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 \(x_m\).

\(\Omega_\mathbb{B_p} \subseteq \Omega_\mathbb{B_u} \subseteq \Omega_\mathbb{B_b} \) with strict inclusions if \(I\) is infinite.

Let \(X\) and \(Y\) be compact spaces. Assume further that \(Y\) is Hausdorff. Let \(f:X \rightarrow Y\) be a continuous surjection. Define the equivalence relation \(\sim\) on \(X\) by \(x_1 \sim x_2\) iff \(f(x_1)=f(x_2)\). Then \(g:X / \sim \rightarrow Y\) given by \(g(|x|)=f(x)\) is a well-defined homeomorphism.

Show that in the ring \(\mathbb{Z}[\sqrt{-3}]\) the gcd of \(4\) and \(2+2\sqrt{-3}\) does not exist.