I'm trying to prove that in a normed vector space the boundary of a set A is a subset of the boundary of the closure of A. I've been using the definition of boundary and concluded that if $x\in boundary(A)$ then the intersection of the ball of radius r>0 centered at $x$ with the closure of A is always non empty, but I'm having trouble proving that the intersection of the ball with the complement of the closure of A is non empty. Any tips? Thanks!

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    $\begingroup$ Why are you trying to prove this? Do you have an external reason to believe this result? $\endgroup$ Oct 3, 2017 at 2:38
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    $\begingroup$ What if $A=\Bbb Q$ as a subset of $\Bbb R$? $\endgroup$
    – user223391
    Oct 3, 2017 at 2:43
  • $\begingroup$ Thank you, I've edited the question to be more clear. This exercise is in my homework and I just can't figure it out. $\endgroup$ Oct 3, 2017 at 2:43

2 Answers 2


Tip: It's not true.

Counterexample: Notice that $\mathbb{R}$ is a normed vector space under the Euclidean norm. Let $A$ be the set $[0,1] \smallsetminus \{1/2\}$, the closed unit interval without its midpoint.

The boundary of $A$ is $\{0,1/2,1\}$. The closure of $A$ is $[0,1]$, the closed unit interval. The boundary of the closure is $\{0,1\}$. Since $1/2 \not \in \{0,1\}$, the boundary of $A$ is not a subset of the boundary of the closure of $A$.


$\partial A \subset \partial \bar{A}$ is not generally true as counterexamples are easy to construct, such as the one Eric provides here. For a more extreme example, consider A to be the set of rationals under the usual topology on the real line. Then $$\partial A = \bar{\mathbb{Q}} \setminus \mbox{int}(\mathbb{Q})=\mathbb{R} \setminus \emptyset =\mathbb{R} $$ whereas for the closure of A, $$\partial \bar{A}= \partial \bar{\mathbb{Q}} =\mathbb{R} \setminus \mathbb{R}=\emptyset$$

The other way around is true, though, namely that $\partial \bar{A} \subset \partial A$ where $\partial A = \bar{A} \setminus \mbox{int}(A)$ denotes the boundary of A because $$\partial \bar{A}=\overline{\bar{A}}\setminus\mbox{int}(\bar{A}) = \bar{A} \setminus \mbox{int}(\bar{A}) \subset \bar{A}\setminus\mbox{int}(A) = \partial A$$

  • $\begingroup$ This doesn't disprove OP's statement, though. $\endgroup$
    – Evpok
    Oct 3, 2017 at 9:58
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    $\begingroup$ @Evpok That's right. It doesn't. Eric Towers had already proved that with a counterexample. I was adding what else would have been useful. Besides, it's possible OP made a mistake in writing the question and this was what OP actually set out to prove. $\endgroup$ Oct 3, 2017 at 10:40
  • $\begingroup$ I just thought it would be better to explicit it :) $\endgroup$
    – Evpok
    Oct 3, 2017 at 12:12

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