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?
Best Answer
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.