# Nested Interval Theorem Supremum and Infimum Proof

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 $$\alpha = \sup\{A_n: n\in N\}$$ $$\text{and let}$$ $$\beta = \inf\{B_n:n\in N\}$$, $$\text{then}\,$$ $$[\alpha , \beta] = \bigcap_{n=1}^\infty I_n$$.

$$\text{Proof}$$

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

• What is $I_n$? ${}$ Nov 7, 2019 at 16:31
• I made the edit, sorry about that. Nov 7, 2019 at 16:35
• Should I assume $I_n = [A_n, B_n]$? Nov 7, 2019 at 16:39
• 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? Nov 7, 2019 at 16:41
• 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. Nov 7, 2019 at 16:48

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.
• @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. Nov 10, 2019 at 4:22