Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 1 down vote accepted

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".

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.