9
$\begingroup$

Consider the real interval $[0,1)$, this is partially ordered set (totally ordered actually). This set has a upper bound like $1$, and according to Zorn's Lemma each partially ordered set with a upper bound should have at least one maximal element. However, in this set there is no maximal element, i.e., element that is greater than every element of the set because you can be as close as to $1$. I am should I don't understand Zorn's Lemma. Please Help!!

$\endgroup$
17
$\begingroup$

As a partial order $[0,1)$ has no upper bound. Sure $1$ is an upper bound of $[0,1)$ in $[0,1]$ or in $\Bbb R$. But that is not the same partial order. You are not allowed to go to larger partial orders when you apply Zorn's lemma.

So $[0,1)$ has many chains without upper bounds. E.g. $[0,1)$ itself.

$\endgroup$
2
  • $\begingroup$ Thanks for the clarification. $\endgroup$
    – user247511
    Aug 15 '15 at 10:35
  • $\begingroup$ You're welcome! $\endgroup$
    – Asaf Karagila
    Aug 15 '15 at 10:37
5
$\begingroup$

You are actually not using Zorn's Lemma which states that if each chain in $[0,1)$ has an upper bound, then $[0,1)$ has a maximal element. However, the chain $\{1- \frac 1 n \}_{n \in \mathbb N}$ has no upper bound in $[0,1)$.

$\endgroup$
1
  • 3
    $\begingroup$ To stress the actual point that the questioner was missing, you should add “in $[0,1)$” to the end of “… states that if each chain in [0,1) has an upper bound”. $\endgroup$
    – k.stm
    Aug 15 '15 at 10:34
1
$\begingroup$

To apply Zorn to a linear order $L$ is pointless: in order the prove the existence of a maximal element (here: also a maximum), you first have to give a maximum for $L$ as the condition says we must have an upper bound for every chain in $L$, including $L$ itself. The conclusion would be weaker than what you need to show in the proof anyway.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy