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 \(\overline{A} \cap Y\).

\(x \in \overline{A}\) iff either \(x \in A\) or \(x\) is a cluster point of \(A\).

If \(X\) is a Hausdorff space then every neighbourhood of a cluster point of \(A\) contains infinitely many points from \(A\).

There is no function \(g:(0,1) \rightarrow \mathbb{R}\) which is continuous on rationals and discontinuous on irrationals.

\(\mathbb{R^n}\) is homeomorphic to \(\mathbb{R}\) iff \(n=1\).

Let \(f: A \rightarrow \Pi_{i=1}^m X_i\) be given by \(f(a)=(f_1(a), \cdots , f_m(a))\), where the functions \(f_i :A \rightarrow X_i (i=1,\cdots ,m)\) are given. Then \(f\) is continuous iff each \(f_i\) is continuous.

Let \(f:A \rightarrow \Pi_\alpha X_\alpha\) be given by \(f(a)=(f_\alpha(a))\), where the functions \(f_\alpha :A \rightarrow X_\alpha\) are given. Suppose \(\Pi_\alpha X_\alpha\) has product topology. Show that \(f\) is continuous iff each \(f_\alpha\) is continuous.

Let \(\{x_n=(x_{n1},x_{n2}, \cdots,) \}\)be a sequence in the product space \(\Pi_{\alpha \in I} X_\alpha\) with product topology. Then the sequence \(\{x_n\}\) converges to\(x+(x_1,x_2, \cdots,)\in \Pi_\alpha X_\alpha\) iff for every positive integer \(m, \{\pi_m(x_n)=x_{nm} \}\) converges to \(x_m\).