Semi-Totally Semi-Continuous Function Researchers in topology defined the following:
Definition 1. Let $X$ be a topological space and $A⊂X$. Then $A$ is semi-open iff $A⊂cl[int(A)]$ (closure of the interior of $A$). The set $A$ is semi-closed iff its complement is semi-open.
Definition 2. Let $X$ and $Y$ be topological spaces. A function $f:X→Y$  is semi-totally semi-continuous on $X$ if for every semi-open set $B$ in $Y$, the set $f^{−1}(B)$ is both semi-open and semi-closed in $X$.
In standard topology books, the definition of continuous function is given as follows:
Definition 3. Let $X$ and $Y$ be topological spaces. A function $f:X→Y$  is continuous on $X$ if for every open set $B$ in $Y$, the set $f^{−1}(B)$ is open  in $X$.
Question: Do the Definitions 2 and 3 have relationships at all?
Thanks for any help. I tried using the definitions, but it seems that they do not have relationship at all.
 A: Example 1. Let $X=[-1,0]\cup\{1/n:n\in\Bbb N\}$ endowed with the topology inherited from the set of reals, $Y=\{0\}\cup \Bbb N$ endowed with the discrete topology, and $f:X\to Y$ be the map such that $f(x)=0$ for each $x\le 0$ and $f(1/n)=n$ for each natural $n$. Since the set $\{0\}$ is open in the space $Y$ and a set $\{f^{-1}(0)\}=[0,1]$ is not open in the space $X$, the map $f$ is not continuous. Now let $A$ be any subset of the space $Y$. Then both sets $f^{-1}(A)$ and its complement is a union of a some (may be empty) family of isolated points or a union of such a family with the set $[0,1]$. It is easy to check that all such sets are semi-open. Thus the map $f$ is semi-totally semi-continuous.
Definition 4. Let $X$ and $Y$ be topological spaces. A function $f:X\to Y$  is irresolute continuous on $X$ if for every semi-open set $B$ in $Y$, the set $f^{−1}(B)$ is semi-open.
Corollary. Each irresolute continuous map is semi-totally semi-continuous.
Example 2. Let $X=[-2,-1]\cup (0,1]$, $Y=[-1,1]$ endowed with standard topology, and $f:X\to Y$ be a map such that $f(x)=x+1$  if $x<0$ and $f(x)=x$ if $x\ge 0$. It is easy to check that the map $f$ is continuous. Let $A=[0,1]\subset Y$. Since 
$A=\operatorname{cl}\operatorname{int} A$, the set $A$ is semi-open. From the other hand, $B=f^{-1}(A)=\{-1\}\cup (0,1]\not\subset \operatorname{cl}\operatorname{int} B=(0,1]$. Therefore the set $B$ is not semi-open. Thus
the map $f$ is not irresolute continuous, and hence not semi-totally semi-continuous.
Example 3. Let $X$ be the set of reals endowed with standard topology, $Y$ be the reals endowed with Sorgenfrey arrow topology (generated by the base consisting of all semi-intervals $[a,b)$, $a<b$), and $f:X\to Y$ is the identity map. Since the set $[0,1)$ is open in $Y$ but $f^{-1}([0,1))=[0,1)$ is not open in $X$, the map $f$ is not continuous. Now let $A\subset Y$ be an arbitrary semi-open set. Then 
$$f^{-1}(A)=A\subset \operatorname{cl}_Y\operatorname{int}_Y A = \mbox{ (as easy to check) }
\operatorname{cl}_Y\operatorname{int}_X A\subset \operatorname{cl}_X\operatorname{int}_X A,$$ 
so $f^{-1}(A)$ is a semi-open subset of the space $X$. Thus the map $f$ is irresolute continuous.
