Thread: Basic facts about Exterior of sets

Originally Posted by Matt Westwood

I'm minutely studying "Counterexamples in Topology" by Steen and Seebach (great book! Recommended) and have reached a statement on Page 7 which I can't prove.

"The exterior ext(A) of a set A is the complement of the closure of A, or equivalently, the interior of the complement of A. ..."

(The closure is the set together with its limit points, the interior is conveniently the complement of the closure of the complement.)

"... the exterior of the intersection is contained in the union of the exteriors ... equality holds for finite unions ..."

So my approach is as follows:

Let T be a topological space and let \mathbb H \subseteq P \left({T}\right) where P \left({T}\right) is the powerset of T.

We want to show ext\left({\bigcap_{H \in \mathbb H} H}\right) \subseteq \bigcup_{H \in \mathbb H} ext(H).

We use cl(H) as the closure of H.

We have by definition of exterior:

ext \left({\bigcap_{H \in \mathbb H} H}\right) = T - cl \left({\bigcap_{H \in \mathbb H} H}\right)

But we have that:
cl \left({\bigcap_{H \in \mathbb H} H}\right) \subseteq \bigcap_{H \in \mathbb H} cl \left({H}\right)

from Closure of Intersection Subset of Intersection of Closures - ProofWiki

But then A \subseteq B \Longrightarrow T - B \subseteq T - A and so:

T - cl \left({\bigcap_{H \in \mathbb H} H}\right) \supseteq T - \bigcap_{H \in \mathbb H} cl \left({H}\right)

By de Morgan this gives:

T - \bigcap_{H \in \mathbb H} cl \left({H}\right) = \bigcup_{H \in \mathbb H} cl \left({T - H}\right)

which by definition of exterior equals \bigcup_{H \in \mathbb H} ext (H)

Putting it all together it seems I've just proved:

ext \left({\bigcap_{H \in \mathbb H} H}\right) \supseteq \bigcup_{H \in \mathbb H} ext (H)

which is the antithesis of what I was expecting to get.

Where am I going wrong - or are Steen and Seebach wrong?


BLAST IT - The LaTeX won't work. Help might be needed.

.....

You are right in that the book is mistaking. I also get a LaTeX compilation error, so hopefully this is understandable: As you seem to have proved, it holds that the set

ext( cap K_i )

contains

cup (ext(K_i)),

and not the other way around like the book says.

And you need not even have equality for a finite number of sets. You can take the sets that the wikipage used as a counterexample for closures, i.e.

H = (0,2) cup (3,4), and
K = (1,3).

For these sets, I get

ext( H cap K ) = (-infty,1) cup (2,infty), and
ext(H) cup ext(K) = (-infty,1) cup (2,3) cup (3,infty),

so compared to the first set, the latter set lacks the number 3.

That's what I thought. I am encouraged - I thought I was doing something terminally wrong! Good to get confirmation.