(Co-countable Topology) For a set \(X\), define \(\Omega\) to be the collection of subsets \(U\) of \(X\) such that either \(U=\emptyset\) or \(X \backslash U\) is countable. Show that \(\Omega\) is a topology on \(X\).

Let \(\Omega\) be the collection of subsets \(U\) of \(X:=\mathbb{R}\) such that either \(X \backslash U= \emptyset\) or \(X\backslash U\) is infinite. Show that \(\Omega\) is not a topology on \(X\).

 Show that the usual topology is finer than the co-finite topology on \(\mathbb{R}\).

Show that the usual topology and co-countable topology on \(\mathbb{R}\) are not comparable.

The collection \(\{(a,b) \subseteq \mathbb{R}:a,b \in \mathbb{Q} \}\) is a basis for a topology on \(\mathbb{R}\).

Show that collection of balls (with rational radii) in a metric space forms a basis.

(Arithmetic Progression Basis) Let \(X\) be the set of positive integers and consider the collection \(\mathbb{B}\) of all arithmetic progressions of positive integers. Then \(\mathbb{B}\) is a basis. If \(m \in X\) then \(B:=\{m+(n-1)p\}\) contains m. Next consider two arithmetic progressions \(B_1=\{a_1+(n-1)p_1\}\)and \(B_2=\{a_2+(n-1)p_2\}\)containing an integer \(m\). Then\(B:=\{m+(n-1)(p)\}\) does the job for \(p:=lcm\{p_1,p_2\}\).

Show that the topology \(\Omega{_B}\) generated by the basis \(\mathbb{B}:=\{(a,b) \subseteq \mathbb{R}:a,b \in \mathbb{Q}\}\) is the usual topology on \(\mathbb{R}\).

The collection \(\{[a,b) \subseteq \mathbb{R}:a,b \in\mathbb{R}\}\)is a basis for a topology on \(\mathbb{R}\). The topology generated by it is known as lower limit topology on \(\mathbb{R}\).

Note that \(\mathbb{B}:=\{p\} \cup \{\{p,q\}: q\in X,q\neq p \}\) is a basis. We check that the topology \(\Omega_{B}\) generated by \(\mathbb{B}\) is the VIP topology on \(X\). Let \(U\) be a subset of \(X\) containing \(p\). If \(x \in U\) then choose \(B=\{p\}\)if \(x=p\), and \(B=\{p,x\}\) otherwise. Note further that if \(p\notin U\)then there is no \(B \in \mathbb{B}\) such that \(B \subseteq U\). This shows that \(\Omega_{\mathbb{B}}\) is precisely the VIP topology on \(X\).

Show that the topology generated by the basis \(\mathbb{B}:={X}\cup\{\{q\}:q\in X,q\neq p \}\) is the outcast topology.