Let $X$ be a topological space with subspace $Y$ and $A$ a subset of $Y$. Let $Cl_X(A)$ denote the closure of $A$ in $X$ and $Cl_Y(A)$ denote the closure of $A$ in $Y$. Show that $Cl_Y(A)\subseteq Cl_X(A)$ and in general, $Cl_Y(A)\neq Cl_X(A)$.

My attempt: Since $Cl_X(A)$ is a closed subset of $X$ containing $A$, $Cl_X(A)\cap Y$ is a closed subset in $Y$ containing $A$, but $Cl_Y(A)$ is, by definition, a subset of every closed set in $Y$ containing $A$, so $Cl_Y(A)\subseteq Cl_X(A)\cap Y\subseteq Cl_X(A)$. To show that equality fails in general, consider $X=\mathbb{R}$ and $Y=A=(0,1)$. $Cl_{\mathbb{R}}(0,1)=[0,1]$, but $Cl_{(0,1)}(0,1)=(0,1)$. So equality fails in general.

Do my arguments look ok?

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    $\begingroup$ The arguments are completely correct. $\endgroup$ – 5xum Feb 27 '14 at 7:37
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    $\begingroup$ perfectly fine. $\endgroup$ – Max Feb 27 '14 at 7:40
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    $\begingroup$ Looks great. Perhaps, copy the paragraph after "My attempt" as an answer and select it so that the question is marked as answered. $\endgroup$ – Sammy Black Feb 27 '14 at 7:56

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