0$ there exists a Cantor set $C \subset (x_o - \varepsilon , x_o)$ such that $f(C) \subset K.$ \end{enumerate} \end{df} In an analogous manner we define the notion of the right {\civ} at a point $x_o$ (we write $x_o \in {\rm {\bf {\sf CIVP}}}^+(f)$). The function that has the left and the right {\civ} $\;$at a point $x_o$ is said to have the {\civ} $\;$at the point $x_o.$ Then we write $x_o \in {\rm {\bf {\sf CIVP}}}(f).$ If in the above definitions we assume additionally that $f_{|C}$ is continuous than we obtain definitions the left, the right and the bilateral strong {\civ} $\;$at the point $x_o.$ Then we shall write $x_o \in {\rm {\bf {\sf SCIVP}}}^-(f),$ $x_o \in {\rm {\bf {\sf SCIVP}}}^+(f)$ and $x_o \in {\rm {\bf {\sf SCIVP}}}(f),$ respectively. One can observe that if $f$ is continuous at a point $x_o$ (from one or both sides), then it has the {\civ} and the strong {\civ} at $x_o$ from the same side. Now we shall prove that the local and global properties of the {\civ} are compatible. \begin{tw} A function $f:I \to \re$ has the {\civ} if and only if $f$ has the {\civ} at every point of the interval I. \end{tw} \noindent{\sc Proof. } Let us assume that $f$ has the {\civ} and suppose that there exists a point $x_o$ such that $x_o \not\in {\rm {\bf {\sf CIVP}}}^+(f).$ It is easy to see that $f(x_o) \in C^+(f,x_o).$ Then there exists a Cantor set K such that \begin{enumerate} \item $K \subset (a,b) \subset C^+(f,x_o)$ \item there exists positive number $\varepsilon$ such that $f(C)$ is not contained in $K$ for every Cantor set $C$ included in the interval $(x_o, x_o + \varepsilon).$ \end{enumerate} Let $x_1, x_2 \in (x_o, x_o + \varepsilon)$ be chosen in such a way that $f(x_1) < \inf K$ and $f(x_2) > \sup K.$ Then $K \subset (f(x_1),f(x_2)).$ Therefore there exists a Cantor set $C \subset (x_1, x_2)$ such that $f(C) \subset K$ (because $f$ has the {\civ}). A contradiction. Now we assume that $f$ does not have the {\civ} and $x \in {\rm {\bf {\sf CIVP}}}(f)$ for all $x \in I.$ There exist an interval $(x,y) \subset I$ and a Cantor set $K \subset (f(x),f(y))$ such that $f(C) \not\subset K$ for every Cantor set $C \subset (x,y).$ Let us consider two possibilities. \begin{enumerate} \item There exists a point $x_o \in (x,y)$ such that $ \wn C^+(f,x_o) \cap K \neq \emptyset$ or $\wn C^-(f,x_o) \cap K \neq \emptyset.$ \item $\wn C^+(f,x_o) \cap K=\emptyset$ and $\wn C^-(f,x_o) \cap K = \emptyset$ for each $x_o \in (x,y).$ \end{enumerate} where $\wn A$ denotes the interior of the set $A.$ In the first case (for example let $\wn C^+(f,x_o) \cap K \neq \emptyset$) there exist points $a,b \in C^+(f,x_o)$ such that $K_1=(a,b) \cap K$ is a Cantor set. There exists a Cantor set $C \subset (x_o,y)$ such that $f(C) \subset K_1 \subset K.$ A contradiction. Let us consider the second case. Let $t \in K$ be a point of bilateral accumulation of the set K. By Theorem 1, there is $x_o \in f^{-1}(t) \cap (x,y).$ Then $C(f,x_o)= \{ f(x_o)\};$ so $f$ is continuous at $x_0.$ Let $K_1$ be the set of all points of bilateral accumulation of the set $K.$ Then $f^{-1}(K_1) \cap (x,y) \subset C(f),$ where $C(f)$ denotes the set of all points of continuity of the function $f.$ The set $K_1$ is a Borel set of the power of continuum, $f^{-1}(K_1) \cap (x,y)$ is a Borel set in $C(f)$ and so $f^{-1} (K_1) \cap (x,y)$ is a Borel set of power of continuum. Then there exists a Cantor set $C \subset f^{-1} (K_1) \cap (x,y).$ So $f(C) \subset K_1 \subset K.$\qed \medskip In the similar way we can prove the following. \begin{tw} A function $f : I \to \re$ has the strong \civ $\,$if and only if $f$ has the strong \civ $\,$at every point of the interval $I.$ \end{tw} Now we start to discuss the problem of local characterization of the {\rm {\bf {\sf WCIVP}}} property. First we may say that a function $f:I \to \re$ has a (local) property ${\cal W}$ at a point $x_o ,$ if for each neighborhood $U$ of the point $x_o $ every extension of the function $f_{|U}$ has this property at $x_o .$ We shall write, then $ f \in {\cal W}(x_o ).$ Let us consider the following example. Let $f:I \to (0,1)$ be a function from the class {\rm {\bf {\sf WCIVP}}}, which transforms each subinterval $(a,b)$ of $I$ onto the interval $(0,1).$ This function takes on its supremum at no point of the interval $I.$ Let $x_o$ be a given point from the interval $I.$ Consider the function defined by $$ g(x) = \begin{cases} f(x) & \text{ for } x \neq x_o , \\ 2 & \text{ for } x = x_o . \end{cases}$$ The function $g$ has also {\rm {\bf {\sf WCIVP}}} property. \begin{tw} There exists no local property ${\cal W}(x)$ which characterizes a property ${\cal W},$ i.e. such that $ f \in {\rm {\bf {\sf WCIVP}}} $ if and only if $f \in {\cal W}(x) $ for each $x \in I.$ \end{tw} \noindent{\sc Proof. } If there exists a local property ${\cal W}(x)$ which characterizes functions with {\rm {\bf {\sf WCIVP}}}, then the function $f$ from the above example should have it at each point $x \in I.$ Then the function $g$ from the same example would have the property ${\cal W}(x)$ at each point, in particular at the point $x_o .$ It means that the substitution of any value in the place of $f(x_o )$ has no effect on the local property ${\cal W}(x_0).$ Let $x_1, \; x_2 $ be distinct points from the interval $I.$ The function $h$ defined by $$ h(x) = \begin{cases} f(x) & \text{ for } x \neq x_i , i = 1,2, \\ 2 & \text{ for } x = x_1 , \\ 3 & \text{ for } x = x_2 . \end{cases} $$ has the property ${\cal W}(x)$ at each point of $ x \in I$ and it should have the property {\rm {\bf {\sf WCIVP}}}, which is impossible.\qed \begin{tw} For any function $f:I \to \re$ the set ${\rm {\bf {\sf CIVP}}}^+(f) \triangle {\rm {\bf {\sf CIVP}}}^-(f)$ is at most countable, where $\triangle$ means the symmetric difference of sets. \end{tw} \noindent{\sc Proof. } Since the set of $x \in \re$ at which $C^-(f,x) \ne C^+(f,x)$ or $f(x) \not\in C(f,x)$ is at most countable ([6]), it is enough to consider only the set $A$ of all points of $x \in \re$ at which $f(x)$ belongs to $C(f,x),$ $C^-(f,x) = C^+(f,x)$ and $x \in {\rm {\bf {\sf CIVP}}}^-(f) \setminus {\rm {\bf {\sf CIVP}}}^+(f).$ Let $A_n$ (for each $n \in \na$) denote the set of all those points $x$ of $A$ for which $f(x) \in C^+(f,x)$ and there exists a Cantor set $K \subset \left[\inf C^+(f,x)+ \frac{1}{n} , \sup C^+(f,x)- \frac{1}{n} \right]$ such that $f(C) \not\subset K$ for every Cantor set $C$ contained in $\left[x,x+\frac{1}{n} \right],$ One can see immediately that $A = \bigcup_{n=1}^{\infty}A_n.$ Now we shall show that none of the sets $A_n$ contains its left-sided points of accumulation. Suppose that there exists a sequence $(x_k)$ and a point $x_o$ such that $x_k \in A_n,$ $x_k \to x_o$, $x_k < x_{k+1}