# Open Ball definition of Closure, Interior, and Boundary

First the definitions in question.

Closure: If $A$ is a subset of $\mathbb R ^n$, the closure of $A$ denoted $\bar{A}$, is the set of $x \in \mathbb R ^n$ such that for all $r > 0$ $$B_r (x) \cap A \ne ∅$$

Interior: If $A$ is a subset of $\mathbb R ^n$, the interior of $A$ denoted $\text{int}(A)$ is the set of all $x \in \mathbb R ^n$ such that there exists $r>0$ such that $$B_r (x) \subset A$$

Boundary: The boundary of a subset $A \subset \mathbb R ^n$, denoted $\partial A$ consists of the points $x \in \mathbb R ^n$ such that every neighborhood intesects both $A$ and its compliment.

If we have "for all $r > 0$", could we not pick an $x$ arbitrary outside of $A$ and make our $r$ large enough so that it still would still intersect with $A$? This same idea goes with the definition of boundary. We could pick a point far off into the compliment, and still make it intersect the set $A$ by choosing a large enough $r$.

The equations relating the boundary, interior, and closure to each other make intuitive sense, but I can't see how it could make sense from these definitions.

• For all $r>0$ aims at $r$ being really, really small. You're thinking "for some large $r$" not "for all $r$".
– Pedro
Commented Sep 23, 2014 at 20:44
• But wouldn't it make more sense to say:... such that for all $y \in A$, where $y$ is referring to the $y$ in the definition of an open ball. Perhaps my problem is with the phrase "for all $r$". It doesn't say "for all $r$ such that, but rather "for all $r$". Commented Sep 23, 2014 at 20:54

To reiterate what Pedro had already said, sure you can pick $r$ large enough such that the $B_r(x) \cap A \neq \emptyset$, but this is not true for all $r$. In fact, you specified an $r$ -- a very large one.

Intuitively, you should think of the points in the closure $\overline{A}$ as points that have infinitely many points of $A$ around them so that no matter how small of a radius your ball is, you still have points from $A$.

• Would it make sense to say "for some $r > 0$"? Commented Sep 23, 2014 at 21:07
• Please elaborate. Commented Sep 23, 2014 at 21:08
• It seems like we're saying that we need to pick an $r$ that "works". Almost like "for infinity small $r$". When we say "all $r$" I think that we can pick an arbitrary number and everything will work. Commented Sep 23, 2014 at 21:11
• Yes for all $r > 0$ means "pick any arbitrary $r>0$" and the condition (in our case $B_r (x) \cap A \neq \emptyset$) needs to be satisfied. Commented Sep 23, 2014 at 21:14
• When you pick a large r, the condition needs to be satisfied (and most likely will). But the condition also needs to be satisfied when you pick a very small r (which is sometimes harder to be satisfied than picking a large $r$). That might be your confusion. You are only considering a single $r$ not and ALL of the possible r's. Commented Sep 23, 2014 at 21:22

i see you are having problems with the definition of closure:

Closure: given: $A \subseteq \mathbb{R^n}$ - the set closure is constructed as follows: $$\bar{A}=\{x \in \mathbb{R^n}\mid \forall r \in \mathbb{R^+}:B_r(x)\ \cap A \neq \phi \}$$

in the definition the inclusion must work for all $r$'s so it must work for very small one's, if you choose a point $\hat x$ "too far" from $A$ then there is a ball centered in $\hat x$ such that the intersection is empty! so the point is not in the closure! if you think about it only the points of $A$ and of the boundary of $A$ can satisfy that condition.

more intuitively: a point $\bar x$ is in $\bar A$, iff $\bar x$ is surrounded by an infinite number of points of $A$ which are arbitrarily close to $\bar x$.