Continuous function and residual sets $A$ is a residual set If the complement of the set $A$ is dense


*

*Show that Boundary of any set is the union of two residual sets


*Show that the union of nowhere set and residual set is residual set.


*In $\mathbb Z^+$ , define $ U$ to be open in $\tau$ if satisfies the condition : $n \in U$ $\Rightarrow$ every divisor of $n$ belongs to $U$ then $(\mathbb Z^+, \tau)$ is a topological space and Show that $ f : \mathbb Z^+ \rightarrow \mathbb Z^+$ is continous iff (m divides n)$\Rightarrow$ $(f(m)$ divides $f(n) )$

I have tried


*Let $R$ be a residual set and $A$ is a nowhere dense , then $A^c$ is dense

we have to show that $(A \cup R)^c$ is dense. Let $x \in X$. For each$r >0$, $S_r(x) \cap A^c \neq \phi$ and $S_r(x) \cap R^c \neq \phi$.
Please tell me how will show that $S_r(x) \cap (R^c\cap A^c) \neq \phi$.


*I can easily prove that $(\mathbb Z^+, \tau)$ is a topological space and if  ($m$ divides $n$)$\Rightarrow$ $(f(m)$ divides $f(n) )$, then $f$ is continuous.

Please tell me how to show that if $f$ is continuous then (m divides n) $\Rightarrow $ $(f(m)$ divides $f(n) )$.
Any help would be appreciated Thank you
 A: For the first problem let $B=\operatorname{bdry}A$, and consider the sets $A\setminus\operatorname{int}A$ and $(\operatorname{cl}A)\setminus A$.
In your approach to the second problem you’re assuming that $X$ is a metric space; this does not seem to be justified by the statement of the problem. Fortunately, it’s not necessary. Note that
$$X\setminus(R\cup A)\supseteq X\setminus(R\cup\operatorname{cl}A)=(X\setminus R)\cap(X\setminus\operatorname{cl}A)\;,$$
where $X\setminus R$ is dense in $X$, and $X\setminus\operatorname{cl}A$ is not only dense in $X$, but also open. To finish the proof, show that the intersection of a dense subset of $X$ with a dense open subset of $X$ is dense in $X$.
In the third problem let $B_n=\{k\in\Bbb Z^+:k\mid n\}$; show that $B_n\in\tau$, and that $$B_n=\bigcap\{V\in\tau:n\in V\}\;,$$ so that $B_n$ is actually the smallest open set containing $n$. Now suppose that $f:\Bbb Z^+\to\Bbb Z^+$, that $m,n\in\Bbb Z^+$ are such that $m\mid n$, and that $f(m)\nmid f(n)$. Then $B_{f(n)}$ is an open nbhd of $f(n)$ not containing $f(m)$. Show that there is no open nbhd $V$ of $n$ such that $f[V]\subseteq B_{f(n)}$, and conclude that $f$ is not continuous at $n$.
