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.

Let X = \(\mathbb{R}\) with usual topology. Then the closure of (0, 1) in \(\mathbb{R}\) equals [0, 1] as [0, 1] is the smallest closed set containing (0, 1). However, the closure of (0, 1) in the subspace topology on [0, 1) equals [0, 1) as [0, 1) is the smallest closed in the subspace topology that contains (0, 1).

Consider the subset A := [0, 1) ∪ {2} of the real line with usual topology. Note that 1 is a cluster point of A but 2 is not a cluster point of A.

What are all cluster points of {q} in VIP topology with base point p on \(\mathbb{R}\) ?

Let \((X,\Omega_i)(i=1,2)\) be two topological spaces and consider the identity function \(id\) from \(X\) onto itself. Show that \(\Omega_1\) is finer than \(\Omega_2\) iff the identity function \(id\) is continuous.