Let $X$ be a first countable space and let $A$ be a subset of $X$. Then the following are true:

(1) (Sequence Lemma) $A$ point $x \in \overline{A}$ if and only if there is a sequence of points of A converging to $x$.

(2) (Continuity Versus Sequential Continuity) Let $f: X \rightarrow Y$ . Then $f$ is continuous if and only if $f$ is sequentially continuous.

A metric space $(X,d)$ is second countable if any one of the following holds true:

(1) $X$ has a countable dense subset,

(2) $X$ is compact.

Show that $\mathbb{R_l}$ is not metrizable.

If $X$ is second countable then every open covering of $X$ contains a countable subcover.

Let $(X,d)$ be a metric space with metric. Let $A$ be a subset of $X$, and for $x \in X$, let $d(x,A):=inf\{d(x,a):a \in A \}$. Show that $d(x,A)$ is a continuous function of $x$.

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 $f|_B=1$.