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I know the proof that $\emptyset$ (empty set) has $\infty$ as infimum. This seems to imply to me that $\emptyset$ is unbounded, I mean a set with $\infty$ as lower bound: how can this be bounded?! But if $\emptyset$ is unbounded then $\forall M>0$ $\exists x \in \emptyset $ s.t. $x > M$, which is impossible because there can't exist an element in $\emptyset$. How to resolve this paradox?

thanks.

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  • $\begingroup$ Recall that the infimum is actually defined as the greatest element that is less or equal than any element in $\emptyset$. So $+\infty$ makes sense as a definition. It is also very helpful in many applications. By the way I do not think there is a proof of this fact as you say in your first line, I think this is more a definition. $\endgroup$ – Jimmy R. Aug 6 '16 at 9:02
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    $\begingroup$ A set that is unbounded from below has an infimum of $-\infty$. Since $\infty\ne-\infty$, everything is well. $\endgroup$ – celtschk Aug 6 '16 at 9:03
  • $\begingroup$ A set is bounded if its infimum is $>-\infty$ and supremum is $<\infty$. But that's all fine for the empyt set. $\endgroup$ – Peter Franek Aug 6 '16 at 9:10
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I love to think of the infimum of a set $S\subset \Bbb R$ as an imaginary point $p$ moving from $-\infty$ toward $\infty$. The point $p$ stops when it hits the lowest point of $S$ then $p=\inf S$.

When $S=\emptyset$, since $p$ never stops because it doesn't hit any thing at all, so it moves toward $\infty$. This justifies $\inf \emptyset=\infty$ for me.

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    $\begingroup$ Recognizable way of thinking. $\endgroup$ – drhab Aug 6 '16 at 9:33
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Something more abstract for intuition:

Let $M$ be a monoid (group if you like). Then we usually define: $\prod_{i=1}^0 a_i$, i.e. the empty product to be the identity of that monoid.

The case of infima is completely analogous. Let: $$a \wedge b:= \min(a,b) = \inf(a,b)$$

Then $\mathbb{R}$ equipped with this operation is a monoid. The identity is of course $\infty$, hence:

$$\inf \emptyset = \bigwedge_{i=1}^0 a_i = \infty$$

Of course $\inf$ is defined for infinite sets too, but the reasoning still holds.

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  • $\begingroup$ Amazing point of view! Are you really a second-year undergraduate because looking from your posts history you seem to know a lot more. $\endgroup$ – BigbearZzz Aug 6 '16 at 10:14
  • $\begingroup$ @BigbearZzz Yes, I am. Unfortunately there is a lot stuff I don't know although I should and I'm a chronic procastinator, so I actually have a hard time studying. My posts here usually have nothing to do with my studies. $\endgroup$ – Stefan Perko Aug 6 '16 at 10:20
  • $\begingroup$ That's impressive! I guess I am around 2 years older than you but I still know nothing about category theory (which I certainly should). Maybe that makes me a bigger procastinator then :( $\endgroup$ – BigbearZzz Aug 6 '16 at 10:36
  • $\begingroup$ @BigbearZzz No need to rush but if you are really interested Leinsters "Basic Category Theory" is the way to go imo. $\endgroup$ – Stefan Perko Aug 6 '16 at 11:07
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As it's been pointed out in the comments, infimum is such that

$$\inf X \le x \forall x \in X \land (y \le x \forall x \in X \implies y \le \inf X) $$

by definition. Of course $k \le x \forall x \in \emptyset$ for every finite $k$ — then $\inf \emptyset = \infty$!

Conversely, and even more surprisingly, you can state that $\sup \emptyset = - \infty$ by a similar argument, so that nice relationship $\inf X \le \sup X$ doesn't even hold for $\emptyset $

You might want to check the concept of vacuous truth.

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    $\begingroup$ Of course, despite $\inf\emptyset>\sup\emptyset$, for any element $x\in\emptyset$ we have $\inf\emptyset < x$ and $x<\sup\emptyset$. Side remark: I wonder if under these conditions it is valid to combine those inequalities to $\inf\emptyset<x<\sup\emptyset$; the latter seems to include the claim that $\inf\emptyset<\sup\emptyset$ which clearly is wrong. $\endgroup$ – celtschk Aug 6 '16 at 9:27
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    $\begingroup$ @celtschk there is a certain risk of confusion in writing it that way, yet strictly I'd say that $a< b < c $ is nothing but short for $a< b$ and $b <c$. There are also arguments that show non-existence by coming to an "impossible" chain of inequalities, so such things are used. $\endgroup$ – quid Aug 6 '16 at 9:34
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    $\begingroup$ @quid: Actually thinking more about it, the full statement is actually: $\forall x: x\in\emptyset \implies \inf\emptyset<x<\sup\emptyset$ and thus is doesn't matter if $\inf\emptyset<\sup\emptyset$ is included in the chain, as the statement $\forall x: x\in\emptyset \implies \inf\emptyset<\sup\emptyset$ is vacuously true. $\endgroup$ – celtschk Aug 6 '16 at 9:54
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A subset $S$, of the real numbers, is bounded when there exists some real number $M$ such that $|x| \le M$ for all $x \in S$.

(The same for bounded from above and bounded from below separately.)

According to this definition the empty set is bounded. You can chose $M$ whatever you like. The assertion is true, it is vacuously true.

Further more, every real number is an upper bound an every real number is a lower bound, e.g., $26$ is a lower bound and $-45$ an upper bound of the empty set.

To resolve the paradox one needs to note that for $m$ a lower and $M$ and upper bound of some set $S$, one can conclude that $m \le M$ provided that $S$ is non-empty.

One needs some element in $S$ to make the obvious argument work. (For any $s \in S$, we have $m \le s$ and $s \le M$. Thus $m \le M$.)

A takeway is that vacuous truth can have some counterintuitive consequence.

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