\(X\)Let X be a topological space with discrete topology and \(Y\) be any topological space. Show that any function \(f:X \rightarrow Y\) is continuous.

Let \((X_i, \Omega_i)\) be two topological spaces with topology \(\Omega_i(i=1,2)\)and let \(x_0 \in X_2\). If \(X_2\) is Hausdorff then show that for any continuous function \(f:X_1 \rightarrow X_2\), the set   \(C:= \{x \in X_1 :f(x)=x_0 \}\)is closed in \(X_1\).

If \(f:X \rightarrow Y\) is continuous and a sequence \(\{x_n\}\) in \(X\) converges to \(x \in X\), show that the sequence \(\{f(x_n)\}\) in \(Y\) converges to \(f(x)\).

Let \(E\) be a non-empty proper subset of the topological space \(X\). Consider the characteristic function \(\chi _E\) of \(E\). Show that \(\chi _E\) is continuous iff \(E\) is a closed and open subset of \(X\).

Consider topological spaces \(X_i=(X,\Omega_i),i=1,2\) such that \(\Omega_1\) is finer than \(\Omega_2\). Show that \(f:X_1 \rightarrow Y\) is continuous if so is\(f:X_2 \rightarrow Y\).

Let \(X\) be a topological space and \(Y\) be a topological space with ordered topology. Let \(f,g:X \rightarrow Y\) be continuous functions. Show that the set \(U:=\{x \in X:f(x)>g(x) \}\) is open in \(X\).

Prove: Composition of continuous functions is continuous.

Let \(X:=[0,1]\) and \(A_0:=\{0\} , A_n:=[1/n,1]\). Then the function \(f:[0,1] \rightarrow \mathbb{R}\)given by \(f(0)=1,f(x)=1/x(0<x\le1)\) is discontinuous at 0. However, \(f|_{A_n}\) is continuous for every \(n \ge 0\).

Show that the function \(g: (0,1) \rightarrow \mathbb{R}\) given below is continuous on irrationals and discontinuous on rationals:

\(g(x) = \frac{1}{q} \; if \; x \in \mathbb{Q} \; \cap \; (0,1)\)and \(x = \frac{p}{q}\) in reduced

form = 0 otherwise.

We say that a set is a \(G_{\delta}\) set if it is countable intersection of open sets.

Show that a bijective continuous map from a compact metric space into a metric space sends closed sets to closed sets, and hence it is a homeomorphism.

Define \(f: \mathbb{R} \rightarrow [0,1]\) by \(f(x) =x\) if \(|x| \le 1,\) and \(f(x)= \frac{1}{|x|}\) if \(|x| \ge1\). Then f is continuous on \(\mathbb{R}\). Note that \(f\) is onto but not one-one.