General topology problems and solutions

Solution To Problems On General Topology[MAT404 Continuous assessment]

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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. 

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