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Here's an excerpt from Spivak's Calculus, 4th Edition, page 96:

If we consider the function

$$ f(x)= \left\{ \matrix{0, x \text{ irrational} \\ 1, x \text{ rational}} \right. $$ then, no matter what $a$ is, $f$ does not approach an number $l$ near $a$. In fact, we cannot make $|f(x)-l|<\frac{1}{4}$ no matter how close we bring $x$ to $a$, because in any interval around $a$ there are number $x$ with $f(x)=0$, and also numbers $x$ with $f(x)=1$, so that we would need $|0-l| < \frac{1}{4}$ and also $|1-l| < \frac{1}{4}$.

How do I show that there is no condition that can meet the inequality $|0-l| < \frac{1}{4}$ and also $|1-l| < \frac{1}{4}$ ?

Thank you in advance for any help provided.

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  • $\begingroup$ Think about it. The first says, $\ell$ is less than $1/4$ away from zero. The other says, $\ell$ is less than $1/4$ away from one. How can they both be true? $\endgroup$ – Gerry Myerson Jul 21 '13 at 9:01
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I’ll give you several approaches; all have their uses, though in this case the first seems much the simplest.

For any real numbers $a$ and $b$, $|a-b|$ is the distance between $a$ and $b$. If $|0-\ell|<\frac14$, then the distance between $\ell$ and $0$ is less than $\frac14$. Similarly, if $|1-\ell|<\frac14$, then the distance between $\ell$ and $1$ is less than $\frac14$. If I had such an $\ell$, $0$ and $1$ would be less than $\frac14+\frac14=\frac12$ units apart!

You can think of this in terms of open intervals: if the distance from $0$ to $\ell$ is less than $\frac14$, then $\ell\in\left(-\frac14,\frac14\right)$, and if the distance from $1$ to $\ell$ is also less than $\frac14$, then $\ell\in\left(\frac34,\frac54\right)$. Since these two intervals have empty intersection, this is impossible.

Alternatively, you can use the triangle inequality directly:

$$1=|1-0|=|(1-\ell)+(\ell-0)|\le|1-\ell|+|\ell-0|=|1-\ell|+|\ell-0|<\frac14+\frac14=\frac12\;,$$

which is clearly absurd.

Finally, you can simply translate the absolute value inequalities into pairs of ordinary inequalities: $|a-b|<r$ says exactly the same thing as $-r<a-b<r$, so your hypotheses are that $$-\frac14<0-\ell<\frac14\quad\text{and}\quad-\frac14<1-\ell<\frac14\;.$$ Solve each for $\ell$:

$$-\frac14<\ell<\frac14\quad\text{and}\quad\frac34<\ell<\frac54\;,$$

which again is clearly impossible.

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Use triangular inequality. Suppose we could find such $l$, then $$1=|0-1|\le |0-l|+|l-1|<1/4+1/4=1/2,$$ which is impossible.

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When dealing with equations including absolute values, it's often convenient to rewrite them as equations without absolute values.

In your case, $$|0-l| < \frac{1}{4}$$ gives: $$-\frac{1}{4} < l < \frac{1}{4}$$

And $$|1-l| < \frac{1}{4}$$ gives: $$\frac{3}{4} < l < \frac{5}{4}$$

And now it's easy to spot the contradiction.

It's easy to sometimes overlook cases when rewriting absolute valued equations, but it's good practice.

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