0
$\begingroup$

Let $(E,\leq)$ a partially ordered set. Zorn lemma says that if all chain of $E$ has a supremum, then $E$ has a maximal element.

So if I consider, $\Big((0,1),\leq \Big)$, it has no maximal element but all chain is upper bounded by $1$, so it doesn't work here, no ?

  • 0
    The chain $(0,1)$ has no upper bound in $(0,1)$. – 2017-02-11
  • 0
    The chain (0, 1-1/n) does not have a supremum. – 2017-02-11
  • 0
    The upper bound must lie in $E$ for Zorn's lemma to apply. – 2017-02-11

1 Answers 1

4

Your partial order $((0,1),{\le})$ is a total order, so $(0,1)$ is itself a chain. It has no upper bound! You propose that $1$ is an upper bound, but $1$ isn't even a member of the partial order you're considering, so that doesn't count.

Note, by the way, that Zorn's Lemma doesn't demand that every chain must have a supremum (which means a least upper bound) -- it is enough that every chain has some upper bound in the partial order.

  • 0
    I see, so every chain of $(E,\leq)$ must have an upper bound in $E$, I thought it was not necessary that it's in $E$. – 2017-02-11
  • 1
    @user330587: Indeed. Otherwise you could always define a larger partial order with an artificial maximal element, and then say that that is an upper bound for every chain. – 2017-02-11