# Least Upper Bounds

This is about Chapter 8 Question 8 in Spivak:

Suppose that $f$ is a function such that $f(a) \leq f(b)$ whenever $a < b$. Prove that $\displaystyle\lim_{x\to a^{-}}f(x)$ and $\displaystyle \lim_{x\to a^{+}}f(x)$ both exists.

Here is my work so far for $\displaystyle\lim_{x\to a^{-}}f(x)$:

Considering the interval $(-\infty, a)$. Since $f(a) \leq f(b)$ whenever $a < b$ it is clear that for every $x\in (-\infty, a)$ $f(x) \leq f(a)$. That is $f(x)$ is bounded above at $(-\infty, a)$ by $f(a)$. Since $f(x)$ is bounded above, $f(x)$ has a least upper bound, let it be $\alpha$.

I want to prove that

$\displaystyle\lim_{x\to a^{-}}f(x) = \alpha$

By definition $\forall \varepsilon > 0, \exists \delta > 0$ st. $\forall x$ if $0 < a-x< \delta$ then $|f(x) - \alpha | < \varepsilon$

How would I choose $\delta$ in this case?

Thanks.

-

Let $\alpha = \sup_{x \in (-\infty,a)} f(x)$. You know $\alpha < \infty$ because if $b>a$, we have $f(x) \leq f(b)$ for all $x \in (-\infty,a)$.
By definition of $sup$, for all $\epsilon>0$, there exists $x' \in (-\infty, a)$ such that $f(x')>\alpha-\epsilon$.
Let $\delta = a-x'$. Then if $0 < a-x < \delta$, we have $x'<x<a$, and hence $f(x')\leq f(x) \leq \alpha$. This gives $\alpha-f(x) \leq \alpha - f(x') < \epsilon$, as desired.