# Is this proof of showing $\mathbb{R}$ is uncountable correct?

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.

• 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 $1$?. – Kavi Rama Murthy Oct 22 '18 at 7:41
• @KaviRamaMurthy Yes. I thought that this is obvious and I took it for granted. So that means I have to prove this result too. – StammeringMathematician Oct 22 '18 at 7:42
• 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 – Olof Rubin Oct 22 '18 at 7:59
• 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. – DanielWainfleet Oct 22 '18 at 10:42