Continuity and Dense Subsets of the Real Numbers

In summary, the conversation discusses a proof for the statement that if f is continuous and f(x)=0 for all x in a dense subset of the real numbers, then f(x)=0 for all x \in \mathbb{R}. The proposed proof is to use the fact that f(x) is continuous, and for every given \varepsilon > 0, there exists a \delta > 0 such that |f(x)-f(a)| < \varepsilon for all |x-a| < \delta. However, this proof assumes uniform continuity, which may not be the case. An alternative proof by contradiction is suggested, where assuming f(x0)=c leads to a contradiction. The conversation also mentions the importance
  • #1
jgens
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Homework Statement



If [itex]f[/itex] is continuous and [itex]f(x)=0[/itex] for all [itex]x[/itex] in a dense subset of the real numbers, then [itex]f(x)=0[/itex] for all [itex]x \in \mathbb{R}[/itex].

Homework Equations



N/A

The Attempt at a Solution



Does this solution work? And if it does, can it be improved in some way?

Proof: From the continuity of [itex]f[/itex], for every [itex]\varepsilon > 0[/itex] we can find a [itex]\delta > 0[/itex], such that if [itex]|x-a| < \delta[/itex], then [itex]|f(x)-f(a)| < \varepsilon[/itex]. Because [itex]f(x)=0[/itex] for all [itex]x[/itex] in a dense subset of the real numbers, it's clearly possible to choose a number [itex]x_0[/itex] from [itex](a-\delta,a+\delta)[/itex] with [itex]f(x_0)=0[/itex]. This means that [itex]|f(x_0)-f(a)| = |f(a)| < \varepsilon[/itex]. Since this is necessarily true for any given [itex]\varepsilon > 0[/itex], it follows that [itex]|f(a)| < \varepsilon[/itex] for all [itex]\varepsilon > 0[/itex]. By the Archimedean property of the real numbers, 0 is the only real number satisfying this criterion, so [itex]f(a)=0[/itex]. Clearly, this holds for any number where [itex]f[/itex] is continuous, so [itex]f(x)=0[/itex] for all [itex]x \in \mathbb{R}[/itex], completing the proof.
 
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  • #2
Looks fine to me.
 
  • #3
Thanks. Is there any way that I can improve my proof then?
 
  • #4
jgens said:
Thanks. Is there any way that I can improve my proof then?

Actually, your proof is assuming f is uniformly continuous, which you don't necessarily have. It's more straightforward to do a proof by contradiction. Assume f(x0)=c where c is not zero. Can you show that leads to a contradiction?
 
  • #5
Dick, can you explain how my proof assumes uniform continuity? And yes, I know that I can prove it that way because that was my first thought, I'm just trying to see if it's possible to do it without using a contradiction.
 
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  • #6
jgens said:
Dick, can you explain how my proof assumes uniform continuity? And yes, I know that I can prove it that way because that was my first thought, I'm just trying to see if it's possible to do it without using a contradiction.

Actually, I take that back. The only reason I thought that is because you mentioned delta before you mentioned 'a'. I would put an "At any point a since f is continuous at a..." at the beginning. Or at least make it clear that delta depends on a. Otherwise a careless reader like me might think you are assuming otherwise.
 
  • #7
Thanks! I'm sorry about that. I'll be more careful in the future.
 

What are real numbers?

Real numbers are numbers that can be represented on a number line and include both rational and irrational numbers. Examples of real numbers include integers, fractions, decimals, and irrational numbers such as π and √2.

What is continuity of real numbers?

Continuity of real numbers refers to the property of a function where small changes in the input result in small changes in the output. In other words, there are no sudden jumps or breaks in the graph of the function.

What is a dense subset of real numbers?

A dense subset of real numbers is a set of numbers that is closely packed together on the number line. In other words, between any two numbers in the subset, there exists an infinite number of other numbers in the subset. An example of a dense subset is the set of rational numbers between 0 and 1.

How do we prove a subset is dense in the real numbers?

To prove that a subset is dense in the real numbers, we must show that for any two numbers in the subset, there exists a third number in the subset between them. This can be done using the Archimedean property of real numbers, which states that for any two real numbers, there exists a natural number n such that n is greater than the second number. By repeatedly dividing the interval between the two numbers into smaller parts, we can find a number in the subset that lies between them.

What is the significance of continuity and dense subsets in mathematics?

Continuity and dense subsets play important roles in various fields of mathematics such as calculus, analysis, and topology. They provide a framework for understanding the behavior of functions and the structure of real numbers. They also have practical applications in areas such as physics, engineering, and economics.

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