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 \(\mathbb{R^n}\) .

Let \(X\) be an ordered set with order topology. Note that the complement of [a, b] in \(X\) is open. Thus \(\overline{(a,b)} \subseteq [a,b]\).