Left-hand, right-hand inequality proof problem Let $f$ be bounded and increasing in an open interval (a,b). Prove that, $$\lim_{x\to x_0-}{f(x)}$$ and $$\lim_{x\to x_0+}{f(x)}$$ both exist for $x_0\in (a,b)$ and satisfy the inequality $$\lim_{x\to x_0-}{f(x)}\le\lim_{x\to x_0+}{f(x)}.$$
Also, prove that the number of points $x_0$ where $f$ makes a "jump," i.e. where $$\lim_{x\to x_0-}{f(x)}<\lim_{x\to x_0+}{f(x)}$$ is countable.
For the first part I was going to show that $\lim_{x\to x_0}{f(x)}$ exists, which then follows that both $\lim_{x\to x_0-}{f(x)}$ and $\lim_{x\to x_0+}{f(x)}$ exist. However, I am stuck showing $\lim_{x\to x_0}{f(x)}$ exists by the information given. How would I go about this?
My problem with the second proof is similar as mentioned above but also how would I show that its countable?
 A: Answers to your first question:
Fix $x_0\in(a,b).$ 
If $f$ is a constant function, then it is continuous and hence both limits $\lim_{x\to x_0^-}f(x)$ and $\lim_{x\to x_0^+}f(x)$ exist and equal to $f(x_0).$
Assume that $f$ is not constant. 
Without loss of generality, assume that $f$ is strictly increasing. 
Consider the sets 
$$U = \{ f(x):x<x_0 \} , V = \{ f(x):x>x_0 \}.$$
Since $f$ is bounded, both $U$ and $V$ are nonempty and bounded above and below respectively. 
By the Axiom of Completeness, both $\sup U$ and $\inf V$ exist. 
Let $M = \sup U.$
We claim that $\lim_{x\to x_0^-}f(x) = M.$
Let $\varepsilon>0$ be given. 
By the definition of supremum, there exists $y =f(x_1)$ for some $x_1<x_0$ such that 
$$M - \varepsilon < f(x_1) < M.$$
Let $\delta = x_0-x_1>0.$ 
Then for any $x$ such that $x_0-\delta < x < x_0,$ by monotonicity of $f,$ we have 
$$M - \varepsilon < f(x_1) < f(x) < M.$$
Therefore, 
$$|f(x)-M| = M - f(x)  < M - (M - \varepsilon) = \varepsilon.$$
Hence, the claim is proven. 
Similarly, one can prove that $\lim_{x\to x_0^+}f(x) = \inf V$ by using similar technique. 
Now, we show that $\sup U \leq \inf V.$
Fix $x<x_0.$ 
Then for any $x'>x_0,$ by monotonicity of $f,$ we have 
$$f(x) \leq f(x').$$
Since $x'$ is arbitrary, we have 
$$f(x) \leq \inf V.$$
Again, since $x$ is arbitrary, we have 
$$\sup U \leq \inf V.$$
