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 $\overline{(a,b)} \subseteq [a,b]$.

Consider the topological space $\mathbb{R_l}$ . Then the closure of (non-closed set) (0, √ 2) equals [0, √ 2). This follows from the observation that R \ [0, √ 2) = (−∞, 0) ∪ [ √ 2, ∞) is open in $\mathbb{R_l}$ . If one replaces the lower limit topology by the topology Ω generated by the basis {[a, b) : a, b ∈ Q} then the closure of (0, √ 2) in Ω equals [0, √ 2]. This provides another verification of the fact that Ω and the lower limit topology are different.