# Prove continuous function = 0

• JG89
In summary, the conversation discusses the concept of dense subsets of real numbers and uses it to prove that if a continuous function is equal to 0 for all elements in a dense subset, then it is equal to 0 for all real numbers. This is proven using the concept of continuity and by assuming the function is not equal to 0 for some real number, leading to a contradiction.

## Homework Statement

A subset of $$A \subseteq R$$ of real numbers is called dense if $$\forall \delta > 0 , \forall x \in R , \exists a \in A: |x-a| < \delta$$ .

Suppose $$A \subseteq R$$ is dense. Prove that if g is a continuous function with g(x) = 0 for all $$x \in A$$, then g = 0

## The Attempt at a Solution

From the continuity of g, we know that $$\forall \epsilon > 0, \exists \delta > 0 : |g(x) - g(x_0)| < \epsilon$$ whenever $$0 < |x-x_0| < \delta$$. Now let x_0 be any real number that is not an element of A. We must prove that g(x_0) = 0.

To argue by contradiction, first assume that g(x_0) > 0. Again from g's continuity, $$\forall \epsilon > 0, \exists \delta > 0 : g(x_0) - \epsilon < g(x) < g(x_0) + \epsilon$$ whenever $$0 < |x-x_0| < \delta$$. Putting $$\epsilon = g(x_0)/2$$, we have $$g(x_0) - g(x_0)/2 = g(x_0)/2 < g(x)$$. Now g(x_0)/2 is positive and as a consequence, for all x in the interval $$x_0 - \delta < x < x_0 + \delta$$, g(x) > 0.

Since A is dense, then we can find a value x = x* in our delta-interval that is an element of A. We know that g(x*) = 0, but from our assumption that g(x) > 0 we've proved that g(x*) > 0, and so this is a contradiction.

The proof for assuming g(x) < 0 is similar, and so g(x) = 0 for all x. QED

How does this proof look?

That looks just fine to me.

Thanks for checking :)

## 1. What does it mean for a function to be continuous?

A continuous function is one that has no abrupt changes or breaks in its graph. This means that for any input value, there is a corresponding output value, and the graph of the function is a smooth curve with no gaps or jumps.

## 2. How can a function be proven to be continuous?

A function can be proven to be continuous by showing that it meets the three criteria for continuity: 1) the function is defined at the point in question, 2) the limit of the function as the input approaches the point exists, and 3) the limit equals the value of the function at that point.

## 3. Why is it important to prove that a function is continuous?

Proving that a function is continuous is important because it allows us to make accurate predictions and calculations based on the function. It also ensures that the function is well-defined and can be used reliably in mathematical models and real-world applications.

## 4. What does it mean for a function to equal 0?

When a function equals 0, it means that the output of the function is 0 for every input value. This can also be represented as the function having a root or zero at that point, where the graph crosses the x-axis.

## 5. How can a function be proven to equal 0?

A function can be proven to equal 0 by finding the point or points where the function intersects the x-axis, or by showing that the output of the function is always 0 for every input value. This can be done through various methods such as algebraic manipulation or graphical analysis.