}\}$


(2)  We generate the topology $\mathcal{T}(S)$ by the members of $S$ by taking the union of the finite intersection of members of $S$, so $\mathcal{T}(S)=\{\{a,b\},\{a,c\},\{a\},X,\phi\}$, now we see $X\in{\mathcal{T}}(S)$.
Hence, we conclude that $S$ is not a basis since not every members of $\mathcal{T}(S)$ are union of members of $S$. Since the set $\{a\}$ can not be written as a union of members of $S$.

Next is to show if $S$ is a basis by using the conditions of basis of topological subspaces, by using this conditions you also prove if $B$ is a basis or not.
Click the link below to see proofs

(3)  Proof: Let $A$ be a closed subset of a compact space $(X,T)$. Let $U_i\in{\mathcal{T}}$,$i\in{I}$ be any open covering of $A$. Then  $X\subseteq(\bigcup_{i\in{I}}{U_i})\cup(X\backslash{A})$.
That is $U_i$,$i\in{I}$ together with the open set $X\backslash{A}$  is an open covering of $X$. Therefore there exists a finite sub-covering $U_1,U_2,...,U_{i_k}$, $X\backslash{A}$, [if $X\backslash{A}$ is not in the finite sub-covering then we can include it and still have a finite subcovering of $X$.]
So $X\subseteq{U_{i_1}\cup{U_{i_2}\cup...\cup{U_{i_k}}\cup(X\backslash{A})}}$ therefore
 $A\subseteq{U_{i_1}\cup{U_{i_2}\cup...\cup{U_{i_k}}\cup(X\backslash{A})}}$.
which clearly implies $A\subseteq{U_{i_1}\cup{U_{i_2}\cup...\cup{U_{i_k}}}}$ Since $A\cap(X\backslash{A})=\phi$. Hence $A$ has a finite subcovering and so is compact. $\Box$



(4)  Proof: Let $f$ is continuous. Let $x\in{\mathbb{R}}$ and let $U$ be any open set containing $f(x)$. Then there exists real numbers $c$ and $d$ such that $f(x)\in(c,d)\subseteq{U}$ . let $\epsilon$ be the smaller of the two numbers $d-f(x)$ and $f(x)-c$, so that $(f(x)-\epsilon,f(x)+\epsilon)\subseteq{U}$. As the mapping $f$ is continuous there exists a $\delta>0$ such that $f(x)\in(f(x)-\epsilon,f(x)+\epsilon)$ for all $x\in(x-\delta,x+\delta)$, Let $V$ be the open set $(x-\delta,x+\delta)$. Then $x\in{V}$ and $f(V)\subseteq{U}$ as required.

Conversely, assume that for each $x\in{\mathbb{R}}$ and each open set $U$ containing $x$ such that $f(V)\subseteq{U}$. Now we have to show that $f$ is continuous. Let $x\in{\mathbb{R}}$ and $\epsilon$ be any positive real number. Put $U=(f(x)-\epsilon,f(x)+\epsilon)$, $x$ such that $f(V)\subseteq{U}$. As $V$ is an open set containing $x$, there exists real numbers  $c$ and $d$ such that $x\in(c,d)\subseteq{U}$, put $\delta$ to be equal to the smaller of the two numbers $d-x$ and $x-c$, so that $(x-\delta,x+\delta)\subseteq{V}$. Then for all $x\in(x-\delta,x+\delta), f(x)\in{f(V)}\subseteq{U}$, as required. Hence $f$ is continuous.$\Box$


NOTE: Please kindly use the comment box or contact form on the homepage of this blog for any comment, corrections and observations, because i cannot guarantee a 100% accuracy due to typographic errors and others errors made during the process of solving. 
For comments on mathematical equations,  please use laTex or MathJax to write equations.