interior

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May 6th 2009, 10:20 PM
john_n82

open/closed/interior

http://img14.imageshack.us/img14/6083/72795644.jpg

For the question above, I can only answer part a)
The set is not open or closed since it doesn't contain the neighborhood at 1 and int(S) = {(x,y) in R2 : x^2 + y^2 < 1} \ {(0,0)}. Is it right?

Also, can someone please help me on part b)

Thank you much.
May 6th 2009, 10:31 PM
Gamma

Quote:

Originally Posted by john_n82

For the question above, I can only answer part a)
The set is not open or closed since it doesn't contain the neighborhood at 1 and int(S) = {(x,y) in R2 : x^2 + y^2 < 1} \ {(0,0)}. Is it right?

Also, can someone please help me on part b)

Thank you much.

Well for part a) you got the right answer, but I am not sure for the right reason. You showed the set is in fact not open because I think what you meant to point out for instance there is no open nbhd about (1,0) which is contained in S, so it is not open. But you must make an argument for why it is not closed too. If it were closed, its complement would be open. But its complement has the origin as an isolated point. There is clearly no open nbhd about the origin which would be contained in the complement of S, so the complement is not open thus S is not closed.

Your interior is correct.

2)
This is just basically a square centered at the origin with side length 4.
That is $[-2,2] \times [-2,2]$ . This set is closed because it's complement is open. Think you can prove this? Pick an arbitrary point outside of the box and show there is an open nbhd contained in the complement of the box. The interior will be $(-2,2)\times(-2,2)$ See why?
May 6th 2009, 11:01 PM
john_n82

Gamma,

Thank you for a quick response.

I don't quite understand this term "...no open nbhd about..." Would you please explain it further more?

For this question, if I was asked to show a proper proof? What would I have to show?
May 6th 2009, 11:10 PM
Gamma

Righto, I assume you are taking a topology course? nbhd is just short for an neighborhood.

For $\mathbb{R}^2$ the basis for the standard topology is just the open balls of positive radius right? (there are lots of basis sets that will give you the standard topology, open cubes, euclidean metric, etc)

When I say there is no open neighborhood about a point I mean you cannot find an open set that contains the point in question. In particular if you show there is not a basis element containing the point inside the set you know there cannot possibly be an open set containing the point inside the set, so that is sufficient to show a set is not open. Make sense?
May 6th 2009, 11:16 PM
john_n82

No, I'm taking Analysis I, and just began to discuss about the Topology of Eclidean Space. This is the first time I've encountered these concepts of Topology.
May 6th 2009, 11:23 PM
Gamma

Okay, then you guys are thinking of $\mathbb{R}^n$ as a metric space. The basis for the metric space is the collection of balls centered at x of radius r>0 $B(x,r)=\{y\in \mathbb{R} | d(x,y) < r\}$ where d(x,y) represents the euclidean distance.

So in this context what I mean by open nbhd about x is simply a ball centered at x of radius positive r.
May 6th 2009, 11:32 PM
john_n82

Yes, for us, R^n is a metric space.

If I was asked to give a proper proof. What do I need to show?
May 6th 2009, 11:38 PM
Gamma

My initial proof works.

A set C is closed iff the complement of C is open.

A set $G\subset \mathbb{R}^2$ is open iff for every point $x\in G$ there is an open ball B (as defined above) centered at x such that $x\in B \subset G \subset \mathbb{R}^2$ .

This is what I mean when I say a nbhd about the point.
May 6th 2009, 11:41 PM
john_n82

http://img206.imageshack.us/img206/4306/89991493.jpg

For this problem, I said:

S is not open because it doesnt contain any neighborhood at n = 1.
The complement of S contains 0. But if (-a, a) is any neighborhood of 0, then there exists a very large N that 1/N < a. This neighborhood is not part of the complement, because 1/n is in the complement of S. Therefore the complement of S is not open. That means, however, S is not closed.

is it a proper proof for this problem?
May 6th 2009, 11:48 PM
Gamma

Very very close.

Showing it is not open is correct.

For showing it is not closed you may want to be a little careful.
Say S is not closed because its complement is not open. 0 is in the open, but consider any open neighborhood of 0 is of the form (-a,a). Then what you pointed out about the archimedian property of the reals is great! But from here you need to say, (-a,a) cannot be contained in the complement of S. And this is because $(-a,a) \cap S$ will contain all the points 1/n for n>N, so it is not empty. Thus there can be no open neighborhood about 0 contained in the complement, so the complement of S is not open showing S to be not closed.

But I think you are getting the hang of it beautifully.
May 6th 2009, 11:59 PM
john_n82

From some of the proofs were given in class, my professor proved an open set using esilon and delta. I am wondering that if you could show me a proof using esilon for #4?
May 7th 2009, 12:08 AM
Gamma

Well, you just did that on your last proof you suggested and maybe didn't realize it because you called it a, lol.

Basically to show a set S is open, for every $x \in S$ it needs to be the case that there exists a $\epsilon > 0$ such that $B(x,\epsilon)\subset S$ .

It just so happens that in your last proof, we were working in $\mathbb{R}$ and $B(x,\epsilon )=\{y\in \mathbb{R} | |x-y|< \epsilon \} = (-\epsilon, \epsilon )$
May 7th 2009, 12:21 AM
john_n82

lolz, it takes time to understand all of these concepts . I hope that all of these will make sense when I wake up tomorrow. (Nerd) Thank you much.