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.

Consider the real line \(\mathbb{R}\) with usual topology. Then the set \(\mathbb{Q}\) of rationals is not closed. This follows since any neighbourhood of an irrational number contains rationals. If we enumerate \(\mathbb{Q}\) as a sequence \(\{r_n\}\) then\(\mathbb{Q}=\cup_n\{r_n\}\), which shows that countable union of closed sets need not be closed.

Show that \(X\) is Hausdorff iff the diagonal \(\Delta :=\{(x,x) \in X \times X \}\) is closed in \(X \times X\).

Show that every topological space with order topology is Hausdorff.

What are all closed subsets of R with VIP topology (resp. outcast topology) ?

Show that the only non-empty subset of \(\mathbb{R}\) which is open as well as closed is \(\mathbb{R}\).

Given an example of a non-empty subset of \(Y:= \mathbb{R} \backslash\{0\}\) which is open as well as closed in the subspace topology on \(Y\) .

 Let \(f: \mathbb{R}^n \rightarrow \mathbb{R}\) be a continuous function in the variables \(x_1,...,x_n\). Show that the zero set \(Z(f):=\{x\in \mathbb{R^n}:f(x) =0\}\) is a closed subset of \(\mathbb{R^n}\) .