Help understanding a part of proof of boundedness theorem
I'm studying the extreme value theorem and I'm having trouble understanding a part of the proof. Why is it that
if a function f is not bounded above on the interval [a,b], then, for every natural number n, there exists an x$_n$ in [a,b] such that f(x$_n$) > n
?
The complete proof is available here: .
Thanks very much.
$\endgroup$ 11 Answer
$\begingroup$It's a bit difficult to lay out in text; it'd be better to talk interactively, but I'll give it a try.
This first step in the proof is just laying out what it means if a function is not bounded above.
So let's talk about what that means. What does it mean if a function is bounded above?
If a function $f$ is bounded above on $[a, b]$, it means that there is some value $U$ such that for any $x \in [a,b]$, we have $f(x) <= U$.
If a function $f$ is not bounded above on $[a,b]$, that means that this isn't true.
So how do we take that statement of what boundedness means and make it not true?
Well, we reverse it piece by piece:
"there is some value $U$" becomes "for any value $U$"
"for any $x \in [a,b]$" becomes "there is some $x \in [a,b]$"
"$f(x) <= U$" becomes "$f(x) > U$"
Put it all together, and we get:
If a function $f$ is not bounded above on $[a,b]$, it means that for any value $U$ there exists an $x \in [a,b]$ such that $f(x) > U$
Now, because this holds for every value of $U$, it also holds for every natural number $n$. That's all they're doing here: they're just picking specific values $f$ must exceed if $f$ is unbounded.
Another way to think of it is that the proof is saying: "If $f$ is unbounded on $[a,b]$, then there must be a value for $x \in [a,b]$ so that $f(x) > 1$. Call that value $x_1$. Now there must also be a value for $x \in [a,b]$ so that $f(x) > 2$. Call that value $x_2$. And so on for every natural number."
That second bit of the proof you ask about in a comment is, I think, confusing you because at one point they use the letter $\alpha$ (Greek lowercase alpha) when they already have an $a$ (Roman lowercase "a") in there and it's difficult to see which one they're using.
So let me first rewrite that bit using letters that are visually distinct:
Now look at the sequence ($x_n$). By the Bolzano-Weierstrass theorem, it has a subsequence ($x_{i_j}$) which converges to a point $p \in [a, b]$. By our construction the sequence ($f (x_{i_j})$) is unbounded, but by the continuity of $f$, this sequence should converge to $f (p)$ and we have a contradiction.
So what this bit is saying is that first off, since we have (in the $x_n$) an infinite sequence of values all inside $[a,b]$, then there's some subsequence (that is, some subset of the $x_n$) that converges to a point inside $[a,b]$. Call that point that the convergent subsequence converges to $p$, and define $i_1, i_2, i_3, \mathrm{etc.}$ so that the convergent subset of the $x_n$ is $x_{(i_1)}, x_{(i_2)}, x_{(i_3)}, \mathrm{etc.}$. Note that the $i_j$ is an infinite sequence, and that $$\lim_{j\to\infty} i_j = \infty$$
Now, because of how we initially chose the $x_n$, we know that
\begin{align} f(x_{(i_1)}) &> i_1 \\ f(x_{(i_2)}) &> i_2 \\ f(x_{(i_3)}) &> i_3 \\ & \vdots \end{align}
Therefore, the sequence $f(x_{(i_1)}), f(x_{(i_2)}), f(x_{(i_3)}), ...$ is unbounded. (meaning that for any value $A$, you can find an $n$ so that $f(x_{(i_j)} > A$ for all $j > n$)
But at the same time, because $f$ is continuous and the sequence $x_{(i_1)}, x_{(i_2)}, x_{(i_3)}, ...$ converges to $p$, we must have that $f(x_{(i_1)}), f(x_{(i_2)}), f(x_{(i_3)}), ...$ converges to $f(p)$.
This is the contradiction: we have shown that the sequence $f(x_{(i_j)})$ both must converge to some real value ($f(p)$), and at the same time be unbounded.
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