Is the following proof that the least-upper-bound of an empty set $\phi$ is $ -\infty$ correct? I'm confused because someone told me that every $x \in \mathbb{R}$ is a least-upper-bound of $\phi$.

$\phi$ has no elements so $\forall x \in \mathbb{R}, x$ is an upper-bound for $\phi$. Then, because we want the least upper-bound, we need to look for the smallest element of $\mathbb{R}$ which is $- \infty$. Hence, $\sup{\phi} = - \infty$.

  • $\begingroup$ Notice that every $x \in \mathbb{R}$ is not a least-upper-bound of $\phi$, however it is an upper bound of $\phi$. $\endgroup$ – aortizmena Sep 16 '14 at 17:46

It seems like you have the basic idea, but I think there a few things to discuss. I think it might be a bit perverse to say that $- \infty$ is the smallest element of $\mathbb{R}$, as $- \infty$ isn't traditionally considered to be part of $\mathbb{R}$. Here's a different argument, which also addresses the statement made by your friend.

The element $x \in \mathbb{R}$ is an upper bound of a set $A$ if $\forall a \in A$, $x \geq a$. So, to prove that every $x \in \mathbb{R}$ is an upper bound for $\emptyset$, we consider arbitrary $x \in \mathbb{R}$ and assume (for contradiction), that $x$ is not an upper bound for $\emptyset$. This implies that $\exists y \in \emptyset$ such that $y > x$. But there is no element $y \in \emptyset$, contradiction. Therefore, $\forall x \in \mathbb{R}$, $x$ is an upper bound of $\emptyset$.

But given any $x \in \mathbb{R}$, $x$ cannot be $\sup(\emptyset)$, as, for example, $x-1$ is also an upper bound for $\emptyset$ and $x-1 < x$ (this disproves what you were told). Therefore, there is no element of $\mathbb{R}$ that can be the least upper bound, and so we say that $\sup(\emptyset) = - \infty$. $\quad\quad$

  • $\begingroup$ A nice, concise answer. +1 $\endgroup$ – Rustyn Sep 12 '14 at 7:08

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