All Questions page - 60

 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.

Show that the topological space \(\mathbb{N}\) of positive numbers with topology generated by arithmetic progression basis is Hausdorff.

Let \(X:=\mathbb{R}\). Consider the pairs of bases:

(1)\(\mathbb{B_1}:=\{(a,b) \subseteq \mathbb{R} :a,b \in \mathbb{R} \}\) and \(\mathbb{B_2}:=\{(a,b) \subseteq \mathbb{R} :a,b \in \mathbb{Q} \}\).

(2) \(\mathbb{B_1}:=\{[a,b) \subseteq \mathbb{R} :a,b \in \mathbb{R} \}\) and \(\mathbb{B_2}:=\{[a,b) \subseteq \mathbb{R} :a,b \in \mathbb{Q} \}\).

(3) \(\mathbb{B_1}:=\{[a,b) \subseteq \mathbb{R} :a,b \in \mathbb{R} \}\) and \(\mathbb{B_2}:=\{(a,b) \subseteq \mathbb{R} :a,b \in \mathbb{R} \}\).

Do they generate comparable topologies ? If so then do they generate the same topology ?

([1, H. F¨urstenberg]) Consider \(\mathbb{N}\) with the arithmetic progression topology. Verify the following:

(1) For a prime number p, the basis element {np : n ≥ 1} is closed.

(2) There are infinitely prime numbers.

Show that the open unit disc is open in the product topology on \(\mathbb{R^2}\) . Show further that it is not of form \(U \times V\) for any open subsets \(U\) and \(V\) of \(\mathbb{R}\).

 Consider the space \(\mathbb{R}_l \times \mathbb{R}\) with product topology, where \(\mathbb{R}_l\) denotes  the real line with lower limit topology. The basis for the product topology consists of \(\{(x,y) \in \mathbb{R^2} : a\le x<b ,c<y<d \}.\)​​​​​​​

Let \(X_1\) denote the topological space \(\mathbb{R}\) with discrete topology and let \(X_2\) be \(\mathbb{R}\) with usual topology. Then the product topology \(\Omega\) on \(\mathbb{R} \times \mathbb{R}\) is nothing but the dictionary order topology on \(\mathbb{R}^2\) . Since the basis for the product topology on \(\mathbb{R} \times \mathbb{R}\) is given by \(\{\{x_1\} \times (a,b):x_1,a,b \in \mathbb{R} \}\), any open set in the dictionary order topology is union of open sets in the product topology. We also note that the product topology \(\Omega\) is finer than the usual topology on \(\mathbb{R}^2\) . In fact, any basis element \((a,b) \times (c,d)\) of the usual topology can be  expressed as the union \(\cup_{a<x<b}\{x\} \times \left(c,d \right)\)  of open sets \(\{x\} \times \left( c,d \right)\) in the product toplogy   \(\Omega\) .

Any interval \((x,y)\) in a simply ordered set is convex: If \(a,b \in (x,y)\) then \(x<a<b<y\) and hence \((a,b) \subseteq (x,y)\).

Consider the subset [0, 1]×[0, 1] of \(\mathbb{R^2}\) with dictionary order. Then 0 × 0, 1 × 1 belong to [0, 1] × [0, 1], however, \(\frac{1}{2} \times 2 \in (0 \times 0,1 \times1)\)does not belong to [0, 1] × [0, 1]. The first quadrant is not a convex subset of \(\mathbb{R^2}\) . However, the right half plane is convex.

If \(X\) is Hausdorff then show that the complement of any finite set is open.