Prove that the limit exists of an increasing and bounded function

This was an exam question I had last year but wasn't able to answer it (and still can't).

Suppose $a<b$, and $f : (a,b) \to \mathbb{R}$ is a function that is both increasing and bounded. Prove that $\displaystyle \lim_{x\to b^-} f(x)$ exists.

Could someone construct a proof for this? In the exam I see I attempted to use the sequential definition of a limit but had no luck,

Solution 1:

Since $B = \{f(x): x \in (a, b)\}$ is a bounded set of $\mathbb{R}$, $c = \sup B$ exists. We show that $$\displaystyle \lim_{x \to b^-} f(x) = c.$$ Let $\epsilon > 0$ be given, there exists $x_0 \in (a, b)$ such that $f(x_0) > c - \epsilon$. Choose $\delta = b - x_0 > 0$, then if $|x - b| < \delta$ then $ b - x < b - x_0$ or $x_0 < x < b$ we have: $f(x) > f(x_0) > c - \epsilon$. But $f(x) \leq c < c + \epsilon$. So $c + \epsilon > f(x) > c - \epsilon$ or $|f(x) - c| < \epsilon$. This means: $\displaystyle \lim_{x \to b^-} f(x) = c$.

Solution 2:

This is just elaborating what Kf-Sansoo stated. I couldn't understand his proof without rewriting it, and going through phrase by phrase.

Let $B = \{f(x): x \in (a,b)\}$. As Kf-Sansoo says, $c = \sup B$ exists.

We want to prove the left-sided limit definition for $f$, ie. $\forall \epsilon \gt 0 \exists \delta \gt 0 \forall x \in (a,b), (0 \lt b-x \lt \delta) \implies |f(x) - c| \lt\epsilon$.

Let $\epsilon \gt 0$. Choose $\delta$ to be as Kf-Sansoo said: by definition of $\sup$, if we take a little away from $c$ there must be an element $f(x_0) \in B$ such that $f(x_0) \gt c - \epsilon$ or else $c$ is not the $\sup$. You can prove that by contradiction.

Now if $x \in (a,b)$ and $(0\lt b - x \lt \delta = b - x_0)$, then $x_0 \lt x$ by subtracting $b$ from both sides of the inequality and swapping $x,x_0$ and negating.

Now we have now $c - \epsilon \lt f(x_0) \leq f(x)$ by the increasing property of $f$. But we also have that $f(x) \leq c$ by definition of $\sup$, and trivially $f(x) \lt c + e$. The pair of inequalities $c - \epsilon \lt f(x) \lt c + \epsilon$ is easily proven to be equivalent to the desired inequality involving absolute value.

A similar proof for the right-sided limit should also work.