(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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].

2. Relevant equations

N/A

3. 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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# Homework Help: Continuity and Dense Subsets of the Real Numbers

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