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.
This isn't actually correct. You actually have three statements you need to prove:
If S is a non-empty bounded set, inf S ≤ sup S.
If inf S = sup S then S is a singleton set.
If S is a singleton set then inf S = sup S.
The "iff" part of the question only concerns the case where equality holds (i.e. inf S = sup S).
This means most of your subsequent work is invalid, but I have some comments anyhow....
~snip~
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.
What's all this "set tends towards" nonsense? There's a very informal way in which it might make sense, but I can't believe it's any part of your formal definition of inf(S) and sup(S) and it's certainly too informal to be applicable here. .
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.
If a and b are just things the set "tends towards", I don't see how you can claim this. (Or at least, you definitely need to justify it). .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.
2Using 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.
Again, this whole nonsense about "tending towards" is messing you up. Use the actual definitions for sup and inf and this is trivial..
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?
Integers are real numbers!.
If it is incorrect, I'll just prove the first statement
I don't actually know what you mean by "first statement" here; when you say something like this you should make sure it actually *is* obvious what statement you're talking about: a set where the smallest and the largest element will be the same e.g. a.
This isn't a proper sentence (it's the mathematical equivalent of "a red apple").
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:
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.
It's never the case that inf S > sup S. Did you mean inf S < sup S?
Ignoring that, iff is really 2 statements, so you should treat them both seperately, not try to take the contrapositive of their combinationWould 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.
"the same approach but sort of reversed" is exactly what the contrapositive is. .