Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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 $A\subset\mathbb{R}$ a nonempty set of real numbers bounded above and $u$ be an upper bound of $A$. Prove that if $u\in A$, then $u=\sup A$.

share|cite|improve this question
This is pretty straightforward; where are you stuck? – Brian M. Scott May 1 '12 at 22:43
What have you even tried? – TMM May 1 '12 at 22:51
Assume $v$ is an upper bound and $v > u$. Arrive at a contradiction. Conclude it must be $v \leq u$. – Pedro Tamaroff May 2 '12 at 2:34
Note that Brian Scott and TMM's comments are meant to help us formulate a response. We believe in helping people learn, but you should put in some effort to let us better help you. – Willie Wong May 2 '12 at 8:07
up vote 2 down vote accepted

Recall the definition of supremum: If the non-empty set of real numbers $S$ is bounded above then $u$ is the supremum of $S$ if both the following hold:

$\ \ \ $1) $u$ is an upper bound of $S$


$\ \ \ $2) if $v$ is any upper bound of $S$, then $u\le v$.

A hint for your problem: Suppose $v$ is another upper bound of $A$. Think about condition 2), keeping in mind that $u\in A$.

share|cite|improve this answer

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.