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Let $(X, \mathfrak T)$ be a topological space and supposed that $A$ is a subset of $X$ then the Bd(A) is a closed set.

I am in an introduction to proofs class. I have to decided if this is a true or false statement. If true, prove and if false, give a counterexample.

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 $X−A$.

I think this is a true statement and I would like to show that by showing that the complement of an open set is closed. Showing that the points not in a boundary of a set are an open set. Is this the right track to take?

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    $\begingroup$ Yes, this is the way you have to follow: you have to show that the complement of the boundary is open (if $x$ belongs to the complement of the boundary, then there is an open neighbourhood such that...) $\endgroup$ – Crostul Apr 25 '15 at 16:14
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Yes, that is the correct way. Let $A$ be a subset of $X$, where $(X, \tau)$ is a topological space, then we take a point $x \in X \setminus Bd(A)$, where $Bd(A)$ is the boundary of $A$, then by the negation of the definition of boundary we have that there is an open set $U \in \tau$ such that $U$ does not intersect $A$ or $X \setminus A$. Suppose $U$ does not intersect $X \setminus A$, the other case is analogous, then $U \subset A$. Then every point in $U$ is not in the boundary because there is an open set, $U$ itself, such that $U \cap (X \setminus A) = \emptyset$. Then for every point $x \in X \setminus Bd(A)$there is an open set $U$ such that $x \in U \subset X \setminus Bd(A)$, therefore $X \setminus Bd(A)$ is open and $Bd(A)$ is closed.

Another way is showing that for every $A$ subset of a topological space $X$ we have the equality $Bd(A)=Cl(A) \cap Cl(X \setminus A)$, where $Cl(A)$ is the closure of $A$.

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Hint: Let $X$ be a topological space, and let $B\subset X$ be a subspace. Show that point $x\in X$ is such that any open set containing $x$ intersects $B$ if and only if $x$ is contained in every closed subset of $X$ that contains $B$.

Now try applying this with $B=A$ and $B=X-A$.

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