$g \circ f$ is not continuous If $f: \mathbb{R} \to  \mathbb{R}$ and $g : \mathbb{R} \to  \mathbb{R}$ are functions such that $g \circ f$ is not continuous, then $f$ is not continuous or $g$ is not continuous.
PLEASE ,help me how can I come with counterexample for that ,I try a lot but without useful and If the statement is true how can I prove it?
by using this definition of continuous
f : R → R is said to be continuous if for each open subset V of R , f^-1(v) is an open subset of R .
 A: Your statement is correct. It is equivalent to "if $f$ and $g$ are continuous then $g\circ f$ is continuous".
From the comments I guess the confusion is about the meaning of implication. An implication is considered true even if we have cases where the conclusion holds despite the premises does not, the only thing we can't accept with an implication is cases where the premise is true, but the conclusion isn't.
Your example or attempt using two discontinuous functions $f$ and $g$ that is crafted in such way that $g\circ f$ is continuous can be pulled of, but it isn't a problem. It says that "if $g\circ f$ is not continuous then not both of $f$ and $g$ are", but if we have a case where $g\circ f$ is continuous the premise isn't fulfilled - in such a case the implication is never challenged, the conclusion may or may not hold, in this case the conclusion happens to be valid.
The statement is valid because negations turn an implication. That is $\phi\Rightarrow\psi$ is equivalent to $\neg\psi\rightarrow\neg\phi$.
The proof that if $f$ and $g$ are continuous then $g\circ f$ is also continuous is found in any introductory calculus course. I'll use the epsilon-delta-definition of continuity:

$f$ is said to be continuous if for every $c$ (where $f$ is defined) and every $\epsilon>0$ there exists a $\delta>0$ such that whenever $|x-c|<\delta$ we have that $|f(x)-f(c)|<\epsilon$.

So let's assume that we have a $c$ where $g\circ f$ is defined (which means that $f$ is defined with a value for which $g$ is defined), and $\epsilon>0$. Now since $g$ is continuous and defined at $f(c)$ we have a $\eta>0$ such that $|g(y)-g(f(c))| < \epsilon$  whenever $|y-f(c)|<\eta$, but since $f$ is continuous we have a $\delta>0$ such that $|f(x)-f(c)|<\eta$ whenever $|x-c|<\delta$. That is we have a $\delta>0$ such that $|g(f(x))-g(f(c))| < \epsilon$ whenever $|x-c|<\delta$ which proves that $g\circ f$ is continuous.
A: According to the definition in your question a function $f:\>X\to Y$ is continuous if for any open set $\Omega\subset Y$ its inverse image $f^{-1}(\Omega)=\{x\in X\>|\>f(x)\in\Omega\}$ is open in $X$. Since $(g\circ f)^{-1}(U)=f^{-1}\bigl( g^{-1}(U)\bigr)$ for arbitrary subsets $U\subset Z$ this immediately implies that the composition $g\circ f:\>X\to Z$ of two continuous maps $f:\>X\to Y$ and $g:\>Y\to Z$ is again continuous. Hence, if $g\circ f$ is not continuous this can only happen if at least one of $f$ or $g$ is not continuous.
