Find uncountably many subsets that are neither open nor closed

  • Thread starter Thread starter hnbc1
  • Start date Start date
  • Tags Tags
    Closed Subsets
hnbc1
Messages
5
Reaction score
0

Homework Statement


1. Find an uncountable number of subsets of metric spaces \left(\mathbb{R}^{n},d_{p}\right) and \left(\mathbb{C}^{n},d_{p}\right) that are neither open nor closed.
2. If 1\leq p<q, then the unit ball in \left(\mathbb{R}^{n},d_{p}\right) is contained in the unit ball in \left(\mathbb{R}^{n},d_{q}\right)

Homework Equations


The Attempt at a Solution


For (1), I think I should start with some point and construct balls centered at this point.
For (2), I think it boils down to prove: \left(\sum_{i}^{n}\left|x_{i}\right|^{p}\right)^{\frac{1}{p}}\geq\left(\sum_{i}^{n}\left|x_{i}\right|^{q}\right)^{\frac{1}{q}}
 
Last edited:
Physics news on Phys.org
hnbc1 said:
For (1), I think I should start with some point and construct balls centered at this point.

Can you come up with just one space that is neither open nor closed?? Try to generalize...

For (2), I think it boils down to prove: \left(\sum_{i}^{n}\left|x_{i}\right|^{p}\right)^{\frac{1}{p}}\geq\left(\sum_{i}^{n}\left|x_{i}\right|^{q}\right)^{\frac{1}{q}}

Yes, that is what you need to show. Instead of dealing with general dimension n, try to prove it first for n=1 (this is trivial) and n=2. You'll see easier why the general case holds...
 
micromass said:
Can you come up with just one space that is neither open nor closed?? Try to generalize...
Yes, that is what you need to show. Instead of dealing with general dimension n, try to prove it first for n=1 (this is trivial) and n=2. You'll see easier why the general case holds...

Hi micromass, I'm not required to find spaces that are neither open nor closed, but the subsets of the two metrics spaces mentioned. We can pick any point x in Rn or Cn, construct an open ball which is an open set, and pick another point on the boundary or out of the boundary, and the union of the ball and the point is neither open nor closed. Since the point x uncountable in Rn and Cn, then we are done. I'm not sure this argument is formal enough.

I still don't have a clue about the second question, even when n=2...
 
Last edited:
hnbc1 said:
Hi micromass, I'm not required to find spaces that are neither open nor closed, but the subsets of the two metrics spaces mentioned. We can pick any point x in Rn or Cn, construct an open ball which is an open set, and pick another point on the boundary or out of the boundary, and the union of the ball and the point is neither open nor closed. Since the point x uncountable in Rn and Cn, then we are done. I'm not sure this argument is formal enough.

Well, from your reasing, I could deduce that X=]0,1] is open nor closed. In fact, we could view X as subset of \mathbb{R}^n in general! So this is an example of a subset that is open nor closed. Now, can you make a slight generalization to X such that you have uncountably many spaces??

I still don't have a clue about the second question, even when n=2...

The second question is without doubt the hardest one. Let's do this for n=2, we need to show that

1\leq p\leq q~\Rightarrow~(|x|^q+|y|^q)^{1/q}\leq (|x|^p+|y|^p)^{1/p}

Now, the proof of this is what I called "standard". First, we assume that y=1, we'll deal with the general argument later. Then we need to show that

f(x)=(1+x^p)^{1/p}-(1+x^q)^{1/q}\geq 0~\text{for}~x\geq 0

Try to find the extrema of f (using derivatives), this allows you to find the minima of f on \mathbb{R}^+. If you show that the minimum is \geq 0 (in fact it will =0), then we have proved that the function itself is larger than 0...
 
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