The interior of a set is open and the boundary and closure of a set is closed

Printable View

September 3rd 2010, 03:31 PM
Pinkk

The interior of a set is open and the boundary and closure of a set is closed

So I am pretty sure my proof is correct, just want to verify for correctness and rigor.

Let $a\in int(S)$ . Then there exists $\epsilon > 0$ such that $B(\epsilon , a) \subset S$ . Now suppose $x\in B(\epsilon , a)$ . Since $B(\epsilon, a)$ is open, there exists $\delta > 0$ such that $B(\delta , x) \subset B(\epsilon , a) \subset S$ ; in other words, for any $x\in B(\epsilon , a)$ there is a ball centered at $x$ contained solely in $S$ , which means $x\in int(S)$ . So given $a\in int(S)$ there exists $r > 0$ , namely $r = \epsilon$ , such that $B(r,a) \subset int(S)$ , so $int(S)$ is open.

So now consider $\partial S$ . Then $(\partial S)^{c} = int(S) \cup int(S^{c})$ since if $x$ is not a point such that every ball centered at $x$ intersects $S$ and $S^{c}$ , then there exists some ball centered at $x$ that is either contained solely in $S$ or $S^{c}$ . So we have that $(\partial S)^{c}$ is the union of two open sets, which is open, so $\partial S$ is closed. Similarly, if we consider $cl(S)$ , then $(cl(S))^{c} = S^{c}\cap(int(S)\cup int(S^{c})) = int(S^{c})$ , which is open, so $cl(S)$ is closed.

Q.E.D.

So I think this is a perfectly valid and rigorous proof, but any feedback would be appreciated (I am somewhat unsure on the closed portions of the proof, especially the claims about what the complements of the boundary of S and the closure of S actually equal). Thanks.
September 3rd 2010, 03:58 PM
Plato

I find your proof very hard to follow. That is not saying it is incorrect!
I think it is too complicated.

It is easy to prove that any open set is simply the union of balls.
The interior is just the union of balls in it.
The complement of the closure is just the union of balls in it.
The complement of the boundary is just the union of balls in it.

To follow that last bit, think this way.
If $t\notin \beta(A)$ the there is a ball $\mathcal{B}(t;\delta)$ that is a subset of $\mathcal{I}(A)$ or the complement of $A$
That is it contains only points of $A$ or points not in $A$ .
September 3rd 2010, 04:06 PM
Pinkk

Hm, thanks. I guess I made it more complicated if not for the fact that this is my first time learning about open/closed sets in terms of interior points and balls, so when it asks if a set is open I go back to the basic definition that any point in an open set has some ball centered at that point contained solely in the set. (At least that is how the textbook I am using defines it). And since this is a new area of mathematical study for me, I don't know how to actually show that an open set is the union of balls.
September 3rd 2010, 05:53 PM
Plato

Actually that is an axiom.
The union of any set of open sets is itself an open set.
That should have been stated up front.
Then by definition, any ball is an open set.
So any union of balls is open, by the axiom.
September 3rd 2010, 06:12 PM
Pinkk

Ah, okay. I guess my textbook (Advanced Calculus by Gerald B. Folland) does things differently. Thanks again.
September 4th 2010, 08:58 AM
Iondor

The interior of a set M is the largest open set, which is still a subset of M or equivalently the union of all open subsets of M.