How do I prove the following: Let f be increasing and bounded below on (a,b) with largest lower bound l. Then f(x) tends to l as x tends to a+.

Question

Currently working my way through some proofs for monotone functions and am struggling with the proof stated in the question. I have proved: Let $f$ be increasing and bounded above on $(a,b)$ with smallest upper bound $L$. Then $f(x) \rightarrow L$ as $x \rightarrow b-$. But I am struggling to prove the second case. Does I simply need to find an $x$ where $a<x<a + \delta$ that satisfies $|f(x)-l|< \epsilon$?

Answer 1

You may reason as follows:

• As $f$ is monotone increasing and bounded from below, it exists $\lambda = \lim_{x\to a^+}f(x)$
• Let $l$ be the greatest lower bound of $f$ on $(a,b) \Rightarrow l \leq f(x)$ on $(a,b) \Rightarrow l \leq \lambda$.
• Assuming $l < \lambda$ would result in $l' = \frac{l+\lambda}{2}$ being a lower bound for $f$ but greater than the greates one: $\Rightarrow$ Contradiction!