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?

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    $\begingroup$ You are right it follows imediatelly by a countability argument... For your last try, what happens if $x_k=a$ for some $k$ ? Much better to combine that idea with the nested Theorem: Let $I_0=[a,b]$ and at each step pick some $I_n \subset I_{n-1}$ using exactly the last idea, so that $x_n \notin I_n$ ;) $\endgroup$ – N. S. Jun 22 '11 at 16:32
  • $\begingroup$ What are the topics covered before (and including) the chapter where this exercise appears? btw, the density argument won't work, you can pick $\{x_n\} = \mathbb{Q}$. $\endgroup$ – Aryabhata Jun 22 '11 at 16:32
  • $\begingroup$ @Aryabhata: The Nested Interval Theorem, least upper bounds, completeness, etc.. $\endgroup$ – Damien Jun 22 '11 at 16:33
  • $\begingroup$ @user9176: So pick $I_0 = [a,b]$, $I_1 = [a+ \epsilon, b- \epsilon]$, $I_2 = [a+ 2 \epsilon, b- 2 \epsilon]$ etc...? $\endgroup$ – Damien Jun 22 '11 at 16:39
  • $\begingroup$ I think you should include such information in the question itself (and not just this question). $\endgroup$ – Aryabhata Jun 22 '11 at 18:11

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.

  • $\begingroup$ I remember seeing this in my first year course to prove that the reals are uncountable. $\endgroup$ – Beni Bogosel Jun 22 '11 at 16:50
  • $\begingroup$ So essentially you divide the initial interval in three parts. Then look at the part which doesn't contain $x_1$ and divide this into three parts. Then look at the part which doesn't contain $x_2$ and divide this into three parts, etc...? $\endgroup$ – Damien Jun 22 '11 at 16:56
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    $\begingroup$ Yes, that's the main idea. You need three or more parts to be sure that there is one which does not contain a give point. If you use only two points, you may have problems with the middle point. $\endgroup$ – Beni Bogosel Jun 22 '11 at 17:00
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    $\begingroup$ What you saw was essentially Cantor's first proof of the uncountability of the reals. He was fresh out of grad school in Berlin, where Nested Intervals was hot new stuff. It is natural that he used it in his proof. The familiar Diagonal Argument came a number of years later. $\endgroup$ – André Nicolas Jun 22 '11 at 17:22

The interval [a,b], as a closed subset of R, is a complete metric space . If the union of the terms of ${a_n}$ contained [a,b], that would violate Baire Category Theorem, which states that a complete metric space cannot be the countable union of nowhere-dense sets; any singleton is nowhere-dense in [a,b]; actually, the sequence as a whole--even if it converges--is nowhere-dense in [a,b].

Edit: My answer above was a misuse (or unhelpful use) of big machinery.

Consider these two cases:

1) ${a_n}$ converges, to, say, c. Then there is a small interval (c-r,c+r) that contains all-but-finitely-many points of ${a_n}$. Then there are only finitely-many points in each of [a,c-r] and in [c+r,b]. Define the function d($a_n,b_n)=|a_n-b_n|$, in each of these subintervals and throw- out terms that are equal to each other. The function reaches a minimum m (since a finite set of points is compact, or simply because the min. of a finite set exists and is well-defined ) after having thrown out the terms between each other, the value of m is >0. It follows that there is a gap of width at least m between any two terms in the sequence, and this gap is not filled by any other terms in the sequence.

2)${a_n}$ diverges in [a,b]. Then ${a_n}$ is not Cauchy, so that there is some r>0 with $|a_n-a_m|$>r for all m,n integers. So there is a gap in [a,b] not covered by the sequence $a_n$

  • $\begingroup$ I don't have that theorem yet. $\endgroup$ – Damien Jun 22 '11 at 16:41
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    $\begingroup$ My Bad: I am guilty of what I so often criticize. The big machinery should be used only after the theory behind it has been thought out carefully. $\endgroup$ – gary Jun 22 '11 at 17:01

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