Let \((X,\Omega )\) be a topological space with the property that any two disjoint non empty closed subsets of \(X\) can be separated by a continuous function. Show that for any disjoint non-empty closed subsets \(A\) and \(B\) of \(X\), there exist disjoint open sets \(U\) and \(V\) such that \(A \subseteq U\) and \(B \subseteq V\).

Consider the topological space \(\mathbb{R}_K\) . Show that there are no disjoint open sets \(U\) and \(V\) of \(\mathbb{R}_K\) such that \(\{0\} \subseteq U\) and \(K \subseteq V\). In particular, \(\mathbb{R}_K\) is Hausdorff but not normal.

Whether or not any two disjoint non-empty closed subsets of a normal space \(X\) can be separated by a continuous function ?

Suppose \(X\) admits a family \(\{U_r\}_{r \in \mathbb{Q}}\) of nested neighbourhoods. Define\(f: X \rightarrow [0,1]\) by \(f(x) =inf \mathbb{Q}(x)\). Verify the following:

(1) \(f(a)=0\) for every \(a \in A\).

(2)\(f(b) =1\) for every \(b \in B\).

(3) \(f(x) \le r\)r for any \(x \in \overline{U}_r\),

(4) \(f(x) \ge r\) for any \(x \notin U_r\).

Let \(R\) be a commutative ring. Prove that \(HOM_R(R,R)\)) and \(R\) are isomorphic as rings.

 Given closed non-empty disjoint subsets \(A\) and \(B\) of a normal space \(X\), there exists a continuous function \(f:X \rightarrow [0,1]\) such that \(f|_A=0\) and \(f|_B=1\).

 Let \(X\) be a compact Hausdorff space without any isolated points. Consider the vector space \(C(X)\) of continuous functions \(f:X \rightarrow \mathbb{R}\).

Show that \(C(X)\) is infinite-dimensional.

Let \(X= {\Pi^{\infty}_{n=1}} \{0,1\}\). Show that \(X\) is not compact in the box topology. Whether \(X\) is compact in the product topology ?

 An arbitrary product of compact spaces is compact in the product topology.

If a family \(\mathcal{F}\) of subsets of \(X\) has finite intersection property then for each \(\alpha \in J ,\cap_{A \in \mathcal{F} }{\pi_\alpha} (A)\) is non-empty.

Let \((X,\Omega)\) be a topological space with topology \(\Omega\). Let \(A,Y\) be subsets of \(X\) such that \(A \subseteq Y\). Then

(1) \(A\) is closed in the subspace topology on \(Y\) iff \(A=B \cap Y\) for some closed subset \(B\) of \(X\).

(2) The closure of \(A\) in the subspace topology on \(Y\) equals \(\overline{A} \cap Y\).