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Let $(X,\mathcal{T})$ be a topological space and for $Y\in\mathcal{T}$ let $\mathcal{T}_Y:=\{U\cap Y\mid U\in\mathcal{T}\}=\{U\in\mathcal{T}\mid U\subset Y\}$ be the induced topology on $Y$. Then if $A\subset Y$ is such that $Y\backslash A\in\mathcal{T}_Y$ and $\text{int}_Y(A)=\emptyset$, how can we prove that $\text{int}(\text{adh}(A))=\emptyset$?

I proved that for all $B\subset Y$ we have $\text{int}_Y(B)=\text{int}(B)$ and $A=\text{adh}_Y(A)=\text{adh}(A)\cap Y$ which gives $\text{int}(\text{adh}(A))\cap Y=\emptyset$. Does this help?

I use the notation $$\text{int}(B)=\bigcup_{U\in\mathcal{T},\ U\subset B}U,\qquad\text{int}_Y(B)=\bigcup_{U\in\mathcal{T}_Y,\ U\subset B}U$$ and $$ \text{adh}(B)=\bigcap_{U\in\mathcal{T},\ B\subset X\backslash U}X\backslash U,\qquad\text{adh}_Y(B)=\bigcap_{U\in\mathcal{T}_Y,\ B\subset Y\backslash U}Y\backslash U. $$

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    So you know $\operatorname{int}(\operatorname{adh}(A)) \subset \operatorname{adh}(Y)\setminus Y = \partial Y$. What do you know about the boundaries of open sets? – 2017-01-02
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    Ah, they are empty? Because $\operatorname{int}(\partial Y)=\operatorname{int}(\operatorname{adh}(Y)\cap Y^c)=\operatorname{int}(\operatorname{adh}(Y))\cap \operatorname{int}(Y^c)=\operatorname{int}(\operatorname{adh}(Y))\cap \operatorname{adh}(Y)^c\subset\operatorname{adh}(Y)\cap \operatorname{adh}(Y)^c=\emptyset$. Thanks, I get it now. – 2017-01-02
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    No, they are in general not empty. In $\mathbb{R}$, we have $\partial (0,1) = \{0,1\}$, in $\mathbb{C}$, $\partial \{ z : \lvert z\rvert < 1\} = \{ z : \lvert z\rvert = 1\}$. But they have empty interior. – 2017-01-02
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    Yes of course, thats what I wanted to say :) – 2017-01-02

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