# Disjoint open sets homework

1. Nov 14, 2005

### SomeRandomGuy

1.) Let A and B be subsets of real numbers. Show that if there exist disjoint open sets U,V where A is a subset of U and B is a subset of V, then A and B are seperated.

In doing the proof of this, I have come to the conclusion that any 2 disjoint, open sets are seperated. However, I don't see this in my book or notes anywhere, so I don't think I can use it. I was wondering if someone could offer some guidance as to how to prove this otherwise?

2.) a.) Show that using definition 4.3.1 (I will state it below), that any function f with the integers as the domain is necessarily continuous everywhere.

b.) Show in general that if c is an isolated point of A which is a subset of the reals, then f: A -> R is continuous at c.

Definition 4.3.1 - A function f is continuous at a point c if, for all epsilon > 0, there exists a delta > 0 such that whenever |x-c|<delta, |f(x)-f(c)|<epsilon.

To be completely honest, I don't have even a remote clue on this one.

Thanks for any assistance, I appreciate it greatly.

2. Nov 14, 2005

### AKG

What definition do you have for "separated"? Question 1) should follow directly from definition.

For a function with domain A to be continuous at a point c (which must be an element of A), given any real e > 0 you must be able to find a real d > 0 such that for any x in A satisfying |x - c| < d, you have that |f(x) - f(c)| < e.

If A is just the integers, and c is some point in A, then what can you say about those x in A satisfying |x - c| < d. For example, if we let c = 15, d = 7.214, which x satisfy |x - c| < d?

Next example, let our domain be the integers, and let our function be defined by f(n) = n. Show that it is continuous at 3. First, start by picking an appropriate d when e = 12.2. What is the maximum possible d that will work for e = 12.2 by the way? Then do it for e = 3.2. Then e = 1.1. Then e = 1. Then e = 0.9. Then e arbitrarily small (thereby proving continuity). Generalize from this example what d you can choose that will in fact work for any e. Generalize from this example what you can do for isolated points of an arbitrary domain.

3. Nov 14, 2005

### SomeRandomGuy

Thanks for your quick response. For the second question, that is very helpful.

As for question 1, the definition i have of seperated is as follows:

Two sets A and B are seperated if Closer(A) intersect B is empty and A intersect Closure(B) is empty.

EDIT: Just going to type up a summary of what I have for problem 2 that I listed.

Proof: Let f be a function. Let d = e > 0. If we take x = c for all c, |x-c|<d is always satisfied, thus |f(x)-f(c)|<e for ell e>0. Thus, f is continuous everywhere.
This is just a summary of my proof but I was wondering if this is the right idea.

Last edited: Nov 14, 2005
4. Nov 15, 2005

### AKG

You want to show that A intersect Cl(B) = 0 given that A < U, B < V, U and V are open, and U intersect V = 0. Hint, think about the complement of U, what you know about it, and how it relates to the closure of B.

For your proof of question 2, it doesn't appear to make sense. I don't know what you mean by "if we take x = c". You can't choose x. You have to show that for any e > 0, you can choose some d > 0 such that for all x in A satisfying |x-c|<d, the following holds: |f(x) - f(c)| < e.

I asked a number of questions in my previous post. I suggest you explicitly answer them.

For example, if we let c = 15, d = 7.214, which x satisfy |x - c| < d?

Next example, let our domain be the integers, and let our function be defined by f(n) = n. Show that it is continuous at 3. First, start by picking an appropriate d when e = 12.2. What is the maximum possible d that will work for e = 12.2 by the way? Then do it for e = 3.2. Then e = 1.1. Then e = 1. Then e = 0.9. Then e arbitrarily small (thereby proving continuity). Generalize from this example what d you can choose that will in fact work for any e. Generalize from this example what you can do for isolated points of an arbitrary domain.

Hint, when dealing with the natural numbers as your domain, there is a d that will work for each and every e > 0, AND for each and every function f! No matter what function f happens to be, and no matter how small you choose e, there is a d, in fact you can choose a relatively "large" d, such that every x in your domain satisfying |x-c| < d will satisfy |f(x) - f(c)| < e. The trick is that since the elements of N are so "spread out", you can pick a d such that VERY FEW elements satisfy |x-c| < d. Consider a domain of real numbers, with c = 2, and d = 3. Then you are looking at x that satisfy |x-c| < d. The set of x that satisfy this is the set (-1,5), which contains uncountably many elements. It's, relatively speaking, a huge set. It's got more elements than all of the natural numbers. On the other hand, suppose we have a domain of integers. Then the set we're looking at is not uncountable, not even countably infinite. It is finite, and it only contains 5 elements; it is {0,1,2,3,4}. Can you choose a d such that the set of elements x of your domain satisfying |x-2| < d has exactly 4 elements? (No, you can't). What about exactly 3 elements? (Yes, you can). These hints should be more than enough to help you figure out what to do.