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Let $X$ be a topological space and $N$ a subset of $X$. I want to show that $\partial \bar N\subset \partial N$.

I know that since $\bar N$ is closed then $\partial \bar N\subset \bar N$. By definition $\bar N=N \cup \partial N$, now if $x\in \partial\bar N$ then $x\in \bar N$ but why $x$ must lie precisely in $\partial N\subset \bar N$?

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If $x$ is in $\bar N$, then $x$ need not be in $\partial N$. – Chris Eagle Oct 5 '12 at 20:19
I reformulated the question – palio Oct 5 '12 at 20:24
up vote 4 down vote accepted

Alternatively, $\partial N=\overline N\setminus N^\circ$ is the closure minus interior, and $\partial \overline N=\overline N\setminus (\overline N)^\circ$. Since $N\subseteq\overline N,$ we have $N^\circ\subseteq(\overline N)^\circ,$ from which we see the result.

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By definition, the boundary of a set is the intersection of its closure and its complement's closure: $$ \partial N = \overline{N} \cap \overline{C(N)} $$

Apply the same definition to $\overline{N}$ to get: $$ \partial \overline{N} = \overline{N} \cap \overline{C(\overline{N})} $$

Since $N \subset \overline{N}$, $C(\overline{N}) \subset C(N)$. Thus $\overline{C(\overline{N})} \subset \overline{C(N)}$. Compare with the definition of the boundary to find that: $$ \partial \overline{N} \subset \partial N $$

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Alternatively, you can use the definition: a point is in the boundary of $N$ if and only if every open neighbourhood intersects with both $N$ and the complement $N^c$.

Proof: Let $x \in \partial \overline{N}$. We want to show that if $U$ is an open neighbourhood of $x$ then $U \cap N \neq \varnothing$ and $U \cap N^c \neq \varnothing$. Since $x \in \partial \overline{N}$ we have that $U \cap \overline{N} \neq \varnothing$ and $U \cap \overline{N}^c \neq \varnothing$ and since $U \cap \overline{N}^c \subset U \cap N^c$ we have $U \cap N^c \neq \varnothing$.

To show that $U \cap N \neq \varnothing$, we use that since $x \in \partial \overline{N}$, $U$ contains a point in $\overline{N}$ and a point in $\overline{N}^c$. Let $y \in U \cap \overline{N}$. Then in particular, $y \in U$, and since $U$ is open, $U$ is also a neighbourhood of $y$. Also, since $y \in \overline{N}$, every neighbourhood of $y$ contains a point $z \in N$. Hence $z \in U \cap N \neq \varnothing$ which proves the claim.

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