# Proof about dense set and continuous function

• R_beta.v3
In summary, if f is a continuous function where f(x) = 0 for all x in a dense set A, then f(x) must also equal 0 for all x. This is because the open interval (-δ + a, δ + a) must contain a point of A, and since f(z) = 0 (by hypothesis) and f(x) > 0 for all x in this interval, assuming f(a) > 0 leads to a contradiction. This can also be shown for f(x) < 0. Therefore, f(x) = 0 for all x. It is important to note the domain and codomain of f, which in this case is f:\mathbb{R}\rightarrow
R_beta.v3

## Homework Statement

If f is continuous, and f(x) = 0 for all x in A, where A is a dense set. Then f(x) = 0 for all x.

I am using the following definitions:
A set of real numbers A is dense if every open interval contains a point of A.
And the limit definition for a continuous function.

## The Attempt at a Solution

Suppose there is an a such that f(a) > 0, then since f is continuous, there is an δ > 0, such that, for all x,
if |x - a| < δ, then |f(x) - f(a)| < f(a). So f(x) > 0 for all x in (-δ + a, δ + a).
But since (-δ + a, δ + a) is an open interval it contains a point of A call it z, but that is impossible, because then f(z) = 0 (By hypothesis), and f(z) > 0 because f(x) > 0 for all x in (-δ + a, δ + a). Assuming there is an a such that f(a) > 0 leads to a contradiction.

I did basically the same to show that there are no x with f(x) < 0.
Therefore f(x) = 0 for all x.

Am I correct?

Also, hello.

That would be correct.

As a matter of correctness though, you should really mention somewhere what the codomain and domain of f are. I assume here that you're working with $f:\mathbb{R}\rightarrow \mathbb{R}$.

## 1. What is a dense set?

A dense set is a subset of a larger set where every point in the larger set is either a member or a limit point of the subset. This means that the subset is "spread out" throughout the larger set, and there are no "holes" or gaps in between its elements.

## 2. Why is a dense set important in mathematics?

A dense set is important because it allows us to approximate any point in a larger set with points from the subset. This is particularly useful in analysis and calculus, where we often need to find the limit of a function at a point.

## 3. What is the relationship between a dense set and a continuous function?

A continuous function is a function where small changes in the input result in small changes in the output. In order for a function to be continuous, its domain must be a dense set, meaning that we can get arbitrarily close to any point in the domain by choosing a value from the dense set.

## 4. How do we prove that a set is dense?

To prove that a set is dense, we need to show that every point in the larger set is either a member or a limit point of the subset. This can be achieved by using the definition of limit points or by showing that the closure of the subset is equal to the larger set.

## 5. Can a set be both dense and not dense at the same time?

No, a set cannot be both dense and not dense at the same time. A set is either dense or not dense, depending on whether or not it satisfies the definition of a dense set. It is possible for a set to be neither dense nor not dense, if it does not meet the criteria for either.

• Calculus and Beyond Homework Help
Replies
8
Views
327
• Calculus and Beyond Homework Help
Replies
30
Views
2K
• Calculus and Beyond Homework Help
Replies
2
Views
442
• Calculus and Beyond Homework Help
Replies
5
Views
352
• Calculus and Beyond Homework Help
Replies
4
Views
3K
• Calculus and Beyond Homework Help
Replies
6
Views
378
• Calculus and Beyond Homework Help
Replies
6
Views
2K
• Calculus and Beyond Homework Help
Replies
14
Views
469
• Calculus and Beyond Homework Help
Replies
8
Views
2K
• Calculus and Beyond Homework Help
Replies
22
Views
643