Open and Closed Sets in a Metric Space

[PingPong]

Homework Statement

Let (X,d) be a metric space. Can a set E in X be both open and closed? Can a point in E be both isolated and an interior point?

Homework Equations

I've used the metric defined as d(x,y)=1 for $x\ne y$ and 0 if x=y (we used this in a previous problem). I also used the following definitions in baby Rudin:

2.18
(b) A point p is a limit point of the set E if every neighborhood of p contains another point q such that q is in E
(c) If p is in E and p is not a limit point of E, then p is called an isolated point of E
(d) E is closed if every limit point of E is a point of E
(e) A point p is an interior point of E if there is a neighborhood N of p such that N is a subset of E
(f) E is open if every point of E is an interior point of E

The Attempt at a Solution

For the first question, I said yes, that a set can both be open and closed. For example, any subset E of a metric space that uses the above metric is open (because a neighborhood of 1/2 around any point p, $N_{1/2}(p)$, is just that point which is a subset of E). Likewise, any subset is closed because there are no limit points to contain - the same neighborhood $N_{1/2}(p)$ only contains p and no other points, so p is not a limit point.

For the second question, I again said yes, again using the same metric as above. Since every subset E is open, then every point in E is an interior point. On the other hand, there are no limit points, and so every p is an isolated point as well.

I don't think this stuff is difficult, really, just a different way of looking at things (especially for somebody who started as a physics major). It's just a lot of definitions to remember, and I wanted to make sure that I understand the definitions properly.

 

[morphism]

It's all good.

(Side note: In any metric space, the empty set and the entire space are always going to be both open and closed.)

 

[PingPong]

Thanks for the quick reply!

I actually forgot to mention that the problem specifically excludes the empty set and X itself and I was a bit curious as to why at first. Now I see that they're the trivial open-closed sets. Thanks again!

 

[HallsofIvy]

A set is "closed" if it contains all of its boundary points (points such that every neighborhood contains some points in the set and some points not in the set).

A set is "open" if it contains none of its boundary points.

Under what conditions on the set is all and none of its boundary points the same?

 

[PingPong]

A set is "closed" if it contains all of its boundary points (points such that every neighborhood contains some points in the set and some points not in the set).

A set is "open" if it contains none of its boundary points.

Under what conditions on the set is all and none of its boundary points the same?

Obviously when there are no boundary points to be had!

To further test my understanding, I tried to come up with another example. Suppose that X=Natural numbers, and E is any finite subset of X of cardinality greater than 1. Then E is both open and closed because 1) every point is an interior point (that is, if I pick epsilon $\epsilon=max\{d(x,y):x,y\in E\}$, then a neighborhood of radius epsilon + 1 around any point includes at least one other point) and 2) there are no limit points because a neighborhood of radius 1/2 around any point does not contain any others.

 

[HallsofIvy]

Actually, in the space of positive integers, with the "usual" topology, all subsets are both open and closed.