PROOF: Let ? ? R^n be an open set. Show that ? ? bd(?) = ?.

[duskaholic]

Let ? ? R^n be an open set.
Show that ? cannot contain any of its boundary points
IE. Show that ? ? bd(?) = ?.

OKAY.
All I get is that this is like a "There are points in the circle. I'm trying to prove that the points INSIDE the circle are NOT PART of the circle itself" kinda question.
HOW DO I PROVE THIS?!

I just know and only have this much.
The circ. of a sphere is the boundary.
Anything less the circ. of the set is just the set itself.

 

[daon]

Re: PROOF: show that ? intersections boundaryomega = empty s

Not a circle. You are trying to show that an ARBITRARY open set in R is composed of strictly interior points. To do so you may show that for an arbitrary open interval this proposition holds, and conclude (i.e. prove) that it is true for an finite/infinite union of open intervals.

The problem here is that in some texts, this is essentially the definition of what it means to be interior. Please post what the definitions of "open set" and "boundry." And any other relevant terms such as "limit point" "closure" "interior" etc.

 

[duskaholic]

RE: PROOF: Let ? ? R^n be an open set. Show that ? ? bd(?) =

Well, I'm trying to use the definition of an open ball, where r > 0, such that...
Br(?) = {x in R^n : ||? - x|| < r}.

..and that boundary of ? (denoted as bd(?)) is...
bd(?) = {x in R^n : ||? - x|| = r}.

In my notes, we proved a proposition that a set S is closed <=> bd(S) ? S.
I was thinking about proving that if..
? is open <=> bd(?) > ?

 

[daon]

Re: PROOF: Let ? ? R^n be an open set. Show that ? ? bd(?) =

What do you mean by "<" when speaking of sets?

 

[duskaholic]

Re: PROOF: Let ? ? R^n be an open set. Show that ? ? bd(?) =

The definition of an open ball, where radius r > 0, such that...
the Ball with radius r of (?) = {x in R^n : ||? - x|| < r}.

..and that boundary of ? (denoted as bd(?)) is...
bd(?) = {x in R^n : ||? - x|| = r}.

Since in class we proved that a set S is closed iff bd(S) ? S..
I thought maybe it can also be applied to prove ? is open iff bd(?) > ?.