Skip to main content
added update on compactness, pedagogical remark
Source Link
Russ Woodroofe
  • 3.6k
  • 1
  • 25
  • 22

I haven't seen the following proof mentioned, which I learned from Hai Dang at Mississippi State.

Suppose the reals are countable, and let $a_1, a_2, a_3, \dots$ be an enumeration. For each $j$, let $I_j$ be an interval centered at $a_j$ and having length $1 / 2^j$.

Since the sequence $\{a_j\}$ enumerates the reals, it follows that $\bigcup_{j=1}^\infty I_j = \mathbb{R}$. But since the sum of the lengths of the $I_j$ is the geometric series $$ \sum_{j=1}^\infty \frac{1}{2^j} = 1,$$ this is nonsense.

(Pretty much the same proof yields that $[0,2]$ is uncountable.)

There's probably a relation with the outer measure proof previously posted, but this one seems more concrete.

Update: Two extended comments:

  1. As mentioned below in my comment below, proving that [0,2] can't be covered by open intervals of length summing to 1 is easily done with compactness. I didn't quickly see how to prove it without compactness. And of course compactness of a closed interval uses the Nested Interval Theorem, as the original poster was trying to avoid.

  2. I presented a proof along these lines in an Analysis I class. I liked how it came out a great deal, because it gave me a good reason to show students a typical application of compactness. Compactness is abstract and difficult for beginning students to grasp, and I usually find it difficult to find good applications. Students (or at least some of them) seemed to like this one a fair bit.

I haven't seen the following proof mentioned, which I learned from Hai Dang at Mississippi State.

Suppose the reals are countable, and let $a_1, a_2, a_3, \dots$ be an enumeration. For each $j$, let $I_j$ be an interval centered at $a_j$ and having length $1 / 2^j$.

Since the sequence $\{a_j\}$ enumerates the reals, it follows that $\bigcup_{j=1}^\infty I_j = \mathbb{R}$. But since the sum of the lengths of the $I_j$ is the geometric series $$ \sum_{j=1}^\infty \frac{1}{2^j} = 1,$$ this is nonsense.

(Pretty much the same proof yields that $[0,2]$ is uncountable.)

There's probably a relation with the outer measure proof previously posted, but this one seems more concrete.

I haven't seen the following proof mentioned, which I learned from Hai Dang at Mississippi State.

Suppose the reals are countable, and let $a_1, a_2, a_3, \dots$ be an enumeration. For each $j$, let $I_j$ be an interval centered at $a_j$ and having length $1 / 2^j$.

Since the sequence $\{a_j\}$ enumerates the reals, it follows that $\bigcup_{j=1}^\infty I_j = \mathbb{R}$. But since the sum of the lengths of the $I_j$ is the geometric series $$ \sum_{j=1}^\infty \frac{1}{2^j} = 1,$$ this is nonsense.

(Pretty much the same proof yields that $[0,2]$ is uncountable.)

There's probably a relation with the outer measure proof previously posted, but this one seems more concrete.

Update: Two extended comments:

  1. As mentioned below in my comment below, proving that [0,2] can't be covered by open intervals of length summing to 1 is easily done with compactness. I didn't quickly see how to prove it without compactness. And of course compactness of a closed interval uses the Nested Interval Theorem, as the original poster was trying to avoid.

  2. I presented a proof along these lines in an Analysis I class. I liked how it came out a great deal, because it gave me a good reason to show students a typical application of compactness. Compactness is abstract and difficult for beginning students to grasp, and I usually find it difficult to find good applications. Students (or at least some of them) seemed to like this one a fair bit.

Source Link
Russ Woodroofe
  • 3.6k
  • 1
  • 25
  • 22

I haven't seen the following proof mentioned, which I learned from Hai Dang at Mississippi State.

Suppose the reals are countable, and let $a_1, a_2, a_3, \dots$ be an enumeration. For each $j$, let $I_j$ be an interval centered at $a_j$ and having length $1 / 2^j$.

Since the sequence $\{a_j\}$ enumerates the reals, it follows that $\bigcup_{j=1}^\infty I_j = \mathbb{R}$. But since the sum of the lengths of the $I_j$ is the geometric series $$ \sum_{j=1}^\infty \frac{1}{2^j} = 1,$$ this is nonsense.

(Pretty much the same proof yields that $[0,2]$ is uncountable.)

There's probably a relation with the outer measure proof previously posted, but this one seems more concrete.