$(X,\tau)$ is a topological space and $Y\subset X$.How can I visualize ${A^0}^{Y} \supset A^0$,where $A\subset Y$.The proof is not very difficult but the fact is not seeming very obvious to me.Suppose $x \in A^0$ then $\exists U_x$ open such that $x\in U_x \subset A$.Taking intersection with $Y$,we get $x\in U_x\cap Y \subset A\cap Y$.$U_x\cap Y$ is open in $Y$ as $U_x$ is open in $x$ and in fact $U_x \cap Y=U_x$ and $A\cap Y=A$ because $A\subset Y$.So,$x\in {A^0}^Y$.But how to feel that this should be true,I am talking of an intuitive feeling that should be obvious even without calculation.

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    $\begingroup$ What is $A^0$ and $A^{0^Y}$? $\endgroup$ – supinf Mar 9 at 11:36

You can look at it this way: $A^\circ$ is the largest open (from point of view of $X$) subset of $A$, $A^{\circ^Y}$ is the largset open (now form point of view of $Y$) subset of $A$. Clearly, every $X$-open subset of $Y$ is $Y$-open. An these facts together are enough.

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  • $\begingroup$ how about the diagram I gave in question? $\endgroup$ – Kishalay Sarkar Mar 9 at 11:50
  • $\begingroup$ @KishalaySarkar The diagram shows that it may happen that the $Y$-interior is strictly larger than the $X$-interior. $\endgroup$ – user87690 Mar 9 at 11:52
  • $\begingroup$ It is true that $\partial_Y A\subset \partial A$. $\endgroup$ – Kishalay Sarkar Mar 9 at 12:03
  • $\begingroup$ @KishalaySarkar for the first one – it's the other way round: X-interior point is always in Y-interior, but Y-interior point that is not X-interior is clearly in X-boundary. Fot the second question: yes, it is true. $\endgroup$ – user87690 Mar 9 at 12:36

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