Why Does Spivak Exclude Specific Rational Points in His Limit Proof?

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SUMMARY

The discussion centers on Spivak's proof regarding the limit of the function defined as \( f(x) \) in his Calculus book, specifically addressing the limit \( \lim_{x \rightarrow a} f(x) = 0 \) for all \( a \in (0,1) \). The user expresses confusion over Spivak's exclusion of rational points \( \frac{1}{n}, \ldots, \frac{n-1}{n} \) when determining values of \( x \) for which \( |f(x)| < \epsilon \). The user questions why \( x = \frac{1}{n} \) does not satisfy the condition, indicating a misunderstanding of the proof's intent to select a \( \delta \) that avoids these specific rational points.

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Homework Statement



Consider the function
$$
f(x) := \begin{cases}
0, & if~ x \in (0,1) - \mathbb{Q} \\
\frac{1}{q}, & if~ x = \frac{p}{q} \in (0,1) \cap \mathbb{Q} \mbox{ in lowest terms. } \\
\end{cases}
$$

In his Calculus book, Spivak shows us that ##\lim_{x \rightarrow a} f(x) = 0## for all ##a \in (0,1)##. I am having trouble with a few of his remarks

Homework Equations

The Attempt at a Solution



First, he "let ##n## be a natural number so large that ##\frac{1}{n} \le \epsilon##." This is fine, I suppose; this is just the Archimedean property, although I have always taken ##\frac{1}{n}## strictly less than ##\epsilon##. He then goes on to say "that the only numbers ##x## for which ##|f(x)-0| < \epsilon## could be false are:

$$\frac{1}{n},...,\frac{n-1}{n}$$"

I find this rather perplexing...Why wouldn't ##x = \frac{1}{n}## satisfy ##|f(x)| < \epsilon##, for example? Spivak's aim seems to be choosing a ##\delta## such that any ##x## in ##(a-\delta, a+\delta)## is not anyone of the numbers in the above list but is smaller, so that ##|f(x)| < \epsilon##. But, as I ask, why wouldn't ##x = \frac{1}{n}## work?

PS I can't figure out why there are horizontal lines through my text.
Mod note: Fixed this. For some reason the overstruck text was surrounded by overstrike BB code tags -- [ s] and [ /s]. I removed them.
 
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Bashyboy said:

Homework Statement



Consider the function
$$
f(x) := \begin{cases}
0, & if~ x \in (0,1) - \mathbb{Q} \\
\frac{1}{q}, & if~ x = \frac{p}{q} \in (0,1) \cap \mathbb{Q} \mbox{ in lowest terms. } \\
\end{cases}
$$

In his Calculus book, Spivak shows us that ##\lim_{x \rightarrow a} f(x) = 0## for all ##a \in (0,1)##. I am having trouble with a few of his remarks

Homework Equations

The Attempt at a Solution



First, he "let ##n## be a natural number so large that ##\frac{1}{n} \le \epsilon##." This is fine, I suppose; this is just the Archimedean property, although I have always taken ##\frac{1}{n}## strictly less than ##\epsilon##. He then goes on to say "that the only numbers ##x## for which ##|f(x)-0| < \epsilon## could be false are:

$$\frac{1}{n},...,\frac{n-1}{n}$$"

I find this rather perplexing...Why wouldn't ##x = \frac{1}{n}## satisfy ##|f(x)| < \epsilon##, for example? Spivak's aim seems to be choosing a ##\delta## such that any ##x## in ##(a-\delta, a+\delta)## is not anyone of the numbers in the above list but is smaller, so that ##|f(x)| < \epsilon##. But, as I ask, why wouldn't ##x = \frac{1}{n}## work?

PS I can't figure out why there are horizontal lines through my text.

I had such horizontal lines in one of my posts about two weeks ago. The only cure was to log off and re-boot my computer. It worked for me, maybe not for you.
 
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Ray Vickson said:
I had such horizontal lines in one of my posts about two weeks ago. The only cure was to log off and re-boot my computer. It worked for me, maybe not for you.
See my added note in the first post of this thread.
 
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