  ## Proof of the Boundedness Theorem

If $f(x)$ is continuous on $[a,b]$, then it is also bounded on $[a,b]$.

#### Proof:

Consider the set $B$ of $x$-values in $[a,b]$ such that $f(x)$ is bounded on $[a,x]$.

Note that $a$ is in $B$, as for every $x$ in $[a,a]$ (there is only one such $x$) the value of $f(x)$ is $f(a)$, which then serves as a bound.

Note also that if some $x_1 \gt a$ is in $B$, then all $x$-values between $a$ and $x_1$ must also be in $B$.

Given that $f$ is continuous on the right at $a$, for $\epsilon = 1$ we can find a $\delta \gt 0$ such that $|f(x)-f(a)| \lt 1$ for all $x$ in $[a,a+\delta]$. Consequently, on the interval $[a,a+\delta]$, $f$ is bound between $f(a)-1$ and $f(a)+1$ making all $x$-values in $[a,a+\delta]$ elements of $B$.

At this point, we know that $B$ contains some interval of $x$-values, of non-zero length, and closed at its left end by $a$.

Noting that no element of $B$ can be greater than $b$, consider the supremum of $B$ (i.e., the smallest value that is greater than or equal to every value in $B$); let us call it $s$. We aim to show that $s=b$.

We can immediately deduce that $s \gt a$, given the non-zero length of the closed interval in $B$ mentioned above.

That leaves as possibilities for $s$ the values in $(a,b]$. Consider the implications, however, if $s \lt b$.

Recall $f$ is continuous at $x=s$. As such, for $\epsilon = 1$ we can find a $\delta \gt 0$ such that $|f(x) - f(s)| \lt 1$ whenever $x$ is in $[s-\delta,s+\delta]$. Thus, for $x$ in $[s-\delta,s+\delta]$ $f$ is bound between $f(s)-1$ and $f(s)+1$.

Note, there must exist some $x_2$ in $B$ where $s-\delta \lt x_2 \lt s$. If there wasn't, then $s$ wouldn't be the supremum -- some point to its left would be. But if $x_2$ is in $B$, all $x$ values between $a$ and $x_2$ must also be in $B$. So $f$ is bounded on $[a,x_2]$.

As $[a,x_2]$ overlaps $[s - \delta,s + \delta]$, where $f$ is also bounded, $f$ must be bounded on $[a,s+\delta]$.

This requires the supremum of $B$ to be greater than $s$, a contradiction.

Thus having $s \lt b$ is impossible. That leaves as the only possible case that $s=b$.

With $s=b$ established, we now appeal to the continuity on the left of $f$ at $s$, we know that for $\epsilon = 1$, we can find a $\delta \gt 0$ such that $|f(x) - f(s)| \lt 1$ for all $x$ in $[s-\delta,s]$. Similar to our previous arguments, this means that $f$ is bounded on $[s-\delta,s]$.

There must then be some $x_3 \gt s - \delta/2$ in $B$. Otherwise, $s$ would not be the supremum of $B$ -- some point to its left would be.

This implies that $f$ is bounded on $[a,x_3]$ which overlaps $[s-\delta,s]$, making $f$ bounded on $[a,s]$. As $s = b$, we now have $f$ bounded on $[a,b]$ -- which is what we had hoped to show.