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I'm working on Real Analysis problems, and I'm stuck on the following:

Show that a non degenerate interval of real numbers fails to be finite.

Royden's book has a theorem from the section where this question comes from saying "A non degenerate interval of real numbers is uncountable", but we aren't allowed to use this.

All I know is that a non degenerate interval of real numbers contains more than one member. I thought about just working with the case of at least $x_1,x_2\in \mathbb{R}$, but I'm not sure what to say after:

Let $E=\{a,b\in \mathbb{R}| x_0=a<x_1<x_2<b=x_n\}$. Recursively, we can define $x_{i+j/2}$=$x_i+\frac{x_i-x_j}{2}$ for $j<i$. This produces a midpoint for every subinterval, thus we can find an infinite set inside of E. Thus not countable?

Can anyone give me a hint or an overall idea? Not looking for complete solutions, please.

Thanks!

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  • $\begingroup$ Are you allowed to use the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$? The midpoint approach should also work for showing that the interval is infinite. $\endgroup$ – carmichael561 Sep 20 '16 at 20:33
  • $\begingroup$ I believe so. It was featured in the previous section. $\endgroup$ – User5634 Sep 20 '16 at 20:34
  • $\begingroup$ You could show that being an interval $I$ is equivalent to $x, y \in I$ and $x < z < y \Rightarrow z \in I$. $\endgroup$ – 3-in-441 Sep 20 '16 at 20:34
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Let $I$ be a nondegenerate interval and suppose that $I$ is finite. Then there must be a natural number $n \geq 2$ such that $I$ has $n$ elements. Take arbitrary elements $a_{n-1}$ and $a_{n}$ in the interval and suppose that $a_{n-1} < a_{n}$ . Then $a_{n-1} < \frac{a_{n-1} + a_{n}}{2} < a_{n}$, so $\frac{a_{n-1} + a_{n}}{2} \in I $. This shows that $I$ cannot have $n$ elements, since $n-1$ and $n$ are consecutive natural numbers.

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