1. ## [SOLVED] Empty set

I've just seen a nice problem in Krizalid's website.
Show that the empty set is bounded.
As there is no element in this set, it's quite hard for me to imagine that there exists an element that is greater or equal to any element of the empty set, and that there exists one that is lesser or equal than any other.
So how do you prove the statement above?

2. Because a false statement implies any statement, the proof is quite simple.
If $\left\{ {x,y} \right\} \subseteq \emptyset$ then $d\left( {x,y} \right) \le 1$.

3. So the problem remains to prove a false fact. At least I'm glad I wasn't able to do that!

4. Originally Posted by arbolis
So the problem remains to prove a false fact.
NO!
The point is that $\left\{ {x,y} \right\} \subseteq \emptyset$ is false.

5. NO!
The point is that is false.
Ah!!! Yes I know. That's why I don't understand how to prove the statement. I can't see how to use the definition of a bounded set, since there is no element in the empty set. So we cannot say that any element (in R or whatsoever) is greater/equal or lesser/equal than any element of the empty set, since it doesn't contains a single element!
So how would you prove the statement?

6. Originally Posted by arbolis
Ah!!! Yes I know. That's why I don't understand how to prove the statement. I can't see how to use the definition of a bounded set, since there is no element in the empty set. So we cannot say that any element (in R or whatsoever) is greater/equal or lesser/equal than any element of the empty set, since it doesn't contains a single element!
So how would you prove the statement?
You are missing the point:

The statement $p\Rightarrow q$ is true when: $p$ is true and $q$ is true, when $p$ is false and $q$ is true, and when $p$ is false and $q$ is false, and the statement is false if and only if $p$ is true but $q$ is false.

For example, we may say that the statement $n\text{ is even}\Rightarrow n^2\text{ is even}$ is true for all $n$, even though the left-hand side is clearly false for any odd value of $n$. The point is that you cannot find a value which will make the left side true but the right side false, so the implication is true for all values.

If you want further help, post the exact definition of "bounded" that you are using.

7. the statement is false if and only if is true but is false.
Thanks for this! I knew it was true, but I never realized it exactly till now. I mean by bounded, bounded from below and bounded from above. I still don't know how to prove the statement.

8. Originally Posted by Reckoner
The statement $p\Rightarrow q$ is true when: $p$ is true and $q$ is true, when $p$ is false and $q$ is true, and when $p$ is false and $q$ is false, and the statement is false if and only if $p$ is true but $q$ is false.
Let's note that this is because (or implies?) $p \rightarrow q$ has the same truth table as $\bar{p} \vee q$, so it's "p false" or "q true" or both.

9. Originally Posted by arbolis
Thanks for this! I knew it was true, but I never realized it exactly till now. I mean by bounded, bounded from below and bounded from above. I still don't know how to prove the statement.
Okay, let's make your definitions a little more precise. We can say that a set is bounded if it is bounded above and bounded below.

To be bounded above means that the set has an upper bound, and to be bounded below means that the set has a lower bound. We say that $C$ is an upper bound of a set $S$ if $x\leq C\;\;\forall x\in S$, and similarly for a lower bound. Is this (more or less) the definition you are using? If so, then to show that any set $S$ is bounded, we need to prove that there exists a value $M$ such that $\forall x\in S,\;x\leq M$ and, similarly, that $\exists N,\;\forall x\in S,\;x\geq N$. This is equivalent to proving that $x\in S\Rightarrow x\leq M$ and $x\in S\Rightarrow x\geq N$. This works for any set, as long as you can prove that the implication is true.

So for the empty set, if we can show that $x\in\emptyset\Rightarrow x\leq M\text{ and }x\geq N$ for some particular $M$ and $N$, then the empty set is bounded. But the left side of the implication must be false no matter what we choose as $M$ and $N$, so the implication is therefore true. This is basically what Plato said, only with more concise notation.

The empty said is bounded because you can find a value for which there are no elements in the set greater than said value, and a value for which there are no elements in the set smaller than said value. However, if you define "bounded above" as requiring the set to have an element that is greater than all other elements in the set (which seems to be what you did in your original post), then the result is false because no such element can exist. But the usual definition does not require the bounds to be within the set itself. For example, $[0,\;1)$ has an upper bound (and supremum) of 1, but there is no element within the set that is greater than all other elements within the set. This is why you need to be careful with your definitions.

10. To be bounded above means that the set has an upper bound, and to be bounded below means that the set has a lower bound. We say that is an upper bound of a set if , and similarly for a lower bound. Is this (more or less) the definition you are using?
Sorry, I forgot to precise it here, but yes this is exactly the definition I'm using.
So for the empty set, if we can show that for some particular and , then the empty set is bounded. But the left side of the implication must be false no matter what we choose as and , so the implication is therefore true. This is basically what Plato said, only with more concise notation.
I don't understand why you say that "But the left side of the implication must be false no matter what we choose as and ". Why it must? As there is no element in the Empty Set... It's false to say $x\in\emptyset$. I feel I'm missing the point once again.
And for
The empty said is bounded because you can find a value for which there are no elements in the set greater than said value, and a value for which there are no elements in the set smaller than said value. However, if you define "bounded above" as requiring the set to have an element that is greater than all other elements in the set (which seems to be what you did in your original post), then the result is false because no such element can exist. But the usual definition does not require the bounds to be within the set itself. For example, has an upper bound (and supremum) of 1, but there is no element within the set that is greater than all other elements within the set. This is why you need to be careful with your definitions.
I think I understand what you mean. In fact, I posted the problem as it was written, I assumed the empty set has to be bounded from below and from above in order to be "bounded", but as I just understood, it can be bounded without being bounded from below and from above. And I think this exercise was about this point. Thanks a lot! Afterall I feel I've learned something interesting.

11. Originally Posted by arbolis
Sorry, I forgot to precise it here, but yes this is exactly the definition I'm using.
About : I don't understand why you say that "But the left side of the implication must be false no matter what we choose as and ". Why it must? As there is no element in the Empty Set... It's false to say $x\in\emptyset$. I feel I'm missing the point once again.
And for I think I understand what you mean.
You just said: it is false to say that $x\in\emptyset$. Therefore, the antecedent (left side) of the implication $x\in\emptyset\Rightarrow q$ is false (as you said it was), so the implication is true.

Thus, for example, $x\in\emptyset\Rightarrow\text{you are the pope}$ is a true statement, because the antecedent $x\in\emptyset$ is false. Similarly, the statement $x\in\emptyset\Rightarrow N\leq x\leq M$ for some $M,N$ is true.

Originally Posted by arbolis
In fact, I posted the problem as it was written, I assumed the empty set has to be bounded from below and from above in order to be "bounded", but as I just understood, it can be bounded without being bounded from below and from above. And I think this exercise was about this point. Thanks a lot! Afterall I feel I've learned something interesting.
No! To be bounded, you must have an upper bound and a lower bound. The point is that the bounds do not need to be members of the set itself.

12. No! To be bounded, you must have an upper bound and a lower bound. The point is that the bounds do not need to be members of the set itself.
Ok!! I understood. Still, that was the point of the exercise! Correct me if I'm wrong. Any real number (a for example) can be a bound (upper of lower) for the empty set, since the empty set doesn't contain any element greater than a, or lesser than a.

13. Originally Posted by arbolis
Ok!! I understood. Still, that was the point of the exercise! Correct me if I'm wrong. Any real number (a for example) can be a bound (upper of lower) for the empty set, since the empty set doesn't contain any element greater than a, or lesser than a.
Yes, that is the underlying idea here.

14. Originally Posted by arbolis
Correct me if I'm wrong. Any real number (a for example) can be a bound (upper of lower) for the empty set, since the empty set doesn't contain any element greater than a, or lesser than a.
Precisely. That is why we have $\sup{\O}=-\infty$. Similarly, any real number can be a lower bound for the empty set, and so $\inf{\O}=\infty$.

15. ## I don't agree with the above

I don't believe the fact that the lub of the empty set is negative infinity or the glb is positive infinity. I don't believe it is possible to say that it has a least upper or greatest lower bound.

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