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I'm working through Courant's Calculus book, and in one section he proves that the Nested Intervals Theorem follows from the Monotone Convergence Axiom. Here there is no mention of the Least Upper Bound Axiom, it's only mentioned at the very end of the chapter.

Consider the sequence of nested closed intervals $[a_{1},b_{1}] \supset [a_{2},b_{2}] \supset ... $, where $a_{n}, b_{n}$ are real numbers.

Then by the definition of nested intervals, the sequence $(a_{n})$ is monotone increasing. By the Monotone Convergence Axiom, $ \lim_{n \to \infty}(a_{n}) = a$ exists.

Now consider any $m,n$ with $n>m$. Then we have

$$a_{m} \leq a_{n} \leq b_{m}$$

I'm following the argument so far. But then he says: hence also

$$a_{m} \leq \lim_{n \to \infty}(a_{n})=a \leq b_{m}$$

Why are we allowed to pass to the limit while keeping the inequality intact?

Maxis Jaisi
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2 Answers2

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Holding $m$ fixed, $a_m \leq a_n \leq b_m$ for every $n > m$, so the limit can be taken under inequality. A rigorous way to show this would be to define the constant sequences $(a_m)_{n>m},(b_m)_{n>m}$ and invoke this theorem.

Community
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Henricus V.
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Assume that $\lim_{n \to \infty}a_{n}=a > b_{m}$ for some $m$.

By definition of limit, if $\epsilon=\frac{a-b_m}{2}>0$ then there is $N$ such that $|a_n-a|<\epsilon$ for $n\geq N$.

Hence, for $n\geq N$, $$a_n>a-\epsilon=a-\frac{a-b_m}{2}=\frac{a+b_m}{2}>b_m$$ which contradicts your assumption $a_{n} \leq b_{m}$.

Robert Z
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