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Let $(X, \mathfrak T)$ be a topological space and let $A$ but a subset of $X$ then $Int(Bd(A)) = \emptyset$

I need to decide if this is true or not. I have done a little research and some contemplating and I believe this statement is true for open sets but I am not sure if it is true for closed sets or if I am overthinking and do not need to consider both sets. I am wondering if I am overthinking and this is true and a proof by contradiction would work?

My definition of interior is: Let $(X, \mathfrak T)$ be a topological space and let $A \subset X$ is the set of all points $x \in X$ for which there exists an open set $U$ such that $x \in U \subseteq A$.

My definition of boundary is: Let $(X,\mathfrak T)$ be a topological space and let $A \subseteq X$. A point $x \in X$ is in the boundary of A if every open set containing $x$ intersects both $A$ and $X−A$.

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    $\begingroup$ It might be the case $A$ is neither open nor closed. Consider what happens when $X = \Bbb R$ and $A = \Bbb Q$. $\endgroup$ – Rolf Hoyer May 13 '15 at 1:02
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Take $X=\Bbb R$ and $A= \Bbb Q$, then $\operatorname{Bd}(A)=\operatorname{Bd}(\Bbb Q)=\Bbb R$ so $\operatorname{Int}(\operatorname{Bd}(A))=\operatorname{Int}(\Bbb R)\neq \emptyset$, so the proposition doesn't hold for every set $A$.

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  • $\begingroup$ Is the interior of $\mathbb R$= $\mathbb R$? $\endgroup$ – user219081 May 13 '15 at 1:10
  • $\begingroup$ @AlyssaWallace That's right, any idea of why is that the case? $\endgroup$ – Daniel May 13 '15 at 1:11
  • $\begingroup$ well its an open set so the interior is the open set that contains the set so it would have to be the reals? $\endgroup$ – user219081 May 13 '15 at 1:12
  • $\begingroup$ @AlyssaWallace More generally, the interior of an open set equals the set (in every topological space). $\endgroup$ – Daniel May 13 '15 at 1:13

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