When treated as a limit, 'infinity' is essentially synonymous with 'never' - saying that 'a process terminates after an infinite amount of time' means exactly the same thing as 'the process does not terminate'.

So is there a diffierence between 'unbounded' and 'bounded by infinity'?

Or, more generally, if $S\subset X$ is a partially ordered set and $\exists x\in X\setminus S:\forall s\in S.s\ll x$ ($x$ is infinitely greater than $s$, or $x\notin\text{gal}(s)$, or $\nexists t\in S:\Vert s-x\Vert=\Vert t\Vert$, etc...)$^1$ then is '$S$ bounded by $x$' equivalent to '$S$ not bounded'?

Ex. $\mathbb{C}\subset\mathbb{C}\cup\{\infty\}$ and $\forall z\in\mathbb{C}.z\ll\infty$, so $\mathbb{C}$ is bounded by $\infty$

$^1$ It is assumed that $x$ is infinite.

$\text{gal}(s)$ is the galaxy of $s$ from nonstandard analysis. The galaxy of an element $s\in S$ is the subset of elements of $S$ that differ from $s$ by a finite amount - e.g. $\text{gal}(0)=\mathbb{R}$

Note that $\nexists t\in S:\Vert s-x\Vert=\Vert t\Vert$ does not imply that $S$ is not bounded; however if $S$ is not bounded, then $\nexists t\in S:\Vert s-x\Vert=\Vert t\Vert$.

  • $\begingroup$ I could not understand the loaded notation, although I did not vote to close. I have never heard the phrase "bounded by infinity" but I often write "assume $E[|X|]<\infty$" to mean $E[|X|]$ is a (finite) real number, and I use notation "$5 \leq x <\infty$" to mean that $x$ is in the interval $[5, \infty)$. $\endgroup$ – Michael Apr 9 at 14:51
  • $\begingroup$ @Michael Thanks, which notation are you referring to? $\endgroup$ – R. Burton Apr 9 at 14:53
  • $\begingroup$ $\overline{S}$, $gal(\cdot)$, $||S||$, "immeasurably greater," etc. $\endgroup$ – Michael Apr 9 at 14:54
  • $\begingroup$ @Michael Corrected. I should really try to be less lazy with my notation. $gal(s)$ is galaxy of $s$ from nonstandard analysis. The galaxy of an element $s$ is the set of elements that differ from $s$ by a finite amount - e.g. $\text{gal}(0)=\mathbb{R}$ $\endgroup$ – R. Burton Apr 9 at 15:05
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    $\begingroup$ In the context of a partially ordered set such as $X$, it makes no sense to say "$x \in X$ is infinite", and it makes no sense to subtract two elements of $X$. You cannot simply important operations of numbers into set theory in this fashion. $\endgroup$ – Lee Mosher Apr 9 at 15:27

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