# Simple Proof about Sequences and Unbounded Functions

Let $X$ be a compact interval, let $V$ be a normed vector space and suppose that a map $f:X \rightarrow V$ is unbounded. I'm trying to see why there must exists a sequence $(x_i)$ in $X$ such that $|f(x_i)| \geq i \; \; \forall i \in \mathbb{N}$.

My argument is as follows: Since $f$ is unbounded, $sup\{|f(x)| \; : x \in X\}$ does not exist and therefore $1$ is not a supremum. This means that there is some $x_1 \in X$ such that $f(x_1) \geq 1$. We can select $x_2, x_3$ and so in in a similar manner and by induction conclude that there exists a sequence $(x_n) \in X$ such that $|f(x_i)| \geq i, \; \; i \in \mathbb{N}$

My questions are:

(1) I didn't use compactness and I don't believe that its a necessary hypothesis; is this correct? and

(2) This argument selects an infinite number of elements of $X$, more-or-less simultaneously, which I believe actually requires the axiom of choice. Does the induction principle preclude the need to use the choice axiom here or should it be applied to make the argument correct?

-
Whence $q$? Oh... it's right next to "1" on the keyboard :) –  The Chaz 2.0 Sep 27 '11 at 17:48
Didn't stretch the little finger enough.... –  ItsNotObvious Sep 27 '11 at 17:50
If the interval is compact, the $x_i$ can be made convergent.. which might be needed for wherever this came from. –  Zarrax Sep 27 '11 at 17:54
@Zarrax Yes, you're right; this is part of an argument to show that regulated functions are bounded –  ItsNotObvious Sep 27 '11 at 18:08

Ad (1): If $f:X\to V$ is unbounded on a set $X$ then by definition for any $n\in {\mathbb N}$ there is an $x_n\in X$ with $|f(x_n)|\geq n$.
Ad (2): Since there are no assumptions on $f$ other than it is unbounded there is no handle to select the $x_n$ in your argument. At any rate induction is of no help, because the $x_k$ chosen so far neither limit the future choices nor give any hint where the next $x_n$ could be. My suggestion: Look at the proof of the unboundedness of $f$ and see whether you can make it "constructive".
Not sure I understand your comments. If your "Ad (1)" follows directly from the definition of $f$ being unbounded, then there is nothing to prove. So, are you saying that the existence of the sequence $(x_n)$ follows directly from the definition? If so, then I'm not really sure of what point your trying to make with "Ad (2)" –  ItsNotObvious Sep 27 '11 at 20:32
@3Sphere: For the usual working mathematician the case is closed by what I said Ad (1). But from your question (2) I got the impression that the sentence "there exists an $x_n$ such that$\ldots$" rises doubts with you and that you desire an explicit production scheme for $x_n$, given $n$. Therefore I wrote Ad (2) that unless we know more about $f$ and in particular about the proof of its unboundedness it is impossible to give a "constructive" version of the argument in question. –  Christian Blatter Sep 28 '11 at 18:16