(Original post by **ghostwalker**)
If the first part leads to a contradiction, then it must be false, BUT the second part could still be true.

Only by showing that both parts, call them A and B, are false, can you conclude that "A or B" is false.

Oh yes, thank you so much. That is regarding the truth table for 'OR', didn't recognise it - thank you.

In regards to the question I had (which I didn't post), I understand it is not linked to the thread, so if you can answer it that'd be great and much appreciated

(because you usually know the answer to all my questions lol), else I'll create a separate thread on it.

So the question was:

For any non-empty bounded set S, prove that inf S ≤ sup S, and that the equality holds iff S is a singleton set (that is a set with exactly one element).

So, I may have interpreted this question incorrectly and if I have please let me know

.

So I interpreted this as an if and only if statement where both of these conditions have to be satisfied:

If inf S ≤ sup S, then S is a singleton set

and If S in a singleton set, inf S ≤ sup S.

So I could either prove both of these statements true, or use proof by contradiction in which I would assume the negation of their conjunction and reach a contradiction, right (I think lolz)?

So, proof by contradiction:

**Steps of negation**
Spoiler: Show

negation of : {If inf S ≤ sup S, then S is a singleton set **and** If S in a singleton set, inf S ≤ sup S}

= {negation of (If inf S ≤ sup S, then S is a singleton set) **or** negation of (If S in a singleton set, inf S ≤ sup S)}

= {inf S ≤ sup S and S is not a singleton set **or** S in a singleton set and inf S ≥ sup S}

Now, to reach a contradiction: I have to prove both of the statements wrong:

inf S ≤ sup S and S is not a singleton set

**1**
**and**

S in a singleton set and inf S > sup S.

**2**
I mean, I don't really know how this helps lol! But, I just used it as my method.

It's basically the same approach as I would use with (proving both the original statements to be true).

PROOF:

**1**
In a set, inf(S) and sup(S) are either included in the set or they're not.

In the cases, either the set tends towards a lowest/greatest element, or has a lowest/greatest element.

Let that smallest element that the set tends towards or includes, be a, and let the largest element that the set tends towards or includes, be b.

So lower bound ≤ a (regardless of whether it is included or not), so inf(S) = a. Similarly, upper bound ≥ b (regardless of whether it is included or not), so sup(S) = b.

Now, if there is more than 1 element in the set, the elements have to be distinct, so a ≠ b. As a result, inf (S) ≠ sup (S). As it is not equal to, the equality doesn't hold and therefore inf (S) < sup (S) as it can never be greater than it. We have reached a contradiction, and this statement is false.

**2**
Using the information in the above (since I don't want to make it really long), if the set only contains one element, then the smallest and the largest element will either be the same, or it will have a smallest element and tends towards a largest element, or it will have a largest element and tends towards a smallest element

**unsure about these last two**.

**Is this wrong? **E.g. of it will have a smallest element and tends towards a largest element: Let x = {X is an integer | 1 ≤ x < 2}. It doesn't tend towards 2 right because x is not a real number, it is an integer? So would this be incorrect?

If it is incorrect, I'll just prove the first statement: a set where the smallest and the largest element will be the same e.g. a.

In this case, any element ≤ a or ≥ a will be a lower bound and upper bound respectively. So, inf (S) = sup (S) = a.

In this case, the inequality holds and therefore, the strict inequality isn't satisfied - we have reached a contradiction and therefore this statement is false.

As both statements are false, this leads us to conclude that the original statement is true and therefore inf S ≤ sup S, and that the equality holds iff S is a singleton set (that is a set with exactly one element).

By the way, I know you'll be really tired by the time you read all of that, so if you have time, I'd really appreciate if you can just confirm the next part:

Spoiler: Show

Contrapositive:

I assume what I've written below is the contrapositive, can you please confirm that?

For any non-empty bounded set S, prove that S is not a singleton set (that is a set with exactly one element) iff inf S > sup S, and that the equality doesn't hold, only in this case.

Would this be easier to prove in any way, or is it just the same approach but sort of reversed (like backwards) i.e. proving that if inf S > sup S, then S is not a singleton set ... etc.

Thank you so much for your help as always

.