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Let $U$ be an open subset of a metric space $X$, and let $A \subseteq X$. Show that $U \cap A = \emptyset$ implies $U \cap \overline{A} = \emptyset$.

I think that, since $U$ is an open subset of $X$, then every point of $U$ is an interior point of $X$. I believe that $A$ also an interior point, so both $U$ and $A$ are interior points but do not "overlap", hence the intersection being $\emptyset$. As such, the intersection of $U$ and the closure of $A$ would also be $\emptyset$ since the closure of $A$ consists of points outside of $X$.

I am not sure if I am correct and I also need to show it in a mathematically correct way so to speak.

Let $A$ be a subset of a metric space such that $A \subseteq B (p,r)$ for some $p \in X$ and $r > 0$. Show that $\mathrm{diam}(A) \leq 2r$.

I know that $r$ is the radius and as such, the diameter is $2r$. Since $A$ is a subset of $B(p,r)$ it can't have a diameter greater than $2r$, but how do I prove this?

Rebecca J. Stones
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Sammy Williamson
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3 Answers3

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2.1) Assume there exists $x\in U \cap \overline A$. Since $U$ is an open set there exists a radius $\epsilon>0$, so that $B(x,\epsilon)\subset U$.

Furthermore, there exists a sequence $(x_i)_i \subset A$ converging to $x$ and for some $N \in \mathbb{N}$ we have $x_i\in B(x,\epsilon) \subset U$ for all $i>N$. Hence we get the contradiction $x_i \in U \cap A = \emptyset$ for all $i>N$.

2.2) By definition $\textrm{diam } A := \sup_{x,y\in A} d(x,y)$. Now take any arbitrary $x,y\in A$ and find $d(x,y) \leq d(x,p) + d(p,y) \leq r+r=2r$. Taking the supremum yields the result.

Sven Pistre
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I can give you a rather "topological" solution for 2.1. By assumption, $A\subset \complement U$ and $\complement U$, the complement of $U$, is a closed set. By definition, the closure of $A$ is the smallest closed set containing $A$, and hence $\overline{A} \subset \complement U$. But then $U \cap \overline{A} = \emptyset$.

The solution of 2.2 is immediate: pick any two points $u$ and $v$ from $A$. Then $$ d(u,v) \leq d(u,p)+d(v,p) \leq r+r = 2r $$ because of the assumption. We conclude that $$ \operatorname{diam}A = \sup \{d(u,v) \mid u,v \in A \} \leq 2r. $$

Siminore
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  • Thanks for 2.2, I didn't quite know how to proceed, but it is simple. For 2.1, I understand what you mean and it makes sense, however I am not sure about the last part of the question, where U \cap A = \emptyset implies that U \cap \overline{A} = \emptyset. – Sammy Williamson Aug 18 '15 at 15:28
  • If a set $A$ is contained in a closed set, then also the closure of $A$ is contained in that set: the closure is the smallest closed set containing $A$. – Siminore Aug 18 '15 at 15:35
  • I see now, so since the intercept of U and A is zero, the intercept of U and the closure of A must also be zero, because the points within the closure of A are also within A so to speak? – Sammy Williamson Aug 18 '15 at 15:53
  • Well, no... points of the closure are not "within $A$" (unless $A$ is closed and $\overline{A}=A$). – Siminore Aug 18 '15 at 15:58
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On contrary, suppose that $U \cap \overline{A} \neq \emptyset$.

$\Rightarrow x \in U$ and $x\in \overline{A}$.

$\Rightarrow x \in U$ and $x \in A$ or $\partial A$.

If $x \in A$ then since $U \cap A=\emptyset$. A contradiction.

Now if $x \in \partial A$.

By the definition of a boundary point of $A$, all open balls containing $x$ have a non-empty intersection with $A$ and $A^c$.

Now, since $U$ is an open set and $x \in U$, there exists an open ball, say $M$, such that $x \in M \subset U$.

$\Rightarrow x \in M\subset U\cap A^c \neq \emptyset$ and $x \in M\subset U\cap A \neq \emptyset$. Again, a contradiction, since $U\cap A=\emptyset$.

Rebecca J. Stones
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Yasir Mahsud
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