How can a metrix space be open and closed?

[soopo]

Homework Statement

How can a metrix space be open and closed?

 

[lanedance]

Just getting into this stuff, but i'll have a crack

Do you mean metric space?

Take Rⁿ (in Rⁿ) with normal distance metric, clearly any point in the space has a neighbourhood within Rⁿ impling it is an open set

Definition of closed is that the complement of the set is open.

What is the compmelent of Rⁿ? the empty set, which is both open & closed. As its complement is open Rⁿ is closed

A more geometric example might be the 2 sphere, S², in R³, clearly any point on the sphere has an open neighbourhood implying S² is open. Look at the complement of the set, R³ / S². This is clearly also open (though disjoint) meaning S² in R³ is both open & closed.

 

[Dick]

S^2 in R^3 is closed. It's not open. A neighborhood of a point on the sphere will include points off the sphere as well. For S^2 to be open, the neighborhood would have to be contained in S^2.

 

[soopo]

S^2 in R^3 is closed. It's not open. A neighborhood of a point on the sphere will include points off the sphere as well. For S^2 to be open, the neighborhood would have to be contained in S^2.

Do you mean that empty set is the only one which can be closed and open at the same time?

 

[HallsofIvy]

All metric spaces are both open and closed as subsets of themselves. The most general definition of "topology" requires that it include both the entire space and the empty set and other definitions (for metric spaces for example) satisfy that. A set is closed if and only if its complement is open. Since the complement of the entire space is the empty set, and the empty set is open, the entire space is also closed.

Those are the only sets that are both open and closed ("clopen") if and only if the space is connected.

For example, if X= (-[itex]\infty[/itex], -1] U [1, [itex]\infty[/itex]), the real numbers with the open interval (-1, 1) removed, with the "usual" metric |x- y|, then X is not connected and X itself is both open and closed, the empty set is both open and closed, (-[itex]\infty[/itex], -1] is both open and closed, and [1, [itex]\infty[/itex]) is both open and closed.

 

[lanedance]

cheers - i think i get confused between the set & the set its embedded in, ie the set inducing the topology & so the open sets ...

So if i understand on HallsofIvy comment, it all has to do with which set you choose to induce the topology

So, if we take the toplogy induced by the set X = (-inf, -1] U [1, inf), X is open & closed by definition. This gives us the subsets with the properties as described by Halls of Ivy

But if we take the topology induced by R^1

then
C = (-1,1) is open as every points neighbourhood in R^1 is contained in C

-> X = (-inf, -1] U [1, inf) is closed

But the for the point 1 & -1 in X, the neighbourhood in R^1 is not contained in X, so X is not open

So in R^1 the set X is closed & not open

still a bit light on the fundamentals, so does this sound about right? sounds consistent with what Dick said as well?

 

[Tobias Funke]

You may want to think about whether you actually have trouble believing that sets can have the properties of an open set and a closed set at the same time, or whether you've convinced yourself (quite understandably) that something which is open can't be closed. The choice of names is unfortunate here- if they were called "O-sets" and "C-sets" would you still wonder how an O-set could be a C-set?

 

[lanedance]

by the way sorry for hijacking your forum soopo, but any confirmation on my last post would be good though...

 

[Dick]

It's not so much what the topology of the subset as what is the whole metric space. E.g. S^2 with the topology induced by R^3 IS S^2 with the usual topology. S^2 as a subset of R^3 is not open. If you throw away the R^3/S^2 part, then S^2 is the whole space. So it is open.

 

[lanedance]

thanks for the reply Dick, i know my terminology is a little sloppy, bit of work to do i think...

 

[Dick]

S'ok. Trying to help is good, learning from correction is better. That's how I did it.