TITLE: Sequences and Intervals
QUESTION [6 upvotes]: I came across another real analysis problem in my self study:

Let $[a,b]$ be a closed interval in $\mathbb{R}$ and let $(x_n)$ be any sequence in $\mathbb{R}$. Prove that $[a,b]$ contains a real number not equal to any term of the sequence.  

I think I need to use the nested interval theorem:
Theorem. If $(I_n)$ is a nested sequence of closed intervals, then the intersection of the $I_n$ is nonempty. In other words, if $I_n = [a_n, b_n]$, where $a_n \leq b_n$ and $I_1 \supset I_2 \supset I_3 \supset \dots$ and $a = \sup \{a_n: n \in \mathbb{Z}^{+} \}$, $b = \inf \{b_n: n \in \mathbb{Z}^{+} \}$ then $a \leq b$ and $\bigcap_{n=1}^{\infty} [a_n, b_n] = [a,b]$. 
It seems obvious if we know that the interval is uncountable and the sequence is countable. Or could you do the following: Pick an arbitrary element $x_0$ of $(x_n)$ in $[a,b]$ (if there is none then we are done). By denseness, there is a real number $\alpha$ between $a$ and $x_0$. If $\alpha$ is in the sequence pick another number $\alpha_1$ between $a$ and $\alpha$. Keep doing this until you find a number not in the sequence. 
Would this idea work?

REPLY [4 votes]: To use the nested interval theorem, you divide the interval in three parts (compact intervals of equal length). There is a part $I_1$ which does not contain $x_1$. Next, divide this part in three parts. There is one part $I_2 \subset I_1$ which doesn't contain $x_2$ or $x_1$. Doing this inductively you get a decreasing sequence of intervals $(I_n)$ such that $I_n$ does not contain $x_1,...,x_n$. The intersection of these intervals is nonvoid, and it does not contain any of the elements of the sequence.