Proving continuity for functions with the integers as the domain

In summary, the conversation discusses proving that two subsets of real numbers are separated if there exist disjoint open sets where one subset is contained in one set and the other subset is contained in the other set. The conversation also includes a request for assistance in proving the continuity of a function with the integers as the domain and a discussion on isolated points of a subset of real numbers. The expert suggests thinking about the complement of the open set and its relationship to the closure of the other subset, and provides hints for proving continuity at a point and generalizing from a specific example.
  • #1
SomeRandomGuy
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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.
 
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  • #2
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
Thanks for your quick response. For the second question, that is very helpful.

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

Two sets A and B are separated 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.
 
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  • #4
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.
 

FAQ: Proving continuity for functions with the integers as the domain

What are disjoint open sets in topology?

Disjoint open sets in topology are sets that do not have any elements in common and are also open sets, meaning they contain all of their boundary points. In other words, they are sets that do not overlap with each other and do not include any of their boundary points.

How do you prove that two sets are disjoint open sets?

To prove that two sets are disjoint open sets, you need to show that they have no elements in common and that they are both open sets. This can be done by showing that the intersection of the two sets is empty and that each set contains all of its boundary points.

Can disjoint open sets be non-continuous?

Yes, disjoint open sets can be non-continuous. In topology, continuity refers to a function preserving the topological structure of sets. So while disjoint open sets do not have any elements in common, they can be mapped to each other in a non-continuous way.

What is the importance of disjoint open sets in topology?

Disjoint open sets are important in topology because they help define the concept of separation. In fact, two sets are considered separated if they are disjoint open sets. This concept is crucial in many proofs and theorems in topology.

Are disjoint open sets unique?

No, disjoint open sets are not necessarily unique. In some cases, there may be multiple disjoint open sets that satisfy certain conditions. However, there are also cases where disjoint open sets are unique, such as in the separation axioms of topology.

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