My apologies if this is a duplicate. I am relatively new to participating on here and I could not find this question. That being said, this is an extra problem for an introductory real analysis class I am currently taking, and I would like some critiques on this proof and where I can improve things, as the textbook had no solution or hints. Thanks in advance!

$\text{Show that if}$ $(I_n)_n\in N$ is a set of nested downward sequences of closed intervals and $\alpha = \sup\{A_n: n\in N\}$ $\text{and let}$ $\beta = \inf\{B_n:n\in N\}$, $\text{then}\,$ $[\alpha , \beta] = \bigcap_{n=1}^\infty I_n $.


First we will show that $[\alpha , \beta] \subseteq \bigcap_{n=1}^\infty I_n$

$\text{Let}$ $A = \{A_n:n\in N\}$ $\text{and}$ let $B = \{B_n:n\in N\}$.

$\text{Then let}$ $\alpha = \sup\{A_n: n\in N\}$ $\text{and let}$ $\beta = \inf\{B_n:n\in N\}$

Let $x \in [\alpha, \beta]$.

Then $\alpha \le x\le\beta$.

Because $A_n$ $\le$ $\alpha$ and $\beta$ $\le$ $B_n$ , then $A_n \le \alpha \le x \le \beta \le B_n$.

Therefore $x \in [A_n, B_n]$ and thus $x \in\bigcap_{n=1}^\infty I_n$.

So $[\alpha, \beta]\subseteq\bigcap_{n=1}^\infty I_n$.

Now we will show that $\bigcap_{n=1}^\infty I_n\subseteq [\alpha, \beta]$.

Let $x \in\bigcap_{n=1}^\infty I_n$.

So $A_n \le x \le B_n$.

Because $\alpha = \sup\{A_n\}$ and $\beta = \inf\{B_n\}$, $\,$ $A_n \le \alpha \le \beta \le B_n$, $\forall \text{n}\in N$.

Suppose that $A_n \le x \le \alpha$. Because $\alpha$ is the supremum of $A_n$, $\exists$ an element $A_x$ belonging to $[An: n \in N]$ such that $x \le A_x$, and so $x \notin [A_x, B_x]$ and therefore $x \notin \bigcap_{n=1}^\infty I_n$, which is a contradiction.

Then suppose that $\beta \le x \le B_n$. Because $\beta$ is the infimum of $B_n$, $\exists$ an element $B_x$ belonging to $[B_n: n \in N]$ such that $B_x \le x$, and so $x \notin$ $[A_x, B_x]$ and therefore $x \notin \bigcap_{n=1}^\infty I_n$, which is a contradiction.

Thus we can conclude that $\bigcap_{n=1}^\infty I_n \subseteq [\alpha, \beta]$.

  • $\begingroup$ What is $I_n$? ${}$ $\endgroup$
    – William M.
    Nov 7, 2019 at 16:31
  • $\begingroup$ I made the edit, sorry about that. $\endgroup$ Nov 7, 2019 at 16:35
  • $\begingroup$ Should I assume $I_n = [A_n, B_n]$? $\endgroup$
    – William M.
    Nov 7, 2019 at 16:39
  • $\begingroup$ Yes. The question as it is written in the textbook is this " Prove that $[\alpha, \beta] = \bigcap_{n=1}^\infty I_n$." Do you know a better way of asking or posting this question? $\endgroup$ Nov 7, 2019 at 16:41
  • 1
    $\begingroup$ The general idea is OK, the write up is far from ideal. Notice $\beta \leq B_n$ for every $n;$ also you need strict inequalities $x < \alpha$ or $\beta < x$ to really get a contradiction. Having said that, a simpler proof follows: bearing in mind the definitions of supremum and infinimum as the lowest greater bound and greatest lower bound, respectively, it follows that the relation $\alpha \leq x \leq \beta$ is equivalent to $A_n \leq x \leq B_n$ for all $n.$ Q.E.D. $\endgroup$
    – William M.
    Nov 7, 2019 at 16:48

1 Answer 1


Perhaps you forgot to mention that $I_n=[A_n, B_n] $ and the fact that the intervals are nested implies that sequence $\{A_n\} $ is increasing and bounded above by $B_1$ and sequence $\{B_n\} $ is decreasing and bounded below by $A_1$. It follows that $$\alpha=\sup \, \{A_n\mid n\in\mathbb{N} \}, \beta=\inf\, \{B_n\mid n\in\mathbb {N} \} $$ exist (your proof does not mention the bounded nature which guarantees existence of sup and inf). And since $A_n\leq B_n$ we have $\alpha\leq \beta$.

Note that by definition of inf and sup we have $$A_n\leq \alpha \leq \beta\leq B_n, \forall n\in\mathbb {N} $$ (your proof has some of these inequalities reversed, maybe a typo). From this it is obvious that $[\alpha, \beta] \subseteq [A_n, B_n] =I_n$ and hence $$[\alpha, \beta] \subseteq \bigcap\limits_{n=1}^{\infty} I_n$$ To prove the reverse is not that difficult either. If $x\in\bigcap\limits_{n=1}^{\infty} I_n$ then $x\in[A_n, B_n] $ for all $n\in\mathbb {N} $ ie $A_n\leq x\leq B_n$. By definition of sup and inf it now follows that $\alpha\leq x\leq\beta$ ie $x\in[\alpha, \beta]$. Therefore $\bigcap\limits_{n=1}^{\infty} I_n\subseteq [\alpha, \beta] $. The argument for this part in your proof is needlessly elaborate / complicated.

  • $\begingroup$ Thank you so much, that is a very helpful and insightful answer. Is there any other advice you have one tackling problems similar to this or involving nested intervals? $\endgroup$ Nov 9, 2019 at 15:19
  • $\begingroup$ @Matthew: First of all you need to understand all the definitions in your natural language devoid of any mathematical symbols (this is not so difficult) and next you need to have a firm grasp on completeness of real numbers (difficult). See this answer for more details. $\endgroup$
    – Paramanand Singh
    Nov 10, 2019 at 4:20
  • $\begingroup$ @Matthew: as an example supremum $S$ of a non empty set $A$ is such that no member of $A$ exceeds the supremum $S$ but every number smaller (less) than $S$ is exceeded by some member of $A$. You should be able to understand definitions in this manner. $\endgroup$
    – Paramanand Singh
    Nov 10, 2019 at 4:22

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