Here it goes(Please correct me if I am wrong)

Suppose $\mathbb{R}$ is countable.

Then, $\mathbb{R}=\{x_1,x_2,x_3,\dots \}$.

Take each $x_n$ in $\mathbb{R}$ and enclose it in an open ball of length $\frac{1}{2^n}$, $x_n\in \left(x_n-\frac{1}{2^{n+1}}, x_n+\frac{1}{2^{n+1}}\right)=I_n$

The sum of length of Interval $I_n$ is $1/2+1/4+1/8+\dots$ Now observe that $x_n\in \mathbb{R}$ and $\mathbb{R}=\cup_{n=1}^{\infty}\{x_n\}\subseteq \cup I_n$

This means whole real line is contained in the union of intervals whose length add up to $1$.

This is a contradiction and therefore $\mathbb{R}$ is uncountable.

  • 1
    $\begingroup$ What properties of the real line are you using to say that it can not be covered by a union of intervals whose lengths add up to $1$?. $\endgroup$ – Kavi Rama Murthy Oct 22 '18 at 7:41
  • $\begingroup$ @KaviRamaMurthy Yes. I thought that this is obvious and I took it for granted. So that means I have to prove this result too. $\endgroup$ – StammeringMathematician Oct 22 '18 at 7:42
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    $\begingroup$ This does not prove that $\mathbf{R}$ is uncountable, indeed applying the same reasoning to $\mathbf{Q}$ you show that $\mathbf{Q}$ can be contained in a collection of intervals such that their lengths sum to any $\varepsilon>0$ even tough $\mathbf{Q}$ is not bounded. What you have shown is that any countable set has measure zero $\endgroup$ – Olof Rubin Oct 22 '18 at 7:59
  • $\begingroup$ This would be valid if preceded by some basic results of measure theory, in order to justify the assertion that $\Bbb R $ cannot be covered by a countable family of open intervals whose lengths add to only $1.$ Which can be done. $\endgroup$ – DanielWainfleet Oct 22 '18 at 10:42

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