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If $x$ is isolated point of any $Y\subset X$, there exists $T$ open in Y such that $\{x\}=T\cap Y$. So, $x$ is also isolated point in $X$. Because, $T=U\cap Y$ where $U$ is open in $X$. Then $U\cap Y=\{x\}$. But I think something is wrong in this proof.

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$x$ is an isolated point in $X$ if you can show that $\{x\}$ is open. You have not done that. –  Hagen von Eitzen Dec 19 '12 at 16:02
    
I think you want a fixed subspace here (not "any")? –  David Mitra Dec 19 '12 at 16:05
    
Yes, you are right,But also according to the definition, we have to find some open set in X which meets with Y only single point $\{x\}$.So, I find open set U. What is my mistake? –  ege Dec 19 '12 at 16:06
    
Definition of what? What are the roles of $X,Y,x$? If $X$ is a topological space, then $x\in X$ is an isolated point $:\Leftrightarrow$ the set $\{x\}$ is open. What you describe in the question is the property of being an isolated point of $Y\subset X$ with the induced subspace topology: $\{x\}$ is open in $Y$ iff (per definition of subspace topology) it is of the form $\{x\}=T\cap Y$ with $T$ open in $X$. –  Hagen von Eitzen Dec 19 '12 at 16:09
    
a point $x$ is isolated point of $Y\subset X$ iff there exists an open set $T$ such that $T\cap Y=\{x\}$ So, do you mean this definition give us only isolated point of subspace. –  ege Dec 19 '12 at 16:16
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