Closed and Open Subsets of a Metric Space

gajohnson
Messages
70
Reaction score
0

Homework Statement



Let X be an infinite set. For p\in X and q\in X,

d(p,q)=1 for p\neq q and d(p,q)=0 for p=q

Prove that this is a metric. Find all open subsets of X with this metric. Find all closed subsets of X with this metric.

Homework Equations



NA

The Attempt at a Solution



I showed easily that this is indeed a metric.

On the second part of the question, it seems to be the case that all subsets \left\{x\right\} for all x\in X are open because choosing a radius less than 1 gives a neighborhood around x which only contains x itself.

But then any subset of X should be open, shouldn't it? Because each point of that subset can be shown to be an interior point using the logic above.

Similarly, there should be no closed subsets. Each point in a subset of X obviously has a neighborhood which contains only that point.

Any ideas? Thanks!
 
Physics news on Phys.org
gajohnson said:

Homework Statement



Let X be an infinite set. For p\in X and q\in X,

d(p,q)=1 for p\neq q and d(p,q)=0 for p=q

Prove that this is a metric. Find all open subsets of X with this metric. Find all closed subsets of X with this metric.

Homework Equations



NA

The Attempt at a Solution



I showed easily that this is indeed a metric.

On the second part of the question, it seems to be the case that all subsets \left\{x\right\} for all x\in X are open because choosing a radius less than 1 gives a neighborhood around x which only contains x itself.

But then any subset of X should be open, shouldn't it? Because each point of that subset can be shown to be an interior point using the logic above.

Correct. All sets are open.

Similarly, there should be no closed subsets.

You might want to rethink this.
 
micromass said:
Correct. All sets are open.



You might want to rethink this.

OK, so every subset of X contains no limit points, so every subset of X must be closed.

...so every subset of X is both open and closed?
 
That is correct!
 
micromass said:
That is correct!

Excellent. Many thanks to you!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top